| Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
Exercise 8.4
Question 1
The sum of n terms of an A.P. is $\left(\frac{5 n^{2}}{2}+\frac{3 n}{2}\right)$. Find its 20th term.
Sol :$\mathrm{S}=\left(\frac{5 \mathrm{n}^{2}}{2}+\frac{3 \mathrm{n}}{2}\right)$
Taking n = 1, we get
$\mathrm{S}_{1}=\left(\frac{5(1)^{2}}{2}+\frac{3(1)}{2}\right)$
$\Rightarrow \mathrm{S}_{1}=\left(\frac{5}{2}+\frac{3}{2}\right)$
⇒ S1 = 4
⇒ a1 = 4
Taking n = 2, we get
$\mathrm{S}_{2}=\left(\frac{5(2)^{2}}{2}+\frac{3(2)}{2}\right)$
$\Rightarrow \mathrm{S}_{2}=\left(\frac{20}{2}+\frac{6}{2}\right)$
⇒ S2 = 13
∴ a2 = S2 – S1 = 13 – 4 = 9
$\mathrm{S}_{2}=\left(\frac{5(2)^{2}}{2}+\frac{3(2)}{2}\right)$
$\Rightarrow \mathrm{S}_{2}=\left(\frac{20}{2}+\frac{6}{2}\right)$
⇒ S2 = 13
∴ a2 = S2 – S1 = 13 – 4 = 9
Taking n = 3, we get
$\mathrm{S}_{3}=\left(\frac{5(3)^{2}}{2}+\frac{3(3)}{2}\right)$
$\Rightarrow \mathrm{S}_{3}=\left(\frac{45}{2}+\frac{9}{2}\right)$
⇒ S3 = 27
∴ a3 = S3 – S2 = 27 – 13 = 14
So, a = 4,
d = a2 – a1 = 9 – 4 = 5
Now, we have to find the 20th term
an = a + (n – 1)d
a20 = 4 + (20 – 1)5
a20 = 4 + 19 × 5
a20 = 4 + 95
a20 = 99
Hence, the 20th term is 99.
Sn = 3n2 + 2n
Taking n = 1, we get
S1 = 3(1)2 + 2(1)
⇒ S1 = 3 + 2
⇒ S1 = 5
⇒ a1 = 5
Taking n = 2, we get
S2 = 3(2)2 + 2(2)
⇒ S2 = 12 + 4
⇒ S2 = 16
∴ a2 = S2 – S1 = 16 – 5 = 11
Taking n = 3, we get
S3 = 3(3)2 + 2(3)
⇒ S3 = 27 + 6
⇒ S3 = 33
∴ a3 = S3 – S2 = 33 – 16 = 17
So, a = 5,
d = a2 – a1 = 11 – 5 = 6
Now, we have to find the 15th term
an = a + (n – 1)d
a15 = 5 + (15 – 1)6
a15 = 5 + 14 × 6
a15 = 5 + 84
a15 = 89
Hence, the 15th term is 89 and AP is 5, 11, 17, 23,…
Sn =2n2 + 5n
Taking n = 1, we get
S1 = 2(1)2 + 5(1)
⇒ S1 = 2 + 5
⇒ S1 = 7
⇒ a1 = 7
Taking n = 2, we get
S2 = 2(2)2 + 5(2)
⇒ S2 = 8 + 10
⇒ S2 = 18
∴ a2 = S2 – S1 = 18 – 7 = 11
Taking n = 3, we get
S3 = 2(3)2 + 5(3)
⇒ S3 = 18 + 15
⇒ S3 = 33
∴ a3 = S3 – S2 = 33 – 18 = 15
So, a = 7,
d = a2 – a1 = 11 – 7 = 4
Now, we have to find the 15th term
an = a + (n – 1)d
an = 7 + (n – 1)4
an = 7 + 4n – 4
an = 3 + 4n
Hence, the nth term is 4n + 3.
Sn = 3n2 + 5n
Taking n = 1, we get
S1 = 3(1)2 + 5(1)
⇒ S1 = 3 + 5
⇒ S1 = 8
⇒ a1 = 8
Taking n = 2, we get
S2 = 3(2)2 + 5(2)
⇒ S2 = 12 + 10
⇒ S2 = 22
∴ a2 = S2 – S1 = 22 – 8 = 14
Taking n = 3, we get
S3 = 3(3)2 + 5(3)
⇒ S3 = 27 + 15
⇒ S3 = 42
∴ a3 = S3 – S2 = 42 – 22 = 20
So, a = 8,
d = a2 – a1 = 14 – 8 = 6
Now, we have to find the 15th term
an = a + (n – 1)d
a16 = 8 + (16 – 1)6
a16 = 8 + 15 × 6
a16 = 8 + 90
a16 = 98
Hence, the 16th term is 98.
Sn = 3n2 – n
Taking n = 1, we get
S1 = 3(1)2 - (1)
⇒ S1 = 3 – 1
⇒ S1 = 2
⇒ a1 = 2
Taking n = 2, we get
S2 = 3(2)2 – 2
⇒ S2 = 12 – 2
⇒ S2 = 10
∴ a2 = S2 – S1 = 10 – 2 = 8
Taking n = 3, we get
S3 = 3(3)2 – 3
⇒ S3 = 27 – 3
⇒ S3 = 24
∴ a3 = 24 – 10 = 14
So, a = 1,
d = a2 – a1 = 8 - 2 = 6
Now, we have to find the 15th term
an = a + (n – 1)d
an = 2 + (n – 1)6
an = 2 + 6n – 6
an = - 4 + 6n
Hence, the nth term is 4n - 3.
$\mathrm{S}_{\mathrm{n}}=\left(\frac{3 \mathrm{n}^{2}}{2}+\frac{5 \mathrm{n}}{2}\right)$
$\mathrm{S}_{3}=\left(\frac{5(3)^{2}}{2}+\frac{3(3)}{2}\right)$
$\Rightarrow \mathrm{S}_{3}=\left(\frac{45}{2}+\frac{9}{2}\right)$
⇒ S3 = 27
∴ a3 = S3 – S2 = 27 – 13 = 14
So, a = 4,
d = a2 – a1 = 9 – 4 = 5
Now, we have to find the 20th term
an = a + (n – 1)d
a20 = 4 + (20 – 1)5
a20 = 4 + 19 × 5
a20 = 4 + 95
a20 = 99
Hence, the 20th term is 99.
Question 2
The sum of first n terms of an A.P. is given by Sn = 3n2 + 2n. Determine the A.P. and its 15th term.
Sol :Sn = 3n2 + 2n
Taking n = 1, we get
S1 = 3(1)2 + 2(1)
⇒ S1 = 3 + 2
⇒ S1 = 5
⇒ a1 = 5
Taking n = 2, we get
S2 = 3(2)2 + 2(2)
⇒ S2 = 12 + 4
⇒ S2 = 16
∴ a2 = S2 – S1 = 16 – 5 = 11
Taking n = 3, we get
S3 = 3(3)2 + 2(3)
⇒ S3 = 27 + 6
⇒ S3 = 33
∴ a3 = S3 – S2 = 33 – 16 = 17
So, a = 5,
d = a2 – a1 = 11 – 5 = 6
Now, we have to find the 15th term
an = a + (n – 1)d
a15 = 5 + (15 – 1)6
a15 = 5 + 14 × 6
a15 = 5 + 84
a15 = 89
Hence, the 15th term is 89 and AP is 5, 11, 17, 23,…
Question 3 A
The sum of the first n terms of an A.P. is given by Sn = 2n2 + 5n , find the nth term of the A.P.
Sol :Sn =2n2 + 5n
Taking n = 1, we get
S1 = 2(1)2 + 5(1)
⇒ S1 = 2 + 5
⇒ S1 = 7
⇒ a1 = 7
Taking n = 2, we get
S2 = 2(2)2 + 5(2)
⇒ S2 = 8 + 10
⇒ S2 = 18
∴ a2 = S2 – S1 = 18 – 7 = 11
Taking n = 3, we get
S3 = 2(3)2 + 5(3)
⇒ S3 = 18 + 15
⇒ S3 = 33
∴ a3 = S3 – S2 = 33 – 18 = 15
So, a = 7,
d = a2 – a1 = 11 – 7 = 4
Now, we have to find the 15th term
an = a + (n – 1)d
an = 7 + (n – 1)4
an = 7 + 4n – 4
an = 3 + 4n
Hence, the nth term is 4n + 3.
Question 3 B
The sum of n terms of an A.P. is 3n2+ 5n. Find the A.P. Hence, find its 16th term.
Sol :Sn = 3n2 + 5n
Taking n = 1, we get
S1 = 3(1)2 + 5(1)
⇒ S1 = 3 + 5
⇒ S1 = 8
⇒ a1 = 8
Taking n = 2, we get
S2 = 3(2)2 + 5(2)
⇒ S2 = 12 + 10
⇒ S2 = 22
∴ a2 = S2 – S1 = 22 – 8 = 14
Taking n = 3, we get
S3 = 3(3)2 + 5(3)
⇒ S3 = 27 + 15
⇒ S3 = 42
∴ a3 = S3 – S2 = 42 – 22 = 20
So, a = 8,
d = a2 – a1 = 14 – 8 = 6
Now, we have to find the 15th term
an = a + (n – 1)d
a16 = 8 + (16 – 1)6
a16 = 8 + 15 × 6
a16 = 8 + 90
a16 = 98
Hence, the 16th term is 98.
Question 4
If the sum of the first n terms of an A.P. is given by Sn = (3n2- n), find its
(i) first term (ii) common difference
(iii) nth term.
Sol :(i) first term (ii) common difference
(iii) nth term.
Sn = 3n2 – n
Taking n = 1, we get
S1 = 3(1)2 - (1)
⇒ S1 = 3 – 1
⇒ S1 = 2
⇒ a1 = 2
Taking n = 2, we get
S2 = 3(2)2 – 2
⇒ S2 = 12 – 2
⇒ S2 = 10
∴ a2 = S2 – S1 = 10 – 2 = 8
Taking n = 3, we get
S3 = 3(3)2 – 3
⇒ S3 = 27 – 3
⇒ S3 = 24
∴ a3 = 24 – 10 = 14
So, a = 1,
d = a2 – a1 = 8 - 2 = 6
Now, we have to find the 15th term
an = a + (n – 1)d
an = 2 + (n – 1)6
an = 2 + 6n – 6
an = - 4 + 6n
Hence, the nth term is 4n - 3.
Question 5
If the sum to first n terms of an A.P. is $\left(\frac{3 n^{2}}{2}+\frac{5 n}{2}\right)$, find its 25th term.
Sol :$\mathrm{S}_{\mathrm{n}}=\left(\frac{3 \mathrm{n}^{2}}{2}+\frac{5 \mathrm{n}}{2}\right)$
Taking n = 1, we get
$\mathrm{S}_{1}=\left(\frac{3(1)^{2}}{2}+\frac{5(1)}{2}\right)$
$\Rightarrow \mathrm{S}_{1}=\left(\frac{3}{2}+\frac{5}{2}\right)$
⇒ S1 = 4
⇒ a1 = 4
$\mathrm{S}_{1}=\left(\frac{3(1)^{2}}{2}+\frac{5(1)}{2}\right)$
$\Rightarrow \mathrm{S}_{1}=\left(\frac{3}{2}+\frac{5}{2}\right)$
⇒ S1 = 4
⇒ a1 = 4
Taking n = 2, we get
$\mathrm{S}=\left(\frac{3(2)^{2}}{2}+\frac{5(2)}{2}\right)$
$\Rightarrow \mathrm{S}_{2}=\left(\frac{12}{2}+\frac{10}{2}\right)$
⇒ S2 = 11
∴ a2 = S2 – S1 = 11 – 4 = 7
$\mathrm{S}=\left(\frac{3(2)^{2}}{2}+\frac{5(2)}{2}\right)$
$\Rightarrow \mathrm{S}_{2}=\left(\frac{12}{2}+\frac{10}{2}\right)$
⇒ S2 = 11
∴ a2 = S2 – S1 = 11 – 4 = 7
Taking n = 3, we get
$\mathrm{S}_{3}=\left(\frac{3(3)^{2}}{2}+\frac{5(3)}{2}\right)$
$\Rightarrow \mathrm{S}_{3}=\left(\frac{27}{2}+\frac{15}{2}\right)$
⇒ S3 = 21
∴ a3 = S3 – S2 = 21 – 11 = 10
So, a = 4,
d = a2 – a1 = 7 – 4 = 3
Now, we have to find the 25th term
an = a + (n – 1)d
a25 = 4 + (25 – 1)3
a25 = 4 + 24 × 3
a25 = 4 + 72
a25 = 76
Hence, the 25th term is 76.
Given: an = 2n + 1
Taking n = 1,
a1 = 2(1) + 1 = 2 + 1 = 3
Taking n = 2,
a2 = 2(2) + 1 = 4 + 1 = 5
Taking n = 3,
a3 = 2(3) + 1 = 6 + 1 = 7
Therefore the series is 3, 5, 7, …
So, a = 3, d = a2 – a1 = 5 – 3 = 2
$\mathrm{S}_{3}=\left(\frac{3(3)^{2}}{2}+\frac{5(3)}{2}\right)$
$\Rightarrow \mathrm{S}_{3}=\left(\frac{27}{2}+\frac{15}{2}\right)$
⇒ S3 = 21
∴ a3 = S3 – S2 = 21 – 11 = 10
So, a = 4,
d = a2 – a1 = 7 – 4 = 3
Now, we have to find the 25th term
an = a + (n – 1)d
a25 = 4 + (25 – 1)3
a25 = 4 + 24 × 3
a25 = 4 + 72
a25 = 76
Hence, the 25th term is 76.
Question 6
If the nth term of an A.P. is (2n + 1), find the sum of first n terms of the A.P.
Sol :Given: an = 2n + 1
Taking n = 1,
a1 = 2(1) + 1 = 2 + 1 = 3
Taking n = 2,
a2 = 2(2) + 1 = 4 + 1 = 5
Taking n = 3,
a3 = 2(3) + 1 = 6 + 1 = 7
Therefore the series is 3, 5, 7, …
So, a = 3, d = a2 – a1 = 5 – 3 = 2
Now, we have to find the sum of first n terms of the AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 3+(\mathrm{n}-1) 2]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[6+2 \mathrm{n}-2]$
$\Rightarrow S_{n}=\frac{n}{2}[4+2 n]$
⇒ Sn = 2n + n2
Hence, the sum of n terms is n2 + 2n.
Given: an = 9 – 5n
Taking n = 1,
a1 = 9 – 5(1) = 9 – 5 = 4
Taking n = 2,
a2 = 9 – 5(2) = 9 – 10 = -1
Taking n = 3,
a3 = 9 – 5(3) = 9 – 15 = -6
Therefore the series is 4, -1, -6, …
So, a = 4, d = a2 – a1 = -1 – 4 = -5
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 3+(\mathrm{n}-1) 2]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[6+2 \mathrm{n}-2]$
$\Rightarrow S_{n}=\frac{n}{2}[4+2 n]$
⇒ Sn = 2n + n2
Hence, the sum of n terms is n2 + 2n.
Question 7 A
If the nth term of an A.P. is 9 — 5n, find the sum to first 15 terms.
Sol :Given: an = 9 – 5n
Taking n = 1,
a1 = 9 – 5(1) = 9 – 5 = 4
Taking n = 2,
a2 = 9 – 5(2) = 9 – 10 = -1
Taking n = 3,
a3 = 9 – 5(3) = 9 – 15 = -6
Therefore the series is 4, -1, -6, …
So, a = 4, d = a2 – a1 = -1 – 4 = -5
Now, we have to find the sum of the first 15 terms of the AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{15}=\frac{15}{2}[2 \times 4+(15-1)(-5)]$
$\Rightarrow \mathrm{S}_{15}=\frac{15}{2}[8-70]$
$\Rightarrow \mathrm{S}_{15}=\frac{15}{2}[-62]$
⇒ S15 = 15 × (-31)
⇒ S15 = -465
Hence, the sum of 15 terms is -465.
Given: an = 1 – 4n
Taking n = 1,
a1 = 1 – 4(1) = 1 – 4 = -3
Taking n = 2,
a2 = 1 – 4(2) = 1 – 8 = -7
Taking n = 3,
a3 = 1 – 4(3) = 1 – 12 = -11
Therefore the series is -3, -7, -11, …
So, a = -3, d = a2 – a1 = -7 – (-3) = -7 + 3 = -4
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{15}=\frac{15}{2}[2 \times 4+(15-1)(-5)]$
$\Rightarrow \mathrm{S}_{15}=\frac{15}{2}[8-70]$
$\Rightarrow \mathrm{S}_{15}=\frac{15}{2}[-62]$
⇒ S15 = 15 × (-31)
⇒ S15 = -465
Hence, the sum of 15 terms is -465.
Question 7 B
Find the sum of first 25 terms of an A.P. whose nth term is 1 — 4n.
Sol :Given: an = 1 – 4n
Taking n = 1,
a1 = 1 – 4(1) = 1 – 4 = -3
Taking n = 2,
a2 = 1 – 4(2) = 1 – 8 = -7
Taking n = 3,
a3 = 1 – 4(3) = 1 – 12 = -11
Therefore the series is -3, -7, -11, …
So, a = -3, d = a2 – a1 = -7 – (-3) = -7 + 3 = -4
Now, we have to find the sum of the first 25 terms of the AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[2 \times(-3)+(25-1)(-4)]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[-6-96]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[-102]$
⇒ S25 = 25 × (-51)
⇒ S25 = -1275
Hence, the sum of 25 terms is -1275.
Given: Sn = n2 + 2n …(i)
Sn-1 = (n – 1)2 + 2(n – 1) = n2 + 1 – 2n + 2n – 2 = n2 - 1 …(ii)
Subtracting eq (ii) from (i), we get
tn = Sn – Sn-1 = n2 + 2n – n2 + 1 = 2n + 1
The nth term of an AP is 2n + 1.
As it is given that kth term of the AP = 5k + 1
∴ ak = a + (k – 1)d
⇒ 5k + 1 = a + (k – 1)d
⇒ 5k + 1 = a + kd – d
Now, on comparing the coefficient of k, we get
d = 5
and a – d = 1
⇒ a – 5 = 1
⇒ a = 6
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[2 \times(-3)+(25-1)(-4)]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[-6-96]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[-102]$
⇒ S25 = 25 × (-51)
⇒ S25 = -1275
Hence, the sum of 25 terms is -1275.
Question 8
If the sum to n terms of a sequence be n2 + 2n, then prove that the sequence is an A.P.
Sol :Given: Sn = n2 + 2n …(i)
Sn-1 = (n – 1)2 + 2(n – 1) = n2 + 1 – 2n + 2n – 2 = n2 - 1 …(ii)
Subtracting eq (ii) from (i), we get
tn = Sn – Sn-1 = n2 + 2n – n2 + 1 = 2n + 1
The nth term of an AP is 2n + 1.
Question 9
Find the sum to first n terms of an A.P. whose kth term is 5k + 1.
Sol :As it is given that kth term of the AP = 5k + 1
∴ ak = a + (k – 1)d
⇒ 5k + 1 = a + (k – 1)d
⇒ 5k + 1 = a + kd – d
Now, on comparing the coefficient of k, we get
d = 5
and a – d = 1
⇒ a – 5 = 1
⇒ a = 6
We know that,
$S_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 6+(\mathrm{n}-1) 5]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[12+5 \mathrm{n}-5]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[7+5 \mathrm{n}]$
Sn = 3n2 + 5n
Taking n = 1, we get
S1 = 3(1)2 + 5(1)
⇒ S1 = 3 + 5
⇒ S1 = 8
⇒ a1 = 8
Taking n = 2, we get
S2 = 3(2)2 + 5(2)
⇒ S2 = 12 + 10
⇒ S2 = 22
∴ a2 = S2 – S1 = 22 – 8 = 14
Taking n = 3, we get
S3 = 3(3)2 + 5(3)
⇒ S3 = 27 + 15
⇒ S3 = 42
∴ a3 = S3 – S2 = 42 – 22 = 20
So, a = 8,
d = a2 – a1 = 14 – 8 = 6
Now, we have to find the value of m
an = a + (n – 1)d
⇒ am = 8 + (m – 1)6
⇒ 164 = 8 + 6m – 6
⇒ 164 = 2 + 6m
⇒ 162 = 6m
⇒ m = 27
Sn = qn2 + pn
Taking n = 1, we get
S1 = q(1)2 + p(1)
⇒ S1 = q + p
⇒ a1 = q + p
Taking n = 2, we get
S2 = q(2)2 + p(2)
⇒ S2 = 4q + 2p
∴ a2 = S2 – S1 = 4q + 2p – q - p = 3q + p
Taking n = 3, we get
S3 = q(3)2 + p(3)
⇒ S3 = 9q + 3p
∴ a3 = S3 – S2 = 9q + 3p – 4q – 2p = 5q + p
So, a = q + p,
d = a2 – a1 = 3q + p – (q + p) = 3q + p – q – p = 2q
Hence, the common difference is 2q.
$\mathrm{S}_{\mathrm{n}}=\mathrm{nP}+\frac{1}{2} \mathrm{n}(\mathrm{n}-1) \mathrm{Q}$
$S_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 6+(\mathrm{n}-1) 5]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[12+5 \mathrm{n}-5]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[7+5 \mathrm{n}]$
Question 10
If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
[Hint: tm = Sm — Sm-1= 3m2 + 5m — 3 (m— 1)2 — 5 (m— 1) = 3 (2m — 1) + 5 = 6m + 2]
Sol :[Hint: tm = Sm — Sm-1= 3m2 + 5m — 3 (m— 1)2 — 5 (m— 1) = 3 (2m — 1) + 5 = 6m + 2]
Sn = 3n2 + 5n
Taking n = 1, we get
S1 = 3(1)2 + 5(1)
⇒ S1 = 3 + 5
⇒ S1 = 8
⇒ a1 = 8
Taking n = 2, we get
S2 = 3(2)2 + 5(2)
⇒ S2 = 12 + 10
⇒ S2 = 22
∴ a2 = S2 – S1 = 22 – 8 = 14
Taking n = 3, we get
S3 = 3(3)2 + 5(3)
⇒ S3 = 27 + 15
⇒ S3 = 42
∴ a3 = S3 – S2 = 42 – 22 = 20
So, a = 8,
d = a2 – a1 = 14 – 8 = 6
Now, we have to find the value of m
an = a + (n – 1)d
⇒ am = 8 + (m – 1)6
⇒ 164 = 8 + 6m – 6
⇒ 164 = 2 + 6m
⇒ 162 = 6m
⇒ m = 27
Question 11
If the sum of n terms of an A.P. is pn + qn2, where p and q are constants, find the common difference.
Sol :Sn = qn2 + pn
Taking n = 1, we get
S1 = q(1)2 + p(1)
⇒ S1 = q + p
⇒ a1 = q + p
Taking n = 2, we get
S2 = q(2)2 + p(2)
⇒ S2 = 4q + 2p
∴ a2 = S2 – S1 = 4q + 2p – q - p = 3q + p
Taking n = 3, we get
S3 = q(3)2 + p(3)
⇒ S3 = 9q + 3p
∴ a3 = S3 – S2 = 9q + 3p – 4q – 2p = 5q + p
So, a = q + p,
d = a2 – a1 = 3q + p – (q + p) = 3q + p – q – p = 2q
Hence, the common difference is 2q.
Question 12
If the sum of n terms of an A.P. is nP + 1/2 n( n —1)Q , where P and Q are constants, find the common difference of the A.P.
Sol :$\mathrm{S}_{\mathrm{n}}=\mathrm{nP}+\frac{1}{2} \mathrm{n}(\mathrm{n}-1) \mathrm{Q}$
Taking n = 1, we get
$\mathrm{S}_{1}=(1) \mathrm{P}+\frac{1}{2}(1)(1-1) \mathrm{Q}$
⇒ S1 = P
⇒ a1 = P
$\mathrm{S}_{1}=(1) \mathrm{P}+\frac{1}{2}(1)(1-1) \mathrm{Q}$
⇒ S1 = P
⇒ a1 = P
Taking n = 2, we get
$\mathrm{S}_{2}=(2) \mathrm{P}+\frac{1}{2} \times 2(2-1) \mathrm{Q}$
⇒ S2 = 2P + Q
∴ a2 = S2 – S1 = 2P + Q – P = P + Q
$\mathrm{S}_{2}=(2) \mathrm{P}+\frac{1}{2} \times 2(2-1) \mathrm{Q}$
⇒ S2 = 2P + Q
∴ a2 = S2 – S1 = 2P + Q – P = P + Q
Taking n = 3, we get
$\mathrm{S}_{3}=(3) \mathrm{P}+\frac{1}{2}(3)(3-1) \mathrm{Q}$
⇒ S3 = 3P + 3Q
∴ a3 = S3 – S2 = 3P + 3Q – 2P – Q = P + 2Q
So, a = P,
d = a2 – a1 = P + Q – (P) = Q
= a3 – a2 = P + 2Q – (P + Q) = P + 2Q – P – Q = Q
Hence, the common difference is Q.
Here, a = 25, d = 28 – 25 = 3 and an = 100
We know that,
an = a + (n – 1)d
⇒ 100 = 25 + (n – 1)3
⇒ 75 = (n – 1)3
⇒ 25 = n – 1
⇒ 26 = n
Now,
$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{26}=\frac{26}{2}[2 \times 25+(26-1) 3]$
⇒ S26 = 13[50 + 25 × 3]
⇒ S26 = 13[50 + 75]
⇒ S26 = 13 × 125
⇒ S26 = 1625
Let an = 89
AP = 4, 9, 14, …89
Here, a = 4, d = 14 – 9 = 5
We know that
an = a + (n – 1)d
⇒ 89 = 4 + (n – 1)5
⇒ 85 = (n – 1)5
⇒ 17 = n – 1
⇒ 18 = n
So, 89 is the 18th term of the given AP
Now, we find the sum of 4 + 9 + 14 + … + 89
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{18}=\frac{18}{2}[2 \times 4+(18-1) 5]$
⇒ S18 = 9[8 + 17 × 5]
⇒ S18 = 9[8 + 85]
⇒ S18 = 9 × 93
⇒ S18 = 837
Hence, the sum of the given AP is 837.
Here, a = 1, d = 6 – 1 = 5 and Sn = 148
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 1+(\mathrm{n}-1) 5]$
$\Rightarrow 148=\frac{n}{2}[2+5 n-5]$
$\Rightarrow 148=\frac{n}{2}[5 n-3]$
⇒ 296 = n[5n – 3]
⇒ 5n2 – 3n – 296 = 0
⇒ 5n2 – 40n + 37n – 296 = 0
⇒ 5n(n – 8) + 37(n – 8) = 0
⇒ (5n + 37)(n – 8) = 0
⇒ 5n + 37 = 0 or n – 8 = 0
⇒ $\mathrm{n}=-\frac{37}{5}$ or n = 8
But $\mathrm{n}=-\frac{37}{5}$ is not a positive integer.
∴ n = 8
⇒ x = a8 = a + 7d = 1 + 7 × 5 = 1 + 35 = 36
Hence, x = 36
Here, a = 25, d = 22 – 25 = -3 and Sn = 115
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{n}=\frac{n}{2}[2 \times 25+(n-1)(-3)]$
$\Rightarrow 115=\frac{n}{2}[50-3 n+3]$
$\Rightarrow 115=\frac{n}{2}[53-3 n]$
⇒ 230 = n[53 – 3n]
⇒ 3n2 – 53n + 230 = 0
⇒ 3n2 – 30n - 23n + 230 = 0
⇒ 3n(n – 10) - 23(n – 10) = 0
⇒ (3n – 23)(n – 10) = 0
⇒ 3n – 23 = 0 or n – 10 = 0
⇒ $\mathrm{n}=\frac{23}{3}$ or n = 10
But $\mathrm{n}=\frac{23}{3}$ is not an integer.
∴ n = 10
⇒ x = a10 = a + 9d = 25 + 9 × (-3) = 25 – 27 = -2
Hence, x = -2
AP = 64, 60, 56, …
Here, a = 64, d = 60 – 64 = -4
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 544=\frac{n}{2}[2 \times 64+(n-1)(-4)]$
$\Rightarrow 544=\frac{n}{2}[128-4 n+4]$
$\Rightarrow 544=\frac{n}{2}[132-4 n]$
⇒ 1088 = n[132 – 4n]
⇒ 4n2 – 132n + 1088 = 0
⇒ n2 – 33n + 272= 0
⇒ n2 – 16n - 17n + 272 = 0
⇒ n(n – 16) - 17(n – 16) = 0
⇒ (n – 16)(n – 17) = 0
⇒ n – 16 = 0 or n – 17 = 0
⇒ n = 16 or n = 17
If n = 16, a = 64 and d = -4
a16 = 64 + (16 – 1)(-4)
a16 = 64 + 15 × -4
a16 = 64 – 60
a16 = 4
and If n = 17, a = 64 and d = -4
a17 = 64 + (17 – 1)(-4)
a17 = 64 + 16 × -4
a17 = 64 – 64
a17 = 0
$\mathrm{S}_{3}=(3) \mathrm{P}+\frac{1}{2}(3)(3-1) \mathrm{Q}$
⇒ S3 = 3P + 3Q
∴ a3 = S3 – S2 = 3P + 3Q – 2P – Q = P + 2Q
So, a = P,
d = a2 – a1 = P + Q – (P) = Q
= a3 – a2 = P + 2Q – (P + Q) = P + 2Q – P – Q = Q
Hence, the common difference is Q.
Question 13
Find the sum : 25 + 28 + 31 +… + 100
Sol :Here, a = 25, d = 28 – 25 = 3 and an = 100
We know that,
an = a + (n – 1)d
⇒ 100 = 25 + (n – 1)3
⇒ 75 = (n – 1)3
⇒ 25 = n – 1
⇒ 26 = n
Now,
$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{26}=\frac{26}{2}[2 \times 25+(26-1) 3]$
⇒ S26 = 13[50 + 25 × 3]
⇒ S26 = 13[50 + 75]
⇒ S26 = 13 × 125
⇒ S26 = 1625
Question 14
Which term of the A.P. 4, 9, 14, ... is 89? Also, find the sum 4 + 9 + 14 + + 89.
Sol :Let an = 89
AP = 4, 9, 14, …89
Here, a = 4, d = 14 – 9 = 5
We know that
an = a + (n – 1)d
⇒ 89 = 4 + (n – 1)5
⇒ 85 = (n – 1)5
⇒ 17 = n – 1
⇒ 18 = n
So, 89 is the 18th term of the given AP
Now, we find the sum of 4 + 9 + 14 + … + 89
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{18}=\frac{18}{2}[2 \times 4+(18-1) 5]$
⇒ S18 = 9[8 + 17 × 5]
⇒ S18 = 9[8 + 85]
⇒ S18 = 9 × 93
⇒ S18 = 837
Hence, the sum of the given AP is 837.
Question 15 A
Solve for x
1 + 6+11 + 16 +...+x= 148
Sol :1 + 6+11 + 16 +...+x= 148
Here, a = 1, d = 6 – 1 = 5 and Sn = 148
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \times 1+(\mathrm{n}-1) 5]$
$\Rightarrow 148=\frac{n}{2}[2+5 n-5]$
$\Rightarrow 148=\frac{n}{2}[5 n-3]$
⇒ 296 = n[5n – 3]
⇒ 5n2 – 3n – 296 = 0
⇒ 5n2 – 40n + 37n – 296 = 0
⇒ 5n(n – 8) + 37(n – 8) = 0
⇒ (5n + 37)(n – 8) = 0
⇒ 5n + 37 = 0 or n – 8 = 0
⇒ $\mathrm{n}=-\frac{37}{5}$ or n = 8
But $\mathrm{n}=-\frac{37}{5}$ is not a positive integer.
∴ n = 8
⇒ x = a8 = a + 7d = 1 + 7 × 5 = 1 + 35 = 36
Hence, x = 36
Question 15 B
Solve for x
25+22+19+ 16+...+x= 115
Sol :25+22+19+ 16+...+x= 115
Here, a = 25, d = 22 – 25 = -3 and Sn = 115
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{n}=\frac{n}{2}[2 \times 25+(n-1)(-3)]$
$\Rightarrow 115=\frac{n}{2}[50-3 n+3]$
$\Rightarrow 115=\frac{n}{2}[53-3 n]$
⇒ 230 = n[53 – 3n]
⇒ 3n2 – 53n + 230 = 0
⇒ 3n2 – 30n - 23n + 230 = 0
⇒ 3n(n – 10) - 23(n – 10) = 0
⇒ (3n – 23)(n – 10) = 0
⇒ 3n – 23 = 0 or n – 10 = 0
⇒ $\mathrm{n}=\frac{23}{3}$ or n = 10
But $\mathrm{n}=\frac{23}{3}$ is not an integer.
∴ n = 10
⇒ x = a10 = a + 9d = 25 + 9 × (-3) = 25 – 27 = -2
Hence, x = -2
Question 16
Find the number of terms of the A.P. 64, 60, 56, ... so that their sum is 544. Explain the double answer.
Sol :AP = 64, 60, 56, …
Here, a = 64, d = 60 – 64 = -4
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 544=\frac{n}{2}[2 \times 64+(n-1)(-4)]$
$\Rightarrow 544=\frac{n}{2}[128-4 n+4]$
$\Rightarrow 544=\frac{n}{2}[132-4 n]$
⇒ 1088 = n[132 – 4n]
⇒ 4n2 – 132n + 1088 = 0
⇒ n2 – 33n + 272= 0
⇒ n2 – 16n - 17n + 272 = 0
⇒ n(n – 16) - 17(n – 16) = 0
⇒ (n – 16)(n – 17) = 0
⇒ n – 16 = 0 or n – 17 = 0
⇒ n = 16 or n = 17
If n = 16, a = 64 and d = -4
a16 = 64 + (16 – 1)(-4)
a16 = 64 + 15 × -4
a16 = 64 – 60
a16 = 4
and If n = 17, a = 64 and d = -4
a17 = 64 + (17 – 1)(-4)
a17 = 64 + 16 × -4
a17 = 64 – 64
a17 = 0
Now, we will check at which term the sum of the AP is 544.
$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$
$\Rightarrow \mathrm{S}_{16}=\frac{16}{2}[64+4]$
⇒ S16 = 8[68]
⇒ S16 = 544
and $S_{17}=\frac{17}{2}[64+0]$
⇒ S17 = 17 × 32
⇒ S17 = 544
So, the terms may be either 17 or 16 both holds true.
We get a double answer because the 17th term is zero and when we add this in the sum, the sum remains the same.
AP = 3, 5, 7, 9, …
Here, a = 3, d = 5 – 3 = 2 and Sn = 120
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow 120=\frac{n}{2}[2 \times 3+(n-1)(2)]$
$\Rightarrow 120=\frac{n}{2}[6+2 n-2]$
$\Rightarrow 120=\frac{\mathrm{n}}{2}[4+2 \mathrm{n}]$
⇒ 120 = n[2+n]
⇒ n2 + 2n – 120 = 0
⇒ n2 + 12n – 10n – 120 = 0
⇒ n(n + 12) - 10(n + 12) = 0
⇒ (n – 10)(n + 12) = 0
⇒ n – 10 = 0 or n + 12 = 0
⇒ n = 10 or n = -12
But number of terms can’t be negative. So, n = 10
Hence, for n = 10 the sum is 120 for the given AP.
AP = 63, 60, 57,…
Here, a = 63, d = 60 – 63 = -3 and Sn = 693
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 693=\frac{n}{2}[2 \times 63+(\mathrm{n}-1)(-3)]$
$\Rightarrow 693=\frac{n}{2}[126-3 n+3]$
$\Rightarrow 693=\frac{n}{2}[129-3 n]$
⇒ 1386 = n[129 – 3n]
⇒ 3n2 – 129n + 1386 = 0
⇒ n2 – 43n + 462 = 0
⇒ n2 – 22n – 21n + 462 = 0
⇒ n(n – 22) - 21(n – 22) = 0
⇒ (n – 21)(n – 22) = 0
⇒ n – 21 = 0 or n – 22 = 0
⇒ n = 21 or n = 22
So, n = 21 and 22
If n = 21, a = 63 and d = -3
a21 = 63 + (21 – 1)(-3)
a21 = 63 + 20 × -3
a21 = 63 – 60
a21 = 3
and If n = 22, a = 63 and d = -3
a22 = 63 + (22 – 1)(-3)
a22 = 63 + 21 × -3
a22 = 63 – 63
a22 = 0
Now, we will check at which term the sum of the AP is 693.
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[63+3]$
⇒ S21 = 21 × 33
⇒ S21 = 693
and $\mathrm{S}_{22}=\frac{22}{2}[63+0]$
⇒ S22 = 11 × 63
⇒ S22 = 693
So, the terms may be either 21 or 22 both holds true.
We get the double answer because here the 22nd term is zero and it does not affect the sum.
Here, a = 15, d = 12 – 15 = -3 and Sn = 15
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow 15=\frac{n}{2}[2 \times 15+(n-1)(-3)]$
$\Rightarrow 15=\frac{n}{2}[30-3 n+3]$
$\Rightarrow 15=\frac{n}{2}[33-3 n]$
⇒ 30 = n[33 – 3n ]
⇒ 3n2 – 33n + 30 = 0
⇒ 3n2 – 30n – 3n + 30 = 0
⇒ 3n(n – 10) -3(n – 10) = 0
⇒ (n – 10)(3n – 3) = 0
⇒ n – 10 = 0 or 3n – 3 = 0
⇒ n = 10 or n = 1
The number of terms can be 1 or 10.
Here, the common difference is negative.
∴ The AP starts from a positive term, and its terms are decreasing.
∴ All the terms after 6th term are negative.
We get a double answer because these positive terms from 2nd to 5th term when added to negative terms from 7th to 10th term, they cancel out each other and the sum remains same.
The odd numbers lying between 100 and 200 are
101, 103, 105,…, 199
a2 – a1 = 103 – 101 = 2
a3 – a2 = 105 – 103 = 2
∵ a3 – a2 = a2 – a1 = 2
Therefore, the series is in AP
Here, a = 101, d = 2 and an = 199
We know that,
an = a + (n – 1)d
⇒ 199 = 101 + (n – 1)2
⇒ 199 – 101 = (n – 1)2
⇒ 98 = (n – 1)2
⇒ 49 = (n – 1)
⇒ n = 50
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{50}=\frac{50}{2}[2 \times 101+(50-1) 2]$
⇒ S50 = 25[202 + 49 × 2]
⇒ S50 = 25[300]
⇒ S50 = 7500
Hence, the sum of all odd numbers lying between 100 and 200 is 7500.
The odd numbers lying between 1 and 2001 are
1, 3, 5,…, 2001
a2 – a1 = 3 – 1 = 2
a3 – a2 = 5 – 3 = 2
∵ a3 – a2 = a2 – a1 = 2
Therefore, the series is in AP
Here, a = 1, d = 2 and an = 2001
We know that,
an = a + (n – 1)d
⇒ 2001 = 1 + (n – 1)2
⇒ 2001 – 1 = (n – 1)2
⇒ 2000 = (n – 1)2
⇒ 1000 = (n – 1)
⇒ n = 1001
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{1001}=\frac{1001}{2}[2 \times 1+(1001-1) 2]$
⇒ S1001 = 1001[1 + 1000]
⇒ S1001 = 1001 [1001]
⇒ S1001 = 1002001
Hence, the sum of all odd numbers lying between 1 and 2001 is 1002001.
Given: a2 = 2 and a7 = 22 and n = 35
We know that,
a2 = a + d = 2 …(i)
and a7 = a + 6d = 22 …(ii)
Solving the linear equations (i) and (ii), we get
a + d – a – 6d = 2 – 22
⇒ - 5d = -20
⇒ d =4
Putting the value of d in eq. (i), we get
a + 4 = 2
⇒ a = 2 – 4 = -2
Now, we have to find the sum of first 35 terms.
$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{35}=\frac{35}{2}[2 \times(-2)+(35-1) 4]$
$\Rightarrow \mathrm{S}_{35}=\frac{35}{2}[-4+34 \times 4]$
⇒ S35 = 35 [-2 + 34 × 2]
⇒ S35 = 35 [66]
⇒ S35 = 2310
Given: Sp = q and Sq = p
To find: Sp+q
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{\mathrm{p}}=\frac{\mathrm{p}}{2}[2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d}]$
$\Rightarrow \mathrm{q}=\frac{\mathrm{p}}{2}[2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d}]$
$\Rightarrow \frac{2 \mathrm{q}}{\mathrm{p}}=2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d}$
$\Rightarrow \frac{2 \mathrm{q}}{\mathrm{p}}-(\mathrm{p}-1) \mathrm{d}=2 \mathrm{a}$ …(i)
$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$
$\Rightarrow \mathrm{S}_{16}=\frac{16}{2}[64+4]$
⇒ S16 = 8[68]
⇒ S16 = 544
and $S_{17}=\frac{17}{2}[64+0]$
⇒ S17 = 17 × 32
⇒ S17 = 544
So, the terms may be either 17 or 16 both holds true.
We get a double answer because the 17th term is zero and when we add this in the sum, the sum remains the same.
Question 17
How many terms of the A.P. 3, 5, 7, 9, ... must be added to get the sum 120?
Sol :AP = 3, 5, 7, 9, …
Here, a = 3, d = 5 – 3 = 2 and Sn = 120
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow 120=\frac{n}{2}[2 \times 3+(n-1)(2)]$
$\Rightarrow 120=\frac{n}{2}[6+2 n-2]$
$\Rightarrow 120=\frac{\mathrm{n}}{2}[4+2 \mathrm{n}]$
⇒ 120 = n[2+n]
⇒ n2 + 2n – 120 = 0
⇒ n2 + 12n – 10n – 120 = 0
⇒ n(n + 12) - 10(n + 12) = 0
⇒ (n – 10)(n + 12) = 0
⇒ n – 10 = 0 or n + 12 = 0
⇒ n = 10 or n = -12
But number of terms can’t be negative. So, n = 10
Hence, for n = 10 the sum is 120 for the given AP.
Question 18
Find the number of terms of the A.P. 63, 60, 57, ... so that their sum is 693. Explain the double answer.
Sol :AP = 63, 60, 57,…
Here, a = 63, d = 60 – 63 = -3 and Sn = 693
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow 693=\frac{n}{2}[2 \times 63+(\mathrm{n}-1)(-3)]$
$\Rightarrow 693=\frac{n}{2}[126-3 n+3]$
$\Rightarrow 693=\frac{n}{2}[129-3 n]$
⇒ 1386 = n[129 – 3n]
⇒ 3n2 – 129n + 1386 = 0
⇒ n2 – 43n + 462 = 0
⇒ n2 – 22n – 21n + 462 = 0
⇒ n(n – 22) - 21(n – 22) = 0
⇒ (n – 21)(n – 22) = 0
⇒ n – 21 = 0 or n – 22 = 0
⇒ n = 21 or n = 22
So, n = 21 and 22
If n = 21, a = 63 and d = -3
a21 = 63 + (21 – 1)(-3)
a21 = 63 + 20 × -3
a21 = 63 – 60
a21 = 3
and If n = 22, a = 63 and d = -3
a22 = 63 + (22 – 1)(-3)
a22 = 63 + 21 × -3
a22 = 63 – 63
a22 = 0
Now, we will check at which term the sum of the AP is 693.
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[63+3]$
⇒ S21 = 21 × 33
⇒ S21 = 693
and $\mathrm{S}_{22}=\frac{22}{2}[63+0]$
⇒ S22 = 11 × 63
⇒ S22 = 693
So, the terms may be either 21 or 22 both holds true.
We get the double answer because here the 22nd term is zero and it does not affect the sum.
Question 19
How many terms of the series 15 + 12 + 9 + ... must be taken to make 15? Explain the double answer.
Sol :Here, a = 15, d = 12 – 15 = -3 and Sn = 15
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow 15=\frac{n}{2}[2 \times 15+(n-1)(-3)]$
$\Rightarrow 15=\frac{n}{2}[30-3 n+3]$
$\Rightarrow 15=\frac{n}{2}[33-3 n]$
⇒ 30 = n[33 – 3n ]
⇒ 3n2 – 33n + 30 = 0
⇒ 3n2 – 30n – 3n + 30 = 0
⇒ 3n(n – 10) -3(n – 10) = 0
⇒ (n – 10)(3n – 3) = 0
⇒ n – 10 = 0 or 3n – 3 = 0
⇒ n = 10 or n = 1
The number of terms can be 1 or 10.
Here, the common difference is negative.
∴ The AP starts from a positive term, and its terms are decreasing.
∴ All the terms after 6th term are negative.
We get a double answer because these positive terms from 2nd to 5th term when added to negative terms from 7th to 10th term, they cancel out each other and the sum remains same.
Question 20 A
Find the sum of all the odd numbers lying between 100 and 200.
Sol :The odd numbers lying between 100 and 200 are
101, 103, 105,…, 199
a2 – a1 = 103 – 101 = 2
a3 – a2 = 105 – 103 = 2
∵ a3 – a2 = a2 – a1 = 2
Therefore, the series is in AP
Here, a = 101, d = 2 and an = 199
We know that,
an = a + (n – 1)d
⇒ 199 = 101 + (n – 1)2
⇒ 199 – 101 = (n – 1)2
⇒ 98 = (n – 1)2
⇒ 49 = (n – 1)
⇒ n = 50
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{50}=\frac{50}{2}[2 \times 101+(50-1) 2]$
⇒ S50 = 25[202 + 49 × 2]
⇒ S50 = 25[300]
⇒ S50 = 7500
Hence, the sum of all odd numbers lying between 100 and 200 is 7500.
Question 20 B
Find the sum of all odd integers from 1 to 2001.
Sol :The odd numbers lying between 1 and 2001 are
1, 3, 5,…, 2001
a2 – a1 = 3 – 1 = 2
a3 – a2 = 5 – 3 = 2
∵ a3 – a2 = a2 – a1 = 2
Therefore, the series is in AP
Here, a = 1, d = 2 and an = 2001
We know that,
an = a + (n – 1)d
⇒ 2001 = 1 + (n – 1)2
⇒ 2001 – 1 = (n – 1)2
⇒ 2000 = (n – 1)2
⇒ 1000 = (n – 1)
⇒ n = 1001
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{1001}=\frac{1001}{2}[2 \times 1+(1001-1) 2]$
⇒ S1001 = 1001[1 + 1000]
⇒ S1001 = 1001 [1001]
⇒ S1001 = 1002001
Hence, the sum of all odd numbers lying between 1 and 2001 is 1002001.
Question 21
Determine the sum of first 35 terms of an A.P., if the second term is 2 and the seventh term is 22.
Sol :Given: a2 = 2 and a7 = 22 and n = 35
We know that,
a2 = a + d = 2 …(i)
and a7 = a + 6d = 22 …(ii)
Solving the linear equations (i) and (ii), we get
a + d – a – 6d = 2 – 22
⇒ - 5d = -20
⇒ d =4
Putting the value of d in eq. (i), we get
a + 4 = 2
⇒ a = 2 – 4 = -2
Now, we have to find the sum of first 35 terms.
$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{35}=\frac{35}{2}[2 \times(-2)+(35-1) 4]$
$\Rightarrow \mathrm{S}_{35}=\frac{35}{2}[-4+34 \times 4]$
⇒ S35 = 35 [-2 + 34 × 2]
⇒ S35 = 35 [66]
⇒ S35 = 2310
Question 22
If the sum of the first p terms of an A.P. is q and the sum of first q terms is p, then find the sum of first (p + q) terms.
Sol :Given: Sp = q and Sq = p
To find: Sp+q
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{\mathrm{p}}=\frac{\mathrm{p}}{2}[2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d}]$
$\Rightarrow \mathrm{q}=\frac{\mathrm{p}}{2}[2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d}]$
$\Rightarrow \frac{2 \mathrm{q}}{\mathrm{p}}=2 \mathrm{a}+(\mathrm{p}-1) \mathrm{d}$
$\Rightarrow \frac{2 \mathrm{q}}{\mathrm{p}}-(\mathrm{p}-1) \mathrm{d}=2 \mathrm{a}$ …(i)
Now,
$\Rightarrow \mathrm{S}_{\mathrm{q}}=\frac{\mathrm{q}}{2}[2 \mathrm{a}+(\mathrm{q}-1) \mathrm{d}]$
$\Rightarrow \mathrm{p}=\frac{\mathrm{q}}{2}[2 \mathrm{a}+(\mathrm{q}-1) \mathrm{d}]$
$\Rightarrow \frac{2 \mathrm{p}}{\mathrm{q}}=2 \mathrm{a}+(\mathrm{q}-1) \mathrm{d}$
$\Rightarrow \frac{2 p}{q}-(q-1) d=2 a$ …(ii)
$\Rightarrow \mathrm{S}_{\mathrm{q}}=\frac{\mathrm{q}}{2}[2 \mathrm{a}+(\mathrm{q}-1) \mathrm{d}]$
$\Rightarrow \mathrm{p}=\frac{\mathrm{q}}{2}[2 \mathrm{a}+(\mathrm{q}-1) \mathrm{d}]$
$\Rightarrow \frac{2 \mathrm{p}}{\mathrm{q}}=2 \mathrm{a}+(\mathrm{q}-1) \mathrm{d}$
$\Rightarrow \frac{2 p}{q}-(q-1) d=2 a$ …(ii)
From eq. (i) and (ii), we get
$\frac{2 \mathrm{q}}{\mathrm{p}}-(\mathrm{p}-1) \mathrm{d}=\frac{2 \mathrm{p}}{\mathrm{q}}-(\mathrm{q}-1) \mathrm{d}$
$\Rightarrow \frac{2 q}{p}-\frac{2 p}{q}=(p-1) d-(q-1) d$
$\Rightarrow \frac{2 \mathrm{q}^{2}-2 \mathrm{p}^{2}}{\mathrm{pq}}=\mathrm{d}[\mathrm{p}-1-\mathrm{q}+1]$
$\Rightarrow \frac{2(\mathrm{q}-\mathrm{p})(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}=\mathrm{d}(\mathrm{p}-\mathrm{q})$
[∵, a2 – b2 = (a – b)(a + b)]
$\Rightarrow \frac{2(\mathrm{q}-\mathrm{p})(\mathrm{q}+\mathrm{p})}{-\mathrm{pq}(\mathrm{q}-\mathrm{p})}=\mathrm{d}$
$\Rightarrow \mathrm{d}=\frac{-2(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}$ …(iii)
$\frac{2 \mathrm{q}}{\mathrm{p}}-(\mathrm{p}-1) \mathrm{d}=\frac{2 \mathrm{p}}{\mathrm{q}}-(\mathrm{q}-1) \mathrm{d}$
$\Rightarrow \frac{2 q}{p}-\frac{2 p}{q}=(p-1) d-(q-1) d$
$\Rightarrow \frac{2 \mathrm{q}^{2}-2 \mathrm{p}^{2}}{\mathrm{pq}}=\mathrm{d}[\mathrm{p}-1-\mathrm{q}+1]$
$\Rightarrow \frac{2(\mathrm{q}-\mathrm{p})(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}=\mathrm{d}(\mathrm{p}-\mathrm{q})$
[∵, a2 – b2 = (a – b)(a + b)]
$\Rightarrow \frac{2(\mathrm{q}-\mathrm{p})(\mathrm{q}+\mathrm{p})}{-\mathrm{pq}(\mathrm{q}-\mathrm{p})}=\mathrm{d}$
$\Rightarrow \mathrm{d}=\frac{-2(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}$ …(iii)
Now, putting the value of d in eq. (i), we get
$\Rightarrow \frac{2 \mathrm{q}}{\mathrm{p}}-(\mathrm{p}-1)\left(\frac{-2(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}\right)=2 \mathrm{a}$
$\Rightarrow \frac{\mathrm{q}}{\mathrm{p}}+\frac{(\mathrm{p}-1)(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}=\mathrm{a}$
$\Rightarrow \frac{\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}}{\mathrm{pq}}=\mathrm{a}$ …(iv)
$\Rightarrow \frac{2 \mathrm{q}}{\mathrm{p}}-(\mathrm{p}-1)\left(\frac{-2(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}\right)=2 \mathrm{a}$
$\Rightarrow \frac{\mathrm{q}}{\mathrm{p}}+\frac{(\mathrm{p}-1)(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}=\mathrm{a}$
$\Rightarrow \frac{\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}}{\mathrm{pq}}=\mathrm{a}$ …(iv)
Now, we to find Sp+q
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left[2\left(\frac{\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}}{\mathrm{pq}}\right)+(\mathrm{p}+\mathrm{q}-1)\left(\frac{-2(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}\right)\right]$
[from (iii) & (iv)]
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=(\mathrm{p}+\mathrm{q})\left[\left(\frac{\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}}{\mathrm{pq}}\right)+(\mathrm{p}+\mathrm{q}-1)\left(\frac{-(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}\right)\right]$
$\Rightarrow S_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}\left[\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}+(\mathrm{p}+\mathrm{q}-1)(-\mathrm{q}-\mathrm{p})\right]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}\left[\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}-\mathrm{pq}-\mathrm{p}^{2}-\mathrm{q}^{2}-\mathrm{pq}+\mathrm{q}+\mathrm{p}\right]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}[-\mathrm{pq}]$
⇒ Sp+q = - (p+q)
Hence, the sum of first (p+q) terms is –(p + q)
$\mathrm{AP}=-6, \frac{-11}{2},-5, \ldots$
Here, a = -6,
$\mathrm{d}=-\frac{11}{2}-(-6)$ $=\frac{-11+12}{2}=\frac{1}{2}$
and Sn = -25
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow-25=\frac{\mathrm{n}}{2}\left[2 \times(-6)+(\mathrm{n}-1)\left(\frac{1}{2}\right)\right]$
$\Rightarrow-25=\frac{n}{2}\left[\frac{-24+n-1}{2}\right]$
$\Rightarrow-25=\frac{n}{4}[-25+n]$
⇒ -100 = n[-25 + n]
⇒ n2 – 25n + 100 = 0
⇒ n2 – 20n – 5n + 100 = 0
⇒ n(n – 20) - 5(n – 20) = 0
⇒ (n – 20)(n – 5) = 0
⇒ n – 5 = 0 or n – 20 = 0
⇒ n = 5 or n = 20
So, n = 5 or 20
The numbers lying between 107 and 253 that are multiples of 5 are
110, 115, 120,…, 250
a2 – a1 = 115 – 110 = 5
a3 – a2 = 120 – 115 = 5
∵ a3 – a2 = a2 – a1 = 5
Therefore, the series is in AP
Here, a = 110, d = 5 and an = 250
We know that,
an = a + (n – 1)d
⇒ 250 = 110 + (n – 1)5
⇒ 250 – 110 = (n – 1)5
⇒ 140 = (n – 1)5
⇒ 28 = (n – 1)
⇒ n = 29
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{29}=\frac{29}{2}[2 \times 110+(29-1) 5]$
⇒ S29 = 29[110 + 14 × 5]
⇒ S29 = 29[180]
⇒ S29 = 5220
Hence, the sum of all numbers lying between 107 and 253 is 5220.
The numbers lying between 100 and 1000 that are multiples of 5 are
105, 110, 115, 120,…, 995
a2 – a1 = 110 – 105 = 5
a3 – a2 = 115 – 110 = 5
∵ a3 – a2 = a2 – a1 = 5
Therefore, the series is in AP
Here, a = 105, d = 5 and an = 995
We know that,
an = a + (n – 1)d
⇒ 995 = 105 + (n – 1)5
⇒ 995 – 105 = (n – 1)5
⇒ 890 = (n – 1)5
⇒ 178 = (n – 1)
⇒ n = 179
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{179}=\frac{179}{2}[2 \times 105+(179-1) 5]$
⇒ S179 = 179[105 + 89 × 5]
⇒ S179 = 179 [550]
⇒ S179 = 98450
Hence, the sum of all numbers lying between 100 and 1000 that are multiples of 5 is 98450.
The numbers lying between 300 and 700 which are multiples of 9 are
306, 315, 324,…, 693
a2 – a1 = 315 – 306 = 9
a3 – a2 = 324 – 315 = 9
∵ a3 – a2 = a2 – a1 = 9
Therefore, the series is in AP
Here, a = 306, d = 9 and an = 693
We know that,
an = a + (n – 1)d
⇒ 693 = 306 + (n – 1)9
⇒ 693 - 306 = (n – 1)9
⇒ 387 = (n – 1)9
⇒ 43 = (n – 1)
⇒ n = 44
Now, we have to find the sum of this AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{44}=\frac{44}{2}[2 \times 306+(44-1) 9]$
⇒ S44 = 22[612 + 387]
⇒ S44 = 22[999]
⇒ S44 = 21978
Hence, the sum of all numbers lying between 300 and 700 is 21978.
The three digit natural numbers which are multiples of 7 are
105, 112, 119,…, 994
a2 – a1 = 112 – 105 = 7
a3 – a2 = 112 – 105 = 7
∵ a3 – a2 = a2 – a1 = 7
Therefore, the series is in AP
Here, a = 105, d = 7 and an = 994
We know that,
an = a + (n – 1)d
⇒ 994 = 105 + (n – 1)7
⇒ 994 – 105 = (n – 1)7
⇒ 889 = (n – 1)7
⇒ 127 = (n – 1)
⇒ n = 128
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{128}=\frac{128}{2}[2 \times 105+(128-1) 7]$
⇒ S128 = 64[210 + 127 × 7]
⇒ S128 = 64[1099]
⇒ S128 = 70336
Hence, the sum of all three digit numbers which are multiples of 7 are 70336.
The numbers lying between 100 and 500 which are divisible by 8 are
104, 112, 120, 128, 136,…, 496
a2 – a1 = 112 – 104 = 8
a3 – a2 = 120 – 112 = 8
∵ a3 – a2 = a2 – a1 = 8
Therefore, the series is in AP
Here, a = 120, d = 8 and an = 496
We know that,
an = a + (n – 1)d
⇒ 496 = 104 + (n – 1)8
⇒ 496 – 104 = (n – 1)8
⇒ 392 = (n – 1)8
⇒ 49 = (n – 1)
⇒ n = 50
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{50}=\frac{50}{2}[2 \times 104+(50-1) 8]$
⇒ S50 = 25[208 + 49 × 8]
⇒ S50 = 25[600]
⇒ S50 = 15000
Hence, the sum of all numbers lying between 100 and 500 and divisible by 8 is 15000.
The three digit natural numbers which are divisible by 13 are
104, 117, 130,…, 988
a2 – a1 = 117 – 104 = 13
a3 – a2 = 130 – 117 = 13
∵ a3 – a2 = a2 – a1 = 13
Therefore, the series is in AP
Here, a = 104, d = 13 and an = 988
We know that,
an = a + (n – 1)d
⇒ 988 = 104 + (n – 1)13
⇒ 988 – 104 = (n – 1)13
⇒ 884 = (n – 1)13
⇒ 68 = (n – 1)
⇒ n = 69
Now, we have to find the sum of this AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{69}=\frac{69}{2}[2 \times 104+(69-1) 13]$
$\Rightarrow \mathrm{S}_{69}=\frac{69}{2}[2 \times 104+68 \times 13]$
⇒ S69 = 69[104 + 34 × 13]
⇒ S69 = 69[546]
⇒ S69 = 37674
Hence, the sum of three digit natural numbers which are divisible by 13 are 37674.
Given: a5 = 13 and a15 = -17 and n = 21
We know that,
a5 = a + 4d = 13 …(i)
and a15 = a + 14d = -17 …(ii)
Solving the linear equations (i) and (ii), we get
a + 4d – a – 14d = 13 – (-17)
⇒ -10d = 13 + 17
⇒ -10d = 30
⇒ d = -3
Putting the value of d in eq. (i), we get
a + 4(-3) = 13
⇒ a = 13 + 12 = 25
Now, we have to find the sum of first 21 terms.
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[2 \times(25)+(21-1)(-3)]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[50+20 \times(-3)]$
⇒ S21 = 21 [25 + 10 × (-3)]
⇒ S21 = 21 [-5]
⇒ S21 =-105
Given: a2 = 8 and a4 = 14 and n = 21
We know that,
a2 = a + d = 8 …(i)
and a4 = a + 3d = 14 …(ii)
Solving the linear equations (i) and (ii), we get
a + d – a – 3d = 8 – 14
⇒ -2d = -6
⇒ d = 3
Putting the value of d in eq. (i), we get
a + 3 = 8
⇒ a = 8 – 3 = 5
Now, we have to find the sum of first 21 terms.
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[2 \times(5)+(21-1)(3)]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[10+20 \times(3)]$
⇒ S21 = 21 [5 + 10 × (3)]
⇒ S21 = 21 [35]
⇒ S21 = 735
Given: a2 = 2 and a4 = 8 and n = 51
We know that,
a2 = a + d = 2 …(i)
and a4 = a + 3d = 8 …(ii)
Solving the linear equations (i) and (ii), we get
a + d – a – 3d = 2 – 8
⇒ -2d = -6
⇒ d = 3
Putting the value of d in eq. (i), we get
a + 3 = 2
⇒ a = 2 – 3 = -1
Now, we have to find the sum of first 51 terms.
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{51}=\frac{51}{2}[2 \times(-1)+(51-1)(3)]$
$\Rightarrow \mathrm{S}_{51}=\frac{51}{2}[-1+50 \times(3)]$
⇒ S51 = 51 [-1 + 25 × (3)]
⇒ S51 = 51 [74]
⇒ S51 = 3774
Given: a2 = 9 and a4 = 21 and n = 25
We know that,
a2 = a + d = 9 …(i)
and a4 = a + 3d = 21 …(ii)
Solving the linear equations (i) and (ii), we get
a + d – a – 3d = 9 – 21
⇒ -2d = -12
⇒ d = 6
Putting the value of d in eq. (i), we get
a + 6 = 9
⇒ a = 9 – 6 = 3
Now, we have to find the sum of first 25 terms.
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[2 \times(3)+(25-1)(6)]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[6+24 \times(6)]$
⇒ S25 = 25 [3 + 12 × (6)]
⇒ S25 = 25 [75]
⇒ S25 = 1875
Given: S8 = 64 and S19 = 361
We know that,
$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{8}=\frac{8}{2}[2 \mathrm{a}+(8-1) \mathrm{d}]$
⇒ 64 = 4 [2a +7d]
⇒ 16 = 2a + 7d …(i)
Now,
$\Rightarrow \mathrm{S}_{19}=\frac{19}{2}[2 \mathrm{a}+(19-1) \mathrm{d}]$
$\Rightarrow 361=\frac{19}{2}[2 \mathrm{a}+18 \mathrm{d}]$
⇒ 38 = 2a + 18d …(ii)
Solving linear equations (i) and (ii), we get
2a + 7d – 2a – 18d = 16 – 38
⇒ -11d = -22
⇒ d = 2 …(iii)
Putting the value of d in eq. (i), we get
2a + 7(2) = 16
⇒ 2a = 16 – 14
⇒ 2a = 2 …(iv)
Now, we have to find the Sn
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2+(\mathrm{n}-1) 2]$ [from (iii) and (iv)]
⇒ Sn = n [1 + n – 1]
⇒ Sn = n2
Given: First term, a = 17
Last term, l = 350
common difference, d = 9
We know that,
l = a + (n – 1)d
⇒ 350 = 17 + (n – 1)9
⇒ 333 = (n – 1)9
⇒ 37 = n – 1
⇒ n = 38
So, there are 38 terms in the AP
Now, we have to find the sum of this AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{38}=\frac{38}{2}[2 \times 17+(38-1) 9]$
⇒ S38 = 19 [34 +37×9]
⇒ S38 = 19 [34 + 333]
⇒ S38 = 19 × 367
⇒ S38 = 6973
Hence, the sum of 38 terms is 6973.
Given: a1 = a
a3 = a + 2d = b
⇒ 2d = b – a
$\Rightarrow \mathrm{d}=\frac{\mathrm{b}-\mathrm{a}}{2}$
and an = a + (n – 1)d
$c=a+(n-1)\left(\frac{b-a}{2}\right)$
$\Rightarrow \mathrm{c}-\mathrm{a}=(\mathrm{n}-1)\left(\frac{\mathrm{b}-\mathrm{a}}{2}\right)$
$\Rightarrow \frac{2(\mathrm{c}-\mathrm{a})}{(\mathrm{b}-\mathrm{a})}=(\mathrm{n}-1)$
$\Rightarrow \frac{2(c-a)}{(b-a)}+1=n$
$\Rightarrow \frac{2 c-2 a+b-a}{b-a}=n$
$\Rightarrow \frac{b+2 c-3 a}{b-a}=n$
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[\mathrm{a}+\mathrm{l}]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\frac{\mathrm{b}+2 \mathrm{c}-3 \mathrm{a}}{\mathrm{b}-\mathrm{a}}}{2}[\mathrm{a}+\mathrm{c}]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{(\mathrm{b}+2 \mathrm{c}-3 \mathrm{a})(\mathrm{a}+\mathrm{c})}{2(\mathrm{b}-\mathrm{a})}$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{(\mathrm{c}+\mathrm{a})[\mathrm{b}-\mathrm{a}+2(\mathrm{c}-\mathrm{a})]}{2(\mathrm{b}-\mathrm{a})}$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{c}+\mathrm{a}}{2}+\frac{(\mathrm{c}+\mathrm{a})(\mathrm{c}-\mathrm{a})}{\mathrm{b}-\mathrm{a}}$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{c}+\mathrm{a}}{2}+\frac{\mathrm{c}^{2}-\mathrm{a}^{2}}{\mathrm{b}-\mathrm{a}}$
Given: $a_{m}=\frac{1}{n}$
Now, am = a + (m – 1)d
$\Rightarrow \frac{1}{\mathrm{n}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}$
⇒ an + n(m – 1)d = 1
⇒ an + mnd – nd = 1 …(i)
and $a_{n}=\frac{1}{m}$
$\Rightarrow \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\frac{1}{\mathrm{m}}$
⇒ am + mnd – md = 1 …(ii)
From eq. (i) and (ii), we get
an + mnd – nd = am + mnd – md
⇒ a(n – m) –d (n – m) = 0
⇒ a = d
Now, putting the value of a in eq. (i), we get
dn + mnd – nd = 1
⇒ mnd = 1
$\Rightarrow \mathrm{d}=\frac{1}{\mathrm{mn}}$
Hence, $a=\frac{1}{m n}$
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left[2\left(\frac{\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}}{\mathrm{pq}}\right)+(\mathrm{p}+\mathrm{q}-1)\left(\frac{-2(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}\right)\right]$
[from (iii) & (iv)]
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=(\mathrm{p}+\mathrm{q})\left[\left(\frac{\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}}{\mathrm{pq}}\right)+(\mathrm{p}+\mathrm{q}-1)\left(\frac{-(\mathrm{q}+\mathrm{p})}{\mathrm{pq}}\right)\right]$
$\Rightarrow S_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}\left[\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}+(\mathrm{p}+\mathrm{q}-1)(-\mathrm{q}-\mathrm{p})\right]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}\left[\mathrm{q}^{2}+\mathrm{pq}+\mathrm{p}^{2}-\mathrm{q}-\mathrm{p}-\mathrm{pq}-\mathrm{p}^{2}-\mathrm{q}^{2}-\mathrm{pq}+\mathrm{q}+\mathrm{p}\right]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{\mathrm{pq}}[-\mathrm{pq}]$
⇒ Sp+q = - (p+q)
Hence, the sum of first (p+q) terms is –(p + q)
Question 23
How many terms of the A.P. -6,$-\frac{11}{2}$,-5 ... are needed to get the sum - 25?
Sol :$\mathrm{AP}=-6, \frac{-11}{2},-5, \ldots$
Here, a = -6,
$\mathrm{d}=-\frac{11}{2}-(-6)$ $=\frac{-11+12}{2}=\frac{1}{2}$
and Sn = -25
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow-25=\frac{\mathrm{n}}{2}\left[2 \times(-6)+(\mathrm{n}-1)\left(\frac{1}{2}\right)\right]$
$\Rightarrow-25=\frac{n}{2}\left[\frac{-24+n-1}{2}\right]$
$\Rightarrow-25=\frac{n}{4}[-25+n]$
⇒ -100 = n[-25 + n]
⇒ n2 – 25n + 100 = 0
⇒ n2 – 20n – 5n + 100 = 0
⇒ n(n – 20) - 5(n – 20) = 0
⇒ (n – 20)(n – 5) = 0
⇒ n – 5 = 0 or n – 20 = 0
⇒ n = 5 or n = 20
So, n = 5 or 20
Question 24 A
Find the sum of the numbers lying between 107 and 253 that are multiples of 5.
Sol :The numbers lying between 107 and 253 that are multiples of 5 are
110, 115, 120,…, 250
a2 – a1 = 115 – 110 = 5
a3 – a2 = 120 – 115 = 5
∵ a3 – a2 = a2 – a1 = 5
Therefore, the series is in AP
Here, a = 110, d = 5 and an = 250
We know that,
an = a + (n – 1)d
⇒ 250 = 110 + (n – 1)5
⇒ 250 – 110 = (n – 1)5
⇒ 140 = (n – 1)5
⇒ 28 = (n – 1)
⇒ n = 29
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{29}=\frac{29}{2}[2 \times 110+(29-1) 5]$
⇒ S29 = 29[110 + 14 × 5]
⇒ S29 = 29[180]
⇒ S29 = 5220
Hence, the sum of all numbers lying between 107 and 253 is 5220.
Question 24 B
Find the sum of all natural numbers lying between 100 and 1000 which are multiples of 5.
Sol :The numbers lying between 100 and 1000 that are multiples of 5 are
105, 110, 115, 120,…, 995
a2 – a1 = 110 – 105 = 5
a3 – a2 = 115 – 110 = 5
∵ a3 – a2 = a2 – a1 = 5
Therefore, the series is in AP
Here, a = 105, d = 5 and an = 995
We know that,
an = a + (n – 1)d
⇒ 995 = 105 + (n – 1)5
⇒ 995 – 105 = (n – 1)5
⇒ 890 = (n – 1)5
⇒ 178 = (n – 1)
⇒ n = 179
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{179}=\frac{179}{2}[2 \times 105+(179-1) 5]$
⇒ S179 = 179[105 + 89 × 5]
⇒ S179 = 179 [550]
⇒ S179 = 98450
Hence, the sum of all numbers lying between 100 and 1000 that are multiples of 5 is 98450.
Question 25
Find the sum of all the two digit odd positive integers.
Sol :
The two digit odd positive integers are
11, 13, 15,…, 99
a2 – a1 = 13 – 11 = 2
a3 – a2 = 15 – 13 = 2
∵ a3 – a2 = a2 – a1 = 2
Therefore, the series is in AP
Here, a = 11, d = 2 and an = 99
We know that,
an = a + (n – 1)d
⇒ 99 = 11 + (n – 1)2
⇒ 99 – 11 = (n – 1)2
⇒ 88 = (n – 1)2
⇒ 44 = (n – 1)
⇒ n = 45
Now, we have to find the sum of this AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{45}=\frac{45}{2}[2 \times 11+(45-1) 2]$
⇒ S45 = 45[11 + 44]
⇒ S45 = 45[55]
⇒ S45 = 2475
Hence, the sum of all two digit odd numbers are 2475.
11, 13, 15,…, 99
a2 – a1 = 13 – 11 = 2
a3 – a2 = 15 – 13 = 2
∵ a3 – a2 = a2 – a1 = 2
Therefore, the series is in AP
Here, a = 11, d = 2 and an = 99
We know that,
an = a + (n – 1)d
⇒ 99 = 11 + (n – 1)2
⇒ 99 – 11 = (n – 1)2
⇒ 88 = (n – 1)2
⇒ 44 = (n – 1)
⇒ n = 45
Now, we have to find the sum of this AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{45}=\frac{45}{2}[2 \times 11+(45-1) 2]$
⇒ S45 = 45[11 + 44]
⇒ S45 = 45[55]
⇒ S45 = 2475
Hence, the sum of all two digit odd numbers are 2475.
Question 26
Find the sum of all multiplies of 9 lying between 300 and 700.
Sol :The numbers lying between 300 and 700 which are multiples of 9 are
306, 315, 324,…, 693
a2 – a1 = 315 – 306 = 9
a3 – a2 = 324 – 315 = 9
∵ a3 – a2 = a2 – a1 = 9
Therefore, the series is in AP
Here, a = 306, d = 9 and an = 693
We know that,
an = a + (n – 1)d
⇒ 693 = 306 + (n – 1)9
⇒ 693 - 306 = (n – 1)9
⇒ 387 = (n – 1)9
⇒ 43 = (n – 1)
⇒ n = 44
Now, we have to find the sum of this AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{44}=\frac{44}{2}[2 \times 306+(44-1) 9]$
⇒ S44 = 22[612 + 387]
⇒ S44 = 22[999]
⇒ S44 = 21978
Hence, the sum of all numbers lying between 300 and 700 is 21978.
Question 27
Find the sum of all the three digit natural numbers which are multiples of 7.
Sol :The three digit natural numbers which are multiples of 7 are
105, 112, 119,…, 994
a2 – a1 = 112 – 105 = 7
a3 – a2 = 112 – 105 = 7
∵ a3 – a2 = a2 – a1 = 7
Therefore, the series is in AP
Here, a = 105, d = 7 and an = 994
We know that,
an = a + (n – 1)d
⇒ 994 = 105 + (n – 1)7
⇒ 994 – 105 = (n – 1)7
⇒ 889 = (n – 1)7
⇒ 127 = (n – 1)
⇒ n = 128
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{128}=\frac{128}{2}[2 \times 105+(128-1) 7]$
⇒ S128 = 64[210 + 127 × 7]
⇒ S128 = 64[1099]
⇒ S128 = 70336
Hence, the sum of all three digit numbers which are multiples of 7 are 70336.
Question 28
Find the sum of all natural numbers lying between 100 and 500, which are divisible by 8.
Sol :The numbers lying between 100 and 500 which are divisible by 8 are
104, 112, 120, 128, 136,…, 496
a2 – a1 = 112 – 104 = 8
a3 – a2 = 120 – 112 = 8
∵ a3 – a2 = a2 – a1 = 8
Therefore, the series is in AP
Here, a = 120, d = 8 and an = 496
We know that,
an = a + (n – 1)d
⇒ 496 = 104 + (n – 1)8
⇒ 496 – 104 = (n – 1)8
⇒ 392 = (n – 1)8
⇒ 49 = (n – 1)
⇒ n = 50
Now, we have to find the sum of this AP
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{50}=\frac{50}{2}[2 \times 104+(50-1) 8]$
⇒ S50 = 25[208 + 49 × 8]
⇒ S50 = 25[600]
⇒ S50 = 15000
Hence, the sum of all numbers lying between 100 and 500 and divisible by 8 is 15000.
Question 29
Find the sum of all the 3 digit natural numbers which are divisible by 13.
Sol :The three digit natural numbers which are divisible by 13 are
104, 117, 130,…, 988
a2 – a1 = 117 – 104 = 13
a3 – a2 = 130 – 117 = 13
∵ a3 – a2 = a2 – a1 = 13
Therefore, the series is in AP
Here, a = 104, d = 13 and an = 988
We know that,
an = a + (n – 1)d
⇒ 988 = 104 + (n – 1)13
⇒ 988 – 104 = (n – 1)13
⇒ 884 = (n – 1)13
⇒ 68 = (n – 1)
⇒ n = 69
Now, we have to find the sum of this AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{69}=\frac{69}{2}[2 \times 104+(69-1) 13]$
$\Rightarrow \mathrm{S}_{69}=\frac{69}{2}[2 \times 104+68 \times 13]$
⇒ S69 = 69[104 + 34 × 13]
⇒ S69 = 69[546]
⇒ S69 = 37674
Hence, the sum of three digit natural numbers which are divisible by 13 are 37674.
Question 30
The 5th and 15th terms of an A.P. are 13 and - 17 respectively. Find the sum of first 21 terms of the A.P.
Sol :Given: a5 = 13 and a15 = -17 and n = 21
We know that,
a5 = a + 4d = 13 …(i)
and a15 = a + 14d = -17 …(ii)
Solving the linear equations (i) and (ii), we get
a + 4d – a – 14d = 13 – (-17)
⇒ -10d = 13 + 17
⇒ -10d = 30
⇒ d = -3
Putting the value of d in eq. (i), we get
a + 4(-3) = 13
⇒ a = 13 + 12 = 25
Now, we have to find the sum of first 21 terms.
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[2 \times(25)+(21-1)(-3)]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[50+20 \times(-3)]$
⇒ S21 = 21 [25 + 10 × (-3)]
⇒ S21 = 21 [-5]
⇒ S21 =-105
Question 31
Find the sum of first 21 terms of the A.P. whose 2nd term is 8 and 4th term is 14.
Sol :Given: a2 = 8 and a4 = 14 and n = 21
We know that,
a2 = a + d = 8 …(i)
and a4 = a + 3d = 14 …(ii)
Solving the linear equations (i) and (ii), we get
a + d – a – 3d = 8 – 14
⇒ -2d = -6
⇒ d = 3
Putting the value of d in eq. (i), we get
a + 3 = 8
⇒ a = 8 – 3 = 5
Now, we have to find the sum of first 21 terms.
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[2 \times(5)+(21-1)(3)]$
$\Rightarrow \mathrm{S}_{21}=\frac{21}{2}[10+20 \times(3)]$
⇒ S21 = 21 [5 + 10 × (3)]
⇒ S21 = 21 [35]
⇒ S21 = 735
Question 32
Find the sum of 51 terms of the A.P. whose second term is 2 and the 4th term is 8.
Sol :Given: a2 = 2 and a4 = 8 and n = 51
We know that,
a2 = a + d = 2 …(i)
and a4 = a + 3d = 8 …(ii)
Solving the linear equations (i) and (ii), we get
a + d – a – 3d = 2 – 8
⇒ -2d = -6
⇒ d = 3
Putting the value of d in eq. (i), we get
a + 3 = 2
⇒ a = 2 – 3 = -1
Now, we have to find the sum of first 51 terms.
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{51}=\frac{51}{2}[2 \times(-1)+(51-1)(3)]$
$\Rightarrow \mathrm{S}_{51}=\frac{51}{2}[-1+50 \times(3)]$
⇒ S51 = 51 [-1 + 25 × (3)]
⇒ S51 = 51 [74]
⇒ S51 = 3774
Question 33
Find the sum of the first 25 terms of the A.P. whose 2nd term is 9 and 4th term is 21.
Sol :Given: a2 = 9 and a4 = 21 and n = 25
We know that,
a2 = a + d = 9 …(i)
and a4 = a + 3d = 21 …(ii)
Solving the linear equations (i) and (ii), we get
a + d – a – 3d = 9 – 21
⇒ -2d = -12
⇒ d = 6
Putting the value of d in eq. (i), we get
a + 6 = 9
⇒ a = 9 – 6 = 3
Now, we have to find the sum of first 25 terms.
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[2 \times(3)+(25-1)(6)]$
$\Rightarrow \mathrm{S}_{25}=\frac{25}{2}[6+24 \times(6)]$
⇒ S25 = 25 [3 + 12 × (6)]
⇒ S25 = 25 [75]
⇒ S25 = 1875
Question 34 A
If the sum of 8 terms of an A.P. is 64 and the sum of 19 terms is 361, find the sum of n terms.
Sol :Given: S8 = 64 and S19 = 361
We know that,
$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{8}=\frac{8}{2}[2 \mathrm{a}+(8-1) \mathrm{d}]$
⇒ 64 = 4 [2a +7d]
⇒ 16 = 2a + 7d …(i)
Now,
$\Rightarrow \mathrm{S}_{19}=\frac{19}{2}[2 \mathrm{a}+(19-1) \mathrm{d}]$
$\Rightarrow 361=\frac{19}{2}[2 \mathrm{a}+18 \mathrm{d}]$
⇒ 38 = 2a + 18d …(ii)
Solving linear equations (i) and (ii), we get
2a + 7d – 2a – 18d = 16 – 38
⇒ -11d = -22
⇒ d = 2 …(iii)
Putting the value of d in eq. (i), we get
2a + 7(2) = 16
⇒ 2a = 16 – 14
⇒ 2a = 2 …(iv)
Now, we have to find the Sn
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2+(\mathrm{n}-1) 2]$ [from (iii) and (iv)]
⇒ Sn = n [1 + n – 1]
⇒ Sn = n2
Question 34 B
The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there in the A.P. and what is their sum?
Sol :Given: First term, a = 17
Last term, l = 350
common difference, d = 9
We know that,
l = a + (n – 1)d
⇒ 350 = 17 + (n – 1)9
⇒ 333 = (n – 1)9
⇒ 37 = n – 1
⇒ n = 38
So, there are 38 terms in the AP
Now, we have to find the sum of this AP
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{38}=\frac{38}{2}[2 \times 17+(38-1) 9]$
⇒ S38 = 19 [34 +37×9]
⇒ S38 = 19 [34 + 333]
⇒ S38 = 19 × 367
⇒ S38 = 6973
Hence, the sum of 38 terms is 6973.
Question 35
If a, b, c be the 1st, 3rd and nth terms respectively of an A.P., there prove that the sum to n terms is $\frac{c+a}{2}+\frac{c^{2}-a^{2}}{b-a}$
Sol :Given: a1 = a
a3 = a + 2d = b
⇒ 2d = b – a
$\Rightarrow \mathrm{d}=\frac{\mathrm{b}-\mathrm{a}}{2}$
and an = a + (n – 1)d
$c=a+(n-1)\left(\frac{b-a}{2}\right)$
$\Rightarrow \mathrm{c}-\mathrm{a}=(\mathrm{n}-1)\left(\frac{\mathrm{b}-\mathrm{a}}{2}\right)$
$\Rightarrow \frac{2(\mathrm{c}-\mathrm{a})}{(\mathrm{b}-\mathrm{a})}=(\mathrm{n}-1)$
$\Rightarrow \frac{2(c-a)}{(b-a)}+1=n$
$\Rightarrow \frac{2 c-2 a+b-a}{b-a}=n$
$\Rightarrow \frac{b+2 c-3 a}{b-a}=n$
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[\mathrm{a}+\mathrm{l}]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\frac{\mathrm{b}+2 \mathrm{c}-3 \mathrm{a}}{\mathrm{b}-\mathrm{a}}}{2}[\mathrm{a}+\mathrm{c}]$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{(\mathrm{b}+2 \mathrm{c}-3 \mathrm{a})(\mathrm{a}+\mathrm{c})}{2(\mathrm{b}-\mathrm{a})}$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{(\mathrm{c}+\mathrm{a})[\mathrm{b}-\mathrm{a}+2(\mathrm{c}-\mathrm{a})]}{2(\mathrm{b}-\mathrm{a})}$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{c}+\mathrm{a}}{2}+\frac{(\mathrm{c}+\mathrm{a})(\mathrm{c}-\mathrm{a})}{\mathrm{b}-\mathrm{a}}$
$\Rightarrow \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{c}+\mathrm{a}}{2}+\frac{\mathrm{c}^{2}-\mathrm{a}^{2}}{\mathrm{b}-\mathrm{a}}$
Question 36
If the mth term of an A.P. is $\frac{1}{\mathrm{n}}$ and the nth term is $\frac{1}{\mathrm{m}}$, then prove that the sum to mn terms is $\frac{\mathrm{mn}+1}{2}$, where in m≠ n.
Sol :Given: $a_{m}=\frac{1}{n}$
Now, am = a + (m – 1)d
$\Rightarrow \frac{1}{\mathrm{n}}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}$
⇒ an + n(m – 1)d = 1
⇒ an + mnd – nd = 1 …(i)
and $a_{n}=\frac{1}{m}$
$\Rightarrow \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\frac{1}{\mathrm{m}}$
⇒ am + mnd – md = 1 …(ii)
From eq. (i) and (ii), we get
an + mnd – nd = am + mnd – md
⇒ a(n – m) –d (n – m) = 0
⇒ a = d
Now, putting the value of a in eq. (i), we get
dn + mnd – nd = 1
⇒ mnd = 1
$\Rightarrow \mathrm{d}=\frac{1}{\mathrm{mn}}$
Hence, $a=\frac{1}{m n}$
Sum of mn terms of AP is
$\mathrm{S}_{\mathrm{mn}}=\frac{\mathrm{mn}}{2}[2 \mathrm{a}+(\mathrm{mn}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{mn}}=\frac{\mathrm{mn}}{2}\left[2 \times \frac{1}{\mathrm{mn}}+(\mathrm{mn}-1)\left(\frac{1}{\mathrm{mn}}\right)\right]$
$\Rightarrow \mathrm{S}_{\mathrm{mn}}=\frac{\mathrm{mn}}{2}\left[\frac{2}{\mathrm{mn}}+1-\frac{1}{\mathrm{mn}}\right]$
$\Rightarrow \mathrm{S}_{\mathrm{mn}}=\frac{\mathrm{mn}}{2}\left[\frac{1}{\mathrm{mn}}+1\right]$
$\Rightarrow \mathrm{S}_{\mathrm{mn}}=\frac{1}{2}[\mathrm{mn}+1]$
Hence Proved
Given: a12 = -13
⇒ a + 11d = -13
⇒ a = -13 – 11d …(i)
and S4 = 24
$\frac{4}{2}[2(-13-11 \mathrm{d})+(4-1) \mathrm{d}]=24$ [from(i)]
⇒ 2[-26 -22d + 3d] = 24
⇒ -26 – 19d = 12
⇒ -19d = 12 + 26
⇒ -19d = 38
⇒ d = -2
Putting the value of d in eq. (i), we get
a = -13 – 11(-2) = -13 + 22 = 9
So, a = 9 , d = -2 and n = 10
Now, we have to find the S10
$\mathrm{S}_{10}=\frac{10}{2}[2 \mathrm{a}+(10-1)(-2)]$
⇒ S10 = 5[2×9 + 9(-2)]
⇒ S10 = 5[18 – 18]
⇒ S10 = 0
Hence, the sum of first 10 terms is 0
Given: Total number of terms = 2n + 3
Let the first term = a
and the common difference = d
Then, ak = a + (k – 1)d …(i)
Let S1 and S2 denote the sum of all odd terms and the sum of all even terms respectively.
Then,
S1 = a1 + a3 + a5 … + a2n+3
$=\frac{n+2}{2}\left\{a_{1}+a_{2 n+3}\right\}$
$=\frac{n+2}{2}\{a+a+(2 n+3-1) d\}$ [using (i)]
$=\frac{n+2}{2}\{2 a+2 n d+2 d\}$
= (n + 2)(a + nd + d) …(ii)
And, S2 = a2 + a4 + a6 … + a2n+2
$=\frac{n+1}{2}\left\{a_{2}+a_{2 n+2}\right\}$
$=\frac{n+1}{2}[(a+d)+\{a+(2 n+2-1) d\}]$ [using (i)]
$=\frac{n+1}{2}\{2 a+2 n d+2 d\}$
= (n+1)(a + nd + d) …(iii)
$\therefore \frac{\mathrm{S}_{1}}{\mathrm{S}_{2}}=\frac{(\mathrm{n}+1)(\mathrm{a}+\mathrm{nd})}{(\mathrm{n})(\mathrm{a}+\mathrm{nd})}=\frac{\mathrm{n}+2}{\mathrm{n}+1}$
Let the first term be a and common difference of the given AP is d.
Given: Sm = Sn
$\Rightarrow \frac{\mathrm{m}}{2}[2 \mathrm{a}+(\mathrm{m}-1) \mathrm{d}]=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
⇒ 2am + md(m – 1) = 2an + nd(n – 1)
⇒ 2am – 2an + m2d – md – n2d + nd = 0
⇒ 2a (m – n) + d[(m2 – n2) – (m – n)] = 0
⇒ 2a (m – n) + d[(m– n)(m + n) – (m – n)] = 0
⇒ (m – n) [2a + {(m + n) – 1}d] = 0
⇒ 2a + (m + n – 1)d = 0 [∵ m – n ≠ 0]…(i)
Now,
$S_{m+n}=\frac{m+n}{2}[2 a+\{(m+n)-1\} d]=0$
$\Rightarrow \mathrm{S}_{\mathrm{m}+\mathrm{n}}=\frac{\mathrm{m}+\mathrm{n}}{2} \times 0$ [using (i)]
⇒ Sm+n = 0
Hence Proved
Given: first term, a = 2
And
Sum of first five terms$=\frac{1}{4}($ sum of next 5 terms $)$
Sum of next 5 terms$=\frac{1}{4}\left(\right.$ Sum of $6^{\text {th }}$ to $10^{\text {th }}$ terms $)$
$\Rightarrow$ Sum of first 5 terms $=\frac{1}{4}($ Sum of first 10 terms $-$ sum of first five terms)
$\Rightarrow \mathrm{S}_{5}=\frac{1}{4}\left(\mathrm{S}_{10}-\mathrm{S}_{5}\right)$
⇒ 4S5 = S10 – S5
⇒ 5S5 = S10
$\Rightarrow 5 \times \frac{5}{2}[2 \times 2+(5-1) \mathrm{d}]=\frac{10}{2}[2 \times 2+(10-1) \mathrm{d}]$
$\Rightarrow \frac{25}{2}[4+4 \mathrm{d}]=5[4+9 \mathrm{d}]$
⇒ 20 + 20d = 8 + 18d
⇒ 20d – 18d = 8 – 20
⇒ 2d = -12
⇒ d = -6
Thus, a = 2 and d = -6
∴ a20 = a + (n – 1)d
⇒ a20 = 2 + (20 – 1)(-6)
⇒ a20 = 2 + (19)(-6)
⇒ a20 = 2 – 114
⇒ a20 = -112
Hence Proved
Given: Sn be the sum of n terms and d be the common difference.
To Prove: d = Sn - 2Sn-1 + Sn-2
Taking RHS
Sn - 2Sn-1 + Sn-2
$\mathrm{S}_{\mathrm{mn}}=\frac{\mathrm{mn}}{2}[2 \mathrm{a}+(\mathrm{mn}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{mn}}=\frac{\mathrm{mn}}{2}\left[2 \times \frac{1}{\mathrm{mn}}+(\mathrm{mn}-1)\left(\frac{1}{\mathrm{mn}}\right)\right]$
$\Rightarrow \mathrm{S}_{\mathrm{mn}}=\frac{\mathrm{mn}}{2}\left[\frac{2}{\mathrm{mn}}+1-\frac{1}{\mathrm{mn}}\right]$
$\Rightarrow \mathrm{S}_{\mathrm{mn}}=\frac{\mathrm{mn}}{2}\left[\frac{1}{\mathrm{mn}}+1\right]$
$\Rightarrow \mathrm{S}_{\mathrm{mn}}=\frac{1}{2}[\mathrm{mn}+1]$
Hence Proved
Question 37
If the 12th term of an A.P. is - 13 and the sum of the first four terms is 24, what is the sum of the first 10 terms?
Sol :Given: a12 = -13
⇒ a + 11d = -13
⇒ a = -13 – 11d …(i)
and S4 = 24
$\frac{4}{2}[2(-13-11 \mathrm{d})+(4-1) \mathrm{d}]=24$ [from(i)]
⇒ 2[-26 -22d + 3d] = 24
⇒ -26 – 19d = 12
⇒ -19d = 12 + 26
⇒ -19d = 38
⇒ d = -2
Putting the value of d in eq. (i), we get
a = -13 – 11(-2) = -13 + 22 = 9
So, a = 9 , d = -2 and n = 10
Now, we have to find the S10
$\mathrm{S}_{10}=\frac{10}{2}[2 \mathrm{a}+(10-1)(-2)]$
⇒ S10 = 5[2×9 + 9(-2)]
⇒ S10 = 5[18 – 18]
⇒ S10 = 0
Hence, the sum of first 10 terms is 0
Question 38
If the number of terms of an A.P. be 2n + 3, then find the ratio of sum of the odd terms to the sum of even terms.
Sol :Given: Total number of terms = 2n + 3
Let the first term = a
and the common difference = d
Then, ak = a + (k – 1)d …(i)
Let S1 and S2 denote the sum of all odd terms and the sum of all even terms respectively.
Then,
S1 = a1 + a3 + a5 … + a2n+3
$=\frac{n+2}{2}\left\{a_{1}+a_{2 n+3}\right\}$
$=\frac{n+2}{2}\{a+a+(2 n+3-1) d\}$ [using (i)]
$=\frac{n+2}{2}\{2 a+2 n d+2 d\}$
= (n + 2)(a + nd + d) …(ii)
And, S2 = a2 + a4 + a6 … + a2n+2
$=\frac{n+1}{2}\left\{a_{2}+a_{2 n+2}\right\}$
$=\frac{n+1}{2}[(a+d)+\{a+(2 n+2-1) d\}]$ [using (i)]
$=\frac{n+1}{2}\{2 a+2 n d+2 d\}$
= (n+1)(a + nd + d) …(iii)
$\therefore \frac{\mathrm{S}_{1}}{\mathrm{S}_{2}}=\frac{(\mathrm{n}+1)(\mathrm{a}+\mathrm{nd})}{(\mathrm{n})(\mathrm{a}+\mathrm{nd})}=\frac{\mathrm{n}+2}{\mathrm{n}+1}$
Question 39
If the sum of first m terms of an A.P. is the same as the sum of its first n terms, show that the sum of its first (m + n) terms is zero.
Sol :Let the first term be a and common difference of the given AP is d.
Given: Sm = Sn
$\Rightarrow \frac{\mathrm{m}}{2}[2 \mathrm{a}+(\mathrm{m}-1) \mathrm{d}]=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
⇒ 2am + md(m – 1) = 2an + nd(n – 1)
⇒ 2am – 2an + m2d – md – n2d + nd = 0
⇒ 2a (m – n) + d[(m2 – n2) – (m – n)] = 0
⇒ 2a (m – n) + d[(m– n)(m + n) – (m – n)] = 0
⇒ (m – n) [2a + {(m + n) – 1}d] = 0
⇒ 2a + (m + n – 1)d = 0 [∵ m – n ≠ 0]…(i)
Now,
$S_{m+n}=\frac{m+n}{2}[2 a+\{(m+n)-1\} d]=0$
$\Rightarrow \mathrm{S}_{\mathrm{m}+\mathrm{n}}=\frac{\mathrm{m}+\mathrm{n}}{2} \times 0$ [using (i)]
⇒ Sm+n = 0
Hence Proved
Question 40
In an A.P. the first term is 2, and the sum of the first five terms is one-fourth of the next five terms. Show that its 20th term is — 112.
Sol :Given: first term, a = 2
And
Sum of first five terms$=\frac{1}{4}($ sum of next 5 terms $)$
Sum of next 5 terms$=\frac{1}{4}\left(\right.$ Sum of $6^{\text {th }}$ to $10^{\text {th }}$ terms $)$
$\Rightarrow$ Sum of first 5 terms $=\frac{1}{4}($ Sum of first 10 terms $-$ sum of first five terms)
$\Rightarrow \mathrm{S}_{5}=\frac{1}{4}\left(\mathrm{S}_{10}-\mathrm{S}_{5}\right)$
⇒ 4S5 = S10 – S5
⇒ 5S5 = S10
$\Rightarrow 5 \times \frac{5}{2}[2 \times 2+(5-1) \mathrm{d}]=\frac{10}{2}[2 \times 2+(10-1) \mathrm{d}]$
$\Rightarrow \frac{25}{2}[4+4 \mathrm{d}]=5[4+9 \mathrm{d}]$
⇒ 20 + 20d = 8 + 18d
⇒ 20d – 18d = 8 – 20
⇒ 2d = -12
⇒ d = -6
Thus, a = 2 and d = -6
∴ a20 = a + (n – 1)d
⇒ a20 = 2 + (20 – 1)(-6)
⇒ a20 = 2 + (19)(-6)
⇒ a20 = 2 – 114
⇒ a20 = -112
Hence Proved
Question 41
If d be the common difference of an A.P. and Sn be the sum of its n terms, then prove that d = Sn - 2Sn-1 + Sn-2
Sol :Given: Sn be the sum of n terms and d be the common difference.
To Prove: d = Sn - 2Sn-1 + Sn-2
Taking RHS
Sn - 2Sn-1 + Sn-2
$=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]-2 \times \frac{\mathrm{n}-1}{2}[2 \mathrm{a}+(\mathrm{n}-1-1) \mathrm{d}]$ $+\frac{\mathrm{n}-2}{2}[2 \mathrm{a}+(\mathrm{n}-2-1) \mathrm{d}]$
$=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]-\left[\frac{2(\mathrm{n}-1)}{2}[2 \mathrm{a}+(\mathrm{n}-2) \mathrm{d}]\right]+\frac{\mathrm{n}-2}{2}[2 \mathrm{a}+(\mathrm{n}-3) \mathrm{d}]$$=\frac{2 a n+n(n-1) d-4 a(n-1)-2(n-1)(n-2) d+2 a(n-2)+(n-2)(n-3) d}{2}$
$=\frac{1}{2}\left[2 \mathrm{an}+\mathrm{n}^{2} \mathrm{d}-\mathrm{nd}-4 \mathrm{an}+4 \mathrm{a}-2 \mathrm{n}^{2} \mathrm{d}+4 \mathrm{nd}+2 \mathrm{nd}-4 \mathrm{d}+2 \mathrm{an}-4 \mathrm{a}+\mathrm{n}^{2} \mathrm{d}\right.-3 \mathrm{nd}-2 \mathrm{nd}+6 \mathrm{d}]$
$=\frac{1}{2}[2 \mathrm{d}]$= d
=LHS
Hence Proved
Question 42
The sum of the first 7 terms of an A.P. is 10, and that of the next 7 terms is 17. Find the progression.
Sol :Given: Sum of first 7 terms, S7 = 10
and Sum of the next 7 terms = 17
⇒ Sum of 8th to 14th terms = 17
⇒ Sum of first 14 terms – Sum of first 7 terms = 17
⇒ S14 – S7 = 17
⇒ S14 – 10 = 17
⇒ S14 = 27
Sum of 7 terms, $\mathrm{S}_{7}=\frac{7}{2}[2 \mathrm{a}+(7-1) \mathrm{d}]$
$\Rightarrow 10=\frac{7}{2}[2 \mathrm{a}+(7-1) \mathrm{d}$
⇒ 20 = 7[2a + 6d]
⇒ 20 = 14a + 42d …(i)
$\Rightarrow 10=\frac{7}{2}[2 \mathrm{a}+(7-1) \mathrm{d}$
⇒ 20 = 7[2a + 6d]
⇒ 20 = 14a + 42d …(i)
Sum of 14 terms, $S_{14}=\frac{14}{2}[2 a+(14-1) d]$
⇒ 27 = 7[2a + 13d]
⇒ 27 = 14a + 91d …(ii)
Solving the linear equations (i) and (ii), we get
14a + 42d – 14a – 91d = 20 – 27
⇒ -49d = -7
$\Rightarrow \mathrm{d}=\frac{1}{7}$
⇒ 27 = 7[2a + 13d]
⇒ 27 = 14a + 91d …(ii)
Solving the linear equations (i) and (ii), we get
14a + 42d – 14a – 91d = 20 – 27
⇒ -49d = -7
$\Rightarrow \mathrm{d}=\frac{1}{7}$
Putting the value of d in eq. (i), we get
20 = 14a + 42d
$\Rightarrow 20=14 \mathrm{a}+42 \times \frac{1}{7}$
⇒ 20= 14a + 6
⇒ 20 – 6 = 14a
⇒ 14 = 14a
⇒ a = 1
Thus, a = 1 and $\mathrm{d}=\frac{1}{7}$
So, AP is
a1 = 1
a2$=a+d=1+\frac{1}{7}=1 \frac{1}{7}$
a3$=a+2 d=1+2 \times \frac{1}{7}=1 \frac{2}{7}$
Hence, AP is $1,1 \frac{1}{7}, 1 \frac{2}{7}, \ldots$
Given: ap = x and aq = y
We know that,
an = a + (n – 1)d
ap = a + (p – 1)d
⇒ x = a + (p – 1)d …(i)
Now,
aq = a + (q – 1)d
⇒ y = a + (q – 1)d …(ii)
From eq. (i) and (ii), we get
x – (p – 1)d = y – (q – 1)d
⇒ x – y = (p – 1)d – (q – 1)d
⇒ x – y = d [p – 1 – q + 1]
⇒ x – y = d[ p – q]
$\Rightarrow \mathrm{d}=\frac{x-\mathrm{y}}{\mathrm{p}-\mathrm{q}}$ …(iii)
Adding, Eq (i) and (ii), we get
x + y = 2a + (p – 1) + (q – 1)d
⇒ x + y = 2a + d[p + q – 1 – 1]
⇒ x + y = 2a + d (p + q – 1) –d
⇒ x + y + d = 2a + (p + q – 1)d …(iv)
We know that,
$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}[2 \mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}[\mathrm{x}+\mathrm{y}+\mathrm{d}]$ [using (iv)]
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left[\mathrm{x}+\mathrm{y}+\frac{\mathrm{x}-\mathrm{y}}{\mathrm{p}-\mathrm{q}}\right]$ [using (iii)]
Hence Proved
There are two AP with different first term and common difference.
For the First AP
Let first term be a
Common difference = d
Sum of n terms = $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
and nth term = an = a + (n – 1)d
For the second AP
Let first term be A
Common difference = D
Sum of n terms = $S_{n}=\frac{n}{2}[2 A+(n-1) D]$
and nth term = An = A + (n – 1)D
It is given that
$\frac{\text { Sum of } n \text { terms of first } A \cdot P}{\text { Sum of } n \text { terms os second A.P }}=\frac{3 n+8}{7 n+15}$
$\Rightarrow \frac{\frac{n}{2}[2 a+(n-1) d}{\frac{n}{2}[2 A+(n-1) D}=\frac{3 n+8}{7 n+15}$
$\Rightarrow \frac{\mathrm{n}\left[\mathrm{a}+\left(\frac{\mathrm{n}-1}{2}\right) \mathrm{d}\right]}{\mathrm{n}\left[\mathrm{A}+\left(\frac{\mathrm{n}-1}{2}\right) \mathrm{D}\right]}=\frac{3 \mathrm{n}+8}{7 \mathrm{n}+15}$
$\Rightarrow \frac{a+\left(\frac{n-1}{2}\right) d}{A+\left(\frac{n-1}{2}\right) D}=\frac{3 n+8}{7 n+15}$ …(i)
Now, we need to find ratio of their 12th term
i. $e \cdot \frac{12^{\text {th }} \text { term of first } \mathrm{AP}}{12^{\text {th }} \text { term of second } \mathrm{AP}}$
$=\frac{a_{12} \text { of first AP }}{A_{12} \text { of second } A P}$
$=\frac{a+(12-1) d}{A+(12-1) D}$
$=\frac{a+11 d}{A+11 D}$
20 = 14a + 42d
$\Rightarrow 20=14 \mathrm{a}+42 \times \frac{1}{7}$
⇒ 20= 14a + 6
⇒ 20 – 6 = 14a
⇒ 14 = 14a
⇒ a = 1
Thus, a = 1 and $\mathrm{d}=\frac{1}{7}$
So, AP is
a1 = 1
a2$=a+d=1+\frac{1}{7}=1 \frac{1}{7}$
a3$=a+2 d=1+2 \times \frac{1}{7}=1 \frac{2}{7}$
Hence, AP is $1,1 \frac{1}{7}, 1 \frac{2}{7}, \ldots$
Question 43
If the pth term of an A.P. is x and qth term is y, show that the sum of (p + q) terms is $\frac{p+q}{2}\left[x+y+\left(\frac{x-y}{p-q}\right)\right]$
Sol :Given: ap = x and aq = y
We know that,
an = a + (n – 1)d
ap = a + (p – 1)d
⇒ x = a + (p – 1)d …(i)
Now,
aq = a + (q – 1)d
⇒ y = a + (q – 1)d …(ii)
From eq. (i) and (ii), we get
x – (p – 1)d = y – (q – 1)d
⇒ x – y = (p – 1)d – (q – 1)d
⇒ x – y = d [p – 1 – q + 1]
⇒ x – y = d[ p – q]
$\Rightarrow \mathrm{d}=\frac{x-\mathrm{y}}{\mathrm{p}-\mathrm{q}}$ …(iii)
Adding, Eq (i) and (ii), we get
x + y = 2a + (p – 1) + (q – 1)d
⇒ x + y = 2a + d[p + q – 1 – 1]
⇒ x + y = 2a + d (p + q – 1) –d
⇒ x + y + d = 2a + (p + q – 1)d …(iv)
We know that,
$S_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}[2 \mathrm{a}+(\mathrm{p}+\mathrm{q}-1) \mathrm{d}]$
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}[\mathrm{x}+\mathrm{y}+\mathrm{d}]$ [using (iv)]
$\Rightarrow \mathrm{S}_{\mathrm{p}+\mathrm{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left[\mathrm{x}+\mathrm{y}+\frac{\mathrm{x}-\mathrm{y}}{\mathrm{p}-\mathrm{q}}\right]$ [using (iii)]
Hence Proved
Question 44 A
The sum of 17 terms of two series in A.P. are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms.
Sol :There are two AP with different first term and common difference.
For the First AP
Let first term be a
Common difference = d
Sum of n terms = $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
and nth term = an = a + (n – 1)d
For the second AP
Let first term be A
Common difference = D
Sum of n terms = $S_{n}=\frac{n}{2}[2 A+(n-1) D]$
and nth term = An = A + (n – 1)D
It is given that
$\frac{\text { Sum of } n \text { terms of first } A \cdot P}{\text { Sum of } n \text { terms os second A.P }}=\frac{3 n+8}{7 n+15}$
$\Rightarrow \frac{\frac{n}{2}[2 a+(n-1) d}{\frac{n}{2}[2 A+(n-1) D}=\frac{3 n+8}{7 n+15}$
$\Rightarrow \frac{\mathrm{n}\left[\mathrm{a}+\left(\frac{\mathrm{n}-1}{2}\right) \mathrm{d}\right]}{\mathrm{n}\left[\mathrm{A}+\left(\frac{\mathrm{n}-1}{2}\right) \mathrm{D}\right]}=\frac{3 \mathrm{n}+8}{7 \mathrm{n}+15}$
$\Rightarrow \frac{a+\left(\frac{n-1}{2}\right) d}{A+\left(\frac{n-1}{2}\right) D}=\frac{3 n+8}{7 n+15}$ …(i)
Now, we need to find ratio of their 12th term
i. $e \cdot \frac{12^{\text {th }} \text { term of first } \mathrm{AP}}{12^{\text {th }} \text { term of second } \mathrm{AP}}$
$=\frac{a_{12} \text { of first AP }}{A_{12} \text { of second } A P}$
$=\frac{a+(12-1) d}{A+(12-1) D}$
$=\frac{a+11 d}{A+11 D}$
Hence, $\frac{\mathrm{n}-1}{2}=11$
n – 1 = 11 × 2
⇒ n = 22 + 1
⇒ n = 23
Putting n = 23 in eq. (i), we get
$\frac{a+\left(\frac{23-1}{2}\right) d}{A+\left(\frac{23-1}{2}\right) D}=\frac{3(23)+8}{7(23)+15}$
$\Rightarrow \frac{a+11 d}{A+11 D}=\frac{69+8}{161+15}$
$\Rightarrow \frac{12^{\text {th }} \text { term of first } \mathrm{AP}}{12^{\text {th }} \text { term of second } \mathrm{AP}}=\frac{77}{176}$
$\Rightarrow \frac{12^{\text {th }} \text { term of first } \mathrm{AP}}{12^{\text {th }} \text { term of second } \mathrm{AP}}=\frac{7}{16}$
Hence the ratio of 12th term of 1st AP and 12th term if 2nd AP is 7:16
There are two AP with different first term and common difference.
For the First AP
Let first term be a
Common difference = d
Sum of n terms = $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
and nth term = an = a + (n – 1)d
For the second AP
Let first term be A
Common difference = D
Sum of n terms = $S_{n}=\frac{n}{2}[2 A+(n-1) D]$
and nth term = An = A + (n – 1)D
It is given that
$\frac{\text { Sum of } \mathrm{n} \text { terms of first } \mathrm{A.P}}{\text { Sum of } \mathrm{n} \text { terms os second } \mathrm{A.P}}=\frac{5 \mathrm{n}+4}{9 \mathrm{n}+6}$
$\Rightarrow \frac{\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}}{\frac{\mathrm{n}}{2}[2 \mathrm{A}+(\mathrm{n}-1) \mathrm{D}}=\frac{5 \mathrm{n}+4}{9 \mathrm{n}+6}$
$\Rightarrow \frac{n\left[a+\left(\frac{n-1}{2}\right) d\right]}{n\left[A+\left(\frac{n-1}{2}\right) D\right]}=\frac{5 n+4}{9 n+6}$
$\Rightarrow \frac{a+\left(\frac{n-1}{2}\right) d}{A+\left(\frac{n-1}{2}\right) D}=\frac{5 n+4}{9 n+6}$ …(i)
Now, we need to find ratio of their 18th term
i. $e \cdot \frac{18^{\text {th }} \text { term of first } \mathrm{AP}}{18^{\text {th }} \text { term of second } \mathrm{AP}}$
$=\frac{a_{18} \text { of first } A P}{A_{18} \text { of second } A P}$
$=\frac{a+(18-1) d}{A+(18-1) D}$
$=\frac{a+17 d}{A+17 D}$
n – 1 = 11 × 2
⇒ n = 22 + 1
⇒ n = 23
Putting n = 23 in eq. (i), we get
$\frac{a+\left(\frac{23-1}{2}\right) d}{A+\left(\frac{23-1}{2}\right) D}=\frac{3(23)+8}{7(23)+15}$
$\Rightarrow \frac{a+11 d}{A+11 D}=\frac{69+8}{161+15}$
$\Rightarrow \frac{12^{\text {th }} \text { term of first } \mathrm{AP}}{12^{\text {th }} \text { term of second } \mathrm{AP}}=\frac{77}{176}$
$\Rightarrow \frac{12^{\text {th }} \text { term of first } \mathrm{AP}}{12^{\text {th }} \text { term of second } \mathrm{AP}}=\frac{7}{16}$
Hence the ratio of 12th term of 1st AP and 12th term if 2nd AP is 7:16
Question 44 B
The sum of 11 terms of two A.P.'s are in the ratio (5n + 4) : (9n + 6), find the ratio of their 18th terms.
Sol :There are two AP with different first term and common difference.
For the First AP
Let first term be a
Common difference = d
Sum of n terms = $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
and nth term = an = a + (n – 1)d
For the second AP
Let first term be A
Common difference = D
Sum of n terms = $S_{n}=\frac{n}{2}[2 A+(n-1) D]$
and nth term = An = A + (n – 1)D
It is given that
$\frac{\text { Sum of } \mathrm{n} \text { terms of first } \mathrm{A.P}}{\text { Sum of } \mathrm{n} \text { terms os second } \mathrm{A.P}}=\frac{5 \mathrm{n}+4}{9 \mathrm{n}+6}$
$\Rightarrow \frac{\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}}{\frac{\mathrm{n}}{2}[2 \mathrm{A}+(\mathrm{n}-1) \mathrm{D}}=\frac{5 \mathrm{n}+4}{9 \mathrm{n}+6}$
$\Rightarrow \frac{n\left[a+\left(\frac{n-1}{2}\right) d\right]}{n\left[A+\left(\frac{n-1}{2}\right) D\right]}=\frac{5 n+4}{9 n+6}$
$\Rightarrow \frac{a+\left(\frac{n-1}{2}\right) d}{A+\left(\frac{n-1}{2}\right) D}=\frac{5 n+4}{9 n+6}$ …(i)
Now, we need to find ratio of their 18th term
i. $e \cdot \frac{18^{\text {th }} \text { term of first } \mathrm{AP}}{18^{\text {th }} \text { term of second } \mathrm{AP}}$
$=\frac{a_{18} \text { of first } A P}{A_{18} \text { of second } A P}$
$=\frac{a+(18-1) d}{A+(18-1) D}$
$=\frac{a+17 d}{A+17 D}$
Hence, $\frac{\mathrm{n}-1}{2}=17$
n – 1 = 17 × 2
⇒ n = 34 + 1
⇒ n = 35
Putting n = 35 in eq. (i), we get
$\frac{a+\left(\frac{35-1}{2}\right) d}{A+\left(\frac{35-1}{2}\right) D}=\frac{5(35)+4}{9(35)+6}$
$\Rightarrow \frac{a+17 d}{A+17 D}=\frac{175+4}{315+6}$
$\Rightarrow \frac{18^{\text {th }} \text { term of first } \mathrm{AP}}{18^{\text {th }} \text { term of second } \mathrm{AP}}=\frac{179}{321}$
Hence the ratio of 18th term of 1st AP and 18th term if 2nd AP is 179:321
Given: Sn = n2p and Sm = m2p
To Prove: Sp = p3
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{n}^{2} \mathrm{p}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
⇒ 2np = [2a + (n – 1)d]
⇒ 2np – (n – 1)d = 2a …(i)
and $\mathrm{S}_{\mathrm{m}}=\frac{\mathrm{m}}{2}[2 \mathrm{a}+(\mathrm{m}-1) \mathrm{d}]$
$\Rightarrow \mathrm{m}^{2} \mathrm{p}=\frac{\mathrm{m}}{2}[2 \mathrm{a}+(\mathrm{m}-1) \mathrm{d}]$
⇒ 2mp = 2a + (m – 1)d
⇒ 2mp – (m – 1)d = 2a …(ii)
From eq. (i) and (ii), we get
⇒ 2np – (n – 1)d = 2mp – (m – 1)d
⇒ 2np – nd + d = 2mp – md + d
⇒ 2np – nd = 2mp – md
⇒ md – nd = 2mp – 2np
⇒ d(m – n) = 2p(m – n)
⇒ d = 2p …(iii)
Putting the value of d in eq. (i), we get
⇒ 2np – (n – 1)(2p) = 2a
⇒ 2pn – 2pn + 2p = 2a
⇒ 2p = 2a …(iv)
Now, we have to find the Sp
$S_{p}=\frac{p}{2}[2 p+(p-1) 2 p]$ [from (iii) & (iv)]
$\Rightarrow S_{p}=\frac{p}{2}\left[2 p+2 p^{2}-2 p\right]$
$\Rightarrow S_{p}=\frac{p}{2}\left[2 p^{2}\right]$
⇒ Sp = p3
Hence Proved
The income of a person in 1st year = Rs 300000
The income of a person in 2nd year = Rs 300000 + 10000
= Rs 310000
The income of a person in 3rd year = Rs 310000 + 10000
= Rs 320000
and so,on
Therefore, the AP is
300000, 310000, 320000,…
Here a = 300000, d = 310000 – 300000 = 10000
and n = 20
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{20}=\frac{20}{2}[2(300000)+(20-1)(10000)]$
⇒ S20 = 10 [600000 + 190000]
⇒ S20 = 10[790000]
⇒ S20 = 7900000
Hence, the total amount he received in 20 years is Rs 7900000.
The 1st installment of the loan = Rs. 100
the 2nd installment of the loan = Rs 100 + 5 = Rs 105
The 3rd installment of the loan = Rs 105 + 5 = Rs 110
Therefore, the AP is 100, 105, 110, …
Here, a = 100, d = 105 – 100 = 5 and n = 30
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{30}=\frac{30}{2}[2 \times 100+(30-1)(5)]$
⇒ S30 = 15 [200 + 29 × 5]
⇒ S30 = 15 [200+145]
⇒ S30 = 15 [345]
⇒ S30 = 5175
Hence, the amount he will pay in 30th installments is Rs. 5175
Given: The smallest angle is 75°
i.e. a = 75
and common difference = 10°
i.e. d = 10
Therefore, the series is
75, 85, 95, 105, …
and the sum of interior angles of a polygon =(n – 2) 180°
i.e. Sn = 180
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow(n-2) 180=\frac{n}{2}[2 \times(75)+(n-1)(10)]$
$\Rightarrow(n-2) 180=\frac{n}{2}[150+10 n-10]$
⇒ (n – 2)360 = n [140+10n]
⇒ 360n – 720 = 140n + 10n2
⇒ 36n – 72 – 14n – n2 = 0
⇒ n2 – 22n + 72 = 0
⇒ n2 – 18n – 4n + 72 = 0
⇒ n(n – 18) – 4(n – 18) = 0
⇒ (n – 4)(n – 18) = 0
Putting both the factor equal to 0, we get
n – 4 = 0 or n – 18 = 0
⇒ n = 4 or n = 18
Hence, the number of sides of a polygon can be 4 or 18.
n – 1 = 17 × 2
⇒ n = 34 + 1
⇒ n = 35
Putting n = 35 in eq. (i), we get
$\frac{a+\left(\frac{35-1}{2}\right) d}{A+\left(\frac{35-1}{2}\right) D}=\frac{5(35)+4}{9(35)+6}$
$\Rightarrow \frac{a+17 d}{A+17 D}=\frac{175+4}{315+6}$
$\Rightarrow \frac{18^{\text {th }} \text { term of first } \mathrm{AP}}{18^{\text {th }} \text { term of second } \mathrm{AP}}=\frac{179}{321}$
Hence the ratio of 18th term of 1st AP and 18th term if 2nd AP is 179:321
Question 45
In an A.P. Sn denotes the sum to first n terms, if Sn = n2p and Sm = m2p (m
n) prove that Sp = p3.
Sol : n) prove that Sp = p3.Given: Sn = n2p and Sm = m2p
To Prove: Sp = p3
We know that,
$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\Rightarrow \mathrm{n}^{2} \mathrm{p}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
⇒ 2np = [2a + (n – 1)d]
⇒ 2np – (n – 1)d = 2a …(i)
and $\mathrm{S}_{\mathrm{m}}=\frac{\mathrm{m}}{2}[2 \mathrm{a}+(\mathrm{m}-1) \mathrm{d}]$
$\Rightarrow \mathrm{m}^{2} \mathrm{p}=\frac{\mathrm{m}}{2}[2 \mathrm{a}+(\mathrm{m}-1) \mathrm{d}]$
⇒ 2mp = 2a + (m – 1)d
⇒ 2mp – (m – 1)d = 2a …(ii)
From eq. (i) and (ii), we get
⇒ 2np – (n – 1)d = 2mp – (m – 1)d
⇒ 2np – nd + d = 2mp – md + d
⇒ 2np – nd = 2mp – md
⇒ md – nd = 2mp – 2np
⇒ d(m – n) = 2p(m – n)
⇒ d = 2p …(iii)
Putting the value of d in eq. (i), we get
⇒ 2np – (n – 1)(2p) = 2a
⇒ 2pn – 2pn + 2p = 2a
⇒ 2p = 2a …(iv)
Now, we have to find the Sp
$S_{p}=\frac{p}{2}[2 p+(p-1) 2 p]$ [from (iii) & (iv)]
$\Rightarrow S_{p}=\frac{p}{2}\left[2 p+2 p^{2}-2 p\right]$
$\Rightarrow S_{p}=\frac{p}{2}\left[2 p^{2}\right]$
⇒ Sp = p3
Hence Proved
Question 46
The income of a person is Rs. 300000 in the first year and he receives an increase of Rs. 10000 to his income per year for the next 19 years. Find the total amount he received in 20 years.
Sol :The income of a person in 1st year = Rs 300000
The income of a person in 2nd year = Rs 300000 + 10000
= Rs 310000
The income of a person in 3rd year = Rs 310000 + 10000
= Rs 320000
and so,on
Therefore, the AP is
300000, 310000, 320000,…
Here a = 300000, d = 310000 – 300000 = 10000
and n = 20
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{20}=\frac{20}{2}[2(300000)+(20-1)(10000)]$
⇒ S20 = 10 [600000 + 190000]
⇒ S20 = 10[790000]
⇒ S20 = 7900000
Hence, the total amount he received in 20 years is Rs 7900000.
Question 47
A man starts repaying a loan as first installment of Rs. 100. If he increases the installments by Rs. 5 every month, what amount he will pay in 30 installments?
Sol :The 1st installment of the loan = Rs. 100
the 2nd installment of the loan = Rs 100 + 5 = Rs 105
The 3rd installment of the loan = Rs 105 + 5 = Rs 110
Therefore, the AP is 100, 105, 110, …
Here, a = 100, d = 105 – 100 = 5 and n = 30
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow S_{30}=\frac{30}{2}[2 \times 100+(30-1)(5)]$
⇒ S30 = 15 [200 + 29 × 5]
⇒ S30 = 15 [200+145]
⇒ S30 = 15 [345]
⇒ S30 = 5175
Hence, the amount he will pay in 30th installments is Rs. 5175
Question 48
The interior angles of a polygon are in A.P., the smallest angle is 75° and the common difference is 10°. Find the number of sides of the polygon.
Sol :Given: The smallest angle is 75°
i.e. a = 75
and common difference = 10°
i.e. d = 10
Therefore, the series is
75, 85, 95, 105, …
and the sum of interior angles of a polygon =(n – 2) 180°
i.e. Sn = 180
We know that,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow(n-2) 180=\frac{n}{2}[2 \times(75)+(n-1)(10)]$
$\Rightarrow(n-2) 180=\frac{n}{2}[150+10 n-10]$
⇒ (n – 2)360 = n [140+10n]
⇒ 360n – 720 = 140n + 10n2
⇒ 36n – 72 – 14n – n2 = 0
⇒ n2 – 22n + 72 = 0
⇒ n2 – 18n – 4n + 72 = 0
⇒ n(n – 18) – 4(n – 18) = 0
⇒ (n – 4)(n – 18) = 0
Putting both the factor equal to 0, we get
n – 4 = 0 or n – 18 = 0
⇒ n = 4 or n = 18
Hence, the number of sides of a polygon can be 4 or 18.
| S.no | Chapters | Links |
|---|---|---|
| 1 | Real numbers | Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 |
| 2 | Polynomials | Exercise 2.1 Exercise 2.2 Exercise 2.3 |
| 3 | Pairs of Linear Equations in Two Variables | Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5 |
| 4 | Trigonometric Ratios and Identities | Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4 |
| 5 | Triangles | Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5 |
| 6 | Statistics | Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4 |
| 7 | Quadratic Equations | Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5 |
| 8 | Arithmetic Progressions (AP) | Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
| 9 | Some Applications of Trigonometry: Height and Distances | Exercise 9.1 |
| 10 | Coordinates Geometry | Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4 |
| 11 | Circles | Exercise 11.1 Exercise 11.2 |
| 12 | Constructions | Exercise 12.1 |
| 13 | Area related to Circles | Exercise 13.1 |
| 14 | Surface Area and Volumes | Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4 |
| 15 | Probability | Exercise 15.1 |
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