KC Sinha Mathematics Solution Class 10 Chapter 8 Arithmetic Progressions Exercise 8.2

Exercise 8.1
Exercise 8.2
Exercise 8.3
Exercise 8.4

Exercise 8.2

Question 1 A

Find the indicated terms in each of the following arithmetic progression:
1, 6, 11, 16, ..., t₁₆,

Sol :
Given: 1, 6, 11, 16, …

Here, a = 1

d = a₂ – a₁ = 6 – 1 = 5

and n = 16

We have,

t_n = a + (n – 1)d

So, t₁₆ = 1 + (16 – 1)5

= 1 + 15×5

t₁₆ = 1 +75

t₁₆ = 76

Question 1 B

Find the indicated terms in each of the following arithmetic progression:
a = 3, d = 2; ; t_n, t₁₀

Sol :
Given: a = 3 , d = 2

To find: t_n and t₁₀

We have,

t_n = a + (n – 1)d

t_n = 3 + (n – 1) 2

= 3 + 2n – 2

t_n = 2n + 1

Now, n = 10

So, t₁₀ = 3 + (10 – 1)2

= 3 + 9×2

t₁₀ = 3 +18

t₁₀ = 21

Question 1 C

Find the indicated terms in each of the following arithmetic progression:
— 3, — 1/2, 2, ... ; t₁₀,

Sol :
Given: $-3,-\frac{1}{2}, 2, \ldots$

Here, a = –3

$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=-\frac{1}{2}-(-3)$

$=\frac{-1+6}{2}=\frac{5}{2}$

and n = 10

We have,

t_n = a + (n – 1)d

So, $\mathrm{t}_{10}=-3+(10-1) \frac{5}{2}$
$=-3+9 \times \frac{5}{2}$
$=\frac{-6+45}{2}$
$\mathrm{t}_{10}=\frac{39}{2}$

Question 1 D

Find the indicated terms in each of the following arithmetic progression:
a = 21, d = — 5; t_n, t₂₅

Sol :
Given: a = 21 , d = –5

To find: t_n and t₂₅

We have,

t_n = a + (n – 1)d

t_n = 21 + (n – 1)(–5)

= 21 – 5n + 5

t_n = 26 – 5n

Now, n = 25

So, t₂₅ = 21 + (25 – 1)(–5)

= 21 + 24 × (–5)

t₂₅ = 21 – 120

t₂₅ = –99

Question 2

Find the 10th term of the A.P. 10, 5, 0, — 5, — 10, ...

Sol :
Given: 10, 5, 0, –5, –10,…

To find: 10^th term i.e. t₁₀

Here, a = 10

d = a₂ – a₁ = 5 – 10 = –5

and n = 10

We have,

t_n = a + (n – 1)d

t₁₀ = 10 + (10 – 1)(–5)

= 10 + 9 × –5

t₁₀ = 10 – 45

t₁₀ = –35
Therefore, the 10^th term of the given is –35.

Question 3

Find the 10th term of the A.P. $\frac{13}{5}, \frac{7}{5}, \frac{1}{5},-1, \ldots$

Sol :
Given: $\frac{13}{5}, \frac{7}{5}, \frac{1}{5},-1$

Here, $\mathrm{a}=\frac{13}{5}$

$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}$ $=\frac{7}{5}-\frac{13}{5}=-\frac{6}{5}$

and n = 10

We have,
t_n = a + (n – 1)d
$\mathrm{t}_{10}=\frac{13}{5}+(10-1)\left(-\frac{6}{5}\right)$
$\mathrm{t}_{10}=\frac{13}{5}+9 \times \frac{-6}{5}$
$\mathrm{t}_{10}=\frac{13-54}{5}$
$\mathrm{t}_{10}=-\frac{41}{5}$
Therefore, the 10^th term of the given AP is $-\frac{41}{5}$

Question 4

Find the sum of 20th and 25th terms of A.P. 2, 5, 8, 11, ...

Sol :
Given: 2, 5, 8, 11, …

Here, a = 2

d = a₂ – a₁ = 5 – 2 = 3

and n = 20

We have,

t_n = a + (n – 1)d

t₂₀ = 2 + (20 – 1)(3)

t₂₀ = 2 + 19 × 3

= 2 + 57

t₂₀ = 59

Now, n = 25

t₂₅ = 2 + (25 – 1)(3)

t₂₅ = 2 + 24 × 3

t₂₅ = 2 + 72

t₂₅ = 74

The sum of 20^th and 25^th terms of AP = t₂₀ + t₂₅ = 59 + 74 = 133

Question 5 A

Find the number of terms in the following A.P.'s
6, 3, 0, — 3,…..,–36

Sol :
Here, a = 6 , d = 3 – 6 = –3 and l = –36

where l = a + (n – 1)d

⇒ –36 = 6 + (n – 1)(–3)

⇒ –36 = 6 –3n + 3

⇒ –36 = 9 – 3n

⇒ –36 – 9 = –3n

⇒ –45 = –3n
$\Rightarrow \mathrm{n}=\frac{-45}{-3}=15$
Hence, the number of terms in the given AP is 15

Question 5 B

Find the number of terms in the following A.P.'s
$\frac{5}{6}, 1,1 \frac{1}{6}, \ldots, 3 \frac{1}{3}$

Sol :
Here, $a=\frac{5}{6}$

$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}$ $=1-\frac{5}{6}=\frac{6-5}{6}=\frac{1}{6}$
And $1=\frac{10}{3}$

We have,

l = a + (n – 1)d
$\Rightarrow \frac{10}{3}=\frac{5}{6}+(\mathrm{n}-1) \times \frac{1}{6}$
$\Rightarrow \frac{10}{3}-\frac{5}{6}=(\mathrm{n}-1) \times \frac{1}{6}$
$\Rightarrow 6 \times\left(\frac{20-5}{6}\right)=\mathrm{n}-1$
⇒15=n–1
⇒n=16
Hence, the number of terms in the given AP is 16.

Question 6

Determine the number of terms in the A.P. 3, 7, 11, ..., 399. Also, find its 20th term from the end.

Sol :
Here, a = 3, d = 7 – 3 = 4 and l = 399
To find : n and 20^th term from the end

We have,
l = a + (n – 1)d
⇒ 399 = 3 + (n – 1) × 4
⇒ 399 – 3 = 4n – 4
⇒ 396 + 4 = 4n
⇒ 400 = 4n
⇒ n = 100

So, there are 100 terms in the given AP
Last term = 100^th
Second Last term = 100 – 1 = 99^th
Third last term = 100 – 2 = 98^th
And so, on
20^th term from the end = 100 – 19 = 81^st term
The 20^th term from the end will be the 81^st term.
So, t₈₁ = 3 + (81 – 1)(4)
t₈₁ = 3 + 80 × 4
t₈₁ = 3 + 320
t₈₁ = 323
Hence, the number of terms in the given AP is 100, and the 20^th term from the last is 323.

Question 7 A

Which term of the A.P. 5, 9, 13, 17, ... is 81?

Sol :
Here, a = 5, d = 9 – 5 = 4 and a_n = 81

To find : n

We have,

a_n = a + (n – 1)d
⇒ 81 = 5 + (n – 1) × 4
⇒ 81 = 5 + 4n – 4
⇒ 81 = 4n + 1
⇒ 80 = 4n
⇒ n = 20
Therefore, the 20^th term of the given AP is 81.

Question 7 B

Which term of the A.P. 14, 9, 4, – I, – 6, ... is – 41 ?

Sol :
Here, a = 14, d = 9 – 14 = –5 and a_n = –41
To find : n
We have,
a_n = a + (n – 1)d
⇒ –41 = 14 + (n – 1) × (–5)
⇒ –41 = 14 – 5n + 5
⇒ –41 = 19 – 5n
⇒ – 41 – 19 = –5n
⇒ –60 = –5n
⇒ n = 12
Therefore, the 12^th term of the given AP is –41.

Question 7 C

Which term of A.P. 3, 8, 13, 18, ... is 88?

Sol :
Here, a = 3, d = 8 – 3 = 5 and a_n = 88

To find : n

We have,

a_n = a + (n – 1)d

⇒ 88 = 3 + (n – 1) × (5)

⇒ 88 = 3 + 5n – 5

⇒ 88 = –2 + 5n

⇒88 + 2 = 5n

⇒ 90 = 5n

⇒ n = 18

Therefore, the 18^th term of the given AP is 88.

Question 7 D

Which term of A.P. $\frac{5}{6}, 1,1 \frac{1}{6}, 1 \frac{1}{3}, \ldots$ is 3?

Sol :
Here, $a=\frac{5}{6}$

$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=1-\frac{5}{6}$

$=\frac{6-5}{6}=\frac{1}{6}$

and a_n = 3

We have,
a_n = a + (n – 1)d
$\Rightarrow 3=\frac{5}{6}+(\mathrm{n}-1) \times \frac{1}{6}$
$\Rightarrow 3-\frac{5}{6}=(\mathrm{n}-1) \times \frac{1}{6}$
$\Rightarrow 6 \times\left(\frac{18-5}{6}\right)=\mathrm{n}-1$
⇒ 13 = n – 1
⇒ n = 14
Therefore, the 14^th term of a given AP is 3.

Question 7 E

Which term of A.P. 3, 8, 13, 18, ..., is 248 ?

Sol :
Here, a = 3, d = 8 – 3 = 5 and a_n = 248

To find : n

We have,
a_n = a + (n – 1)d
⇒ 248 = 3 + (n – 1) × (5)
⇒ 248 = 3 + 5n – 5
⇒ 248 = –2 + 5n
⇒ 248 + 2 = 5n
⇒ 250 = 5n
⇒ n = 50
Therefore, the 50^th term of the given AP is 248.

Question 8 A

Find the 6th term from end of the A.P. 17, 14, 11,… – 40.

Sol :
Here, a = 17, d = 14 – 17 = –3 and l = –40

where l = a + (n – 1)d

Now, to find the 6^th term from the end, we will find the total number of terms in the AP.

So, –40 = 17 + (n – 1)(–3)

⇒ –40 = 17 –3n + 3

⇒ –40 = 20 – 3n

⇒ –60 = –3n

⇒ n = 20

So, there are 20 terms in the given AP.

Last term = 20^th

Second last term = 20 – 1 = 19^th

Third last term = 20 – 2 = 18^th

And so, on

So, the 6^th term from the end = 20 – 5 = 15^th term

So, a_n = a + (n – 1)d
⇒ a₁₅ = 17 + (15 – 1)(–3)
⇒ a₁₅ = 17 + 14 × –3
⇒ a₁₅ = 17 – 42
⇒ a₁₅ = –25

Question 8 B

Find the 8th term from end of the A.P. 7, 10, 13, ..., 184.

Sol :
Here, a = 7, d = 10 – 7 = 3 and l = 184
where l = a + (n – 1)d
Now, to find the 8^th term from the end, we will find the total number of terms in the AP.

So, 184 = 7 + (n – 1)(3)
⇒ 184 = 7 + 3n – 3
⇒ 184 = 4 + 3n
⇒ 180 = 3n
⇒ n = 60

So, there are 60 terms in the given AP.
Last term = 60^th
Second last term = 60 – 1 = 59^th
Third last term = 60 – 2 = 58^th
And so, on
So, the 8^th term from the end = 60 – 7 = 53^th term

So, a_n = a + (n – 1)d
⇒ a₅₃ = 7 + (53 – 1)(3)
⇒ a₅₃ = 7 + 52 × 3
⇒ a₅₃ = 7 + 156
⇒ a₅₃ = 163

Question 9 A

Find the number of terms of the A.P.
6, 10, 14, 18, ..., 174?

Sol :
Here, a = 6 , d = 10 – 6 = 4 and l = 174

where l = a + (n – 1)d
⇒ 174 = 6 + (n – 1)(4)
⇒ 174 = 6 + 4n – 4
⇒ 174 = 2 + 4n
⇒ 174 – 2 = 4n
⇒ 172 = 4n
$\Rightarrow \mathrm{n}=\frac{172}{4}=43$
Hence, the number of terms in the given AP is 43

Question 9 B

Find the number of terms of the A.P.
7, 11, 15, ..., 139?

Sol :
Here, a = 7 , d = 11 – 7 = 4 and l = 139
where l = a + (n – 1)d
⇒ 139 = 7 + (n – 1)(4)
⇒ 139 = 7 + 4n – 4
⇒ 139 = 3 + 4n
⇒ 139 – 3 = 4n
⇒ 136 = 4n
$\Rightarrow \mathrm{n}=\frac{136}{4}=34$
Hence, the number of terms in the given AP is 34

Question 9 C

Find the number of terms of the A.P.
41, 38, 35, ..., 8?
Sol :
Here, a = 41 , d = 38 – 41 = –3 and l = 8

where l = a + (n – 1)d

⇒ 8 = 41 + (n – 1)(–3)

⇒ 8 = 41 –3n + 3

⇒ 8 = 44 – 3n

⇒ 8 – 44 = –3n

⇒ –36 = –3n

$\Rightarrow \mathrm{n}=\frac{-36}{-3}=12$

Hence, the number of terms in the given AP is 12

Question 10

Find the first negative term of sequence 999, 995, 991, 987, ...

Sol :
AP = 999, 995, 991, 987,…

Here, a = 999, d = 995 – 999 = –4

a_n < 0

⇒ a + (n – 1)d < 0

⇒ 999 + (n – 1)(–4) < 0

⇒ 999 – 4n + 4 < 0

⇒ 1003 – 4n < 0

⇒ 1003 < 4n

$\Rightarrow \frac{1003}{4}<\mathrm{n}$

⇒ n > 250.75

Nearest term greater than 250.75 is 251

So, 251^st term is the first negative term

Now, we will find the 251^st term

a_n = a +(n – 1)d

⇒ a₂₅₁ = 999 + (251 – 1)(–4)

⇒ a₂₅₁ = 999 + 250 × –4

⇒ a₂₅₁ = 999 – 1000

⇒ a₂₅₁ = – 1

∴, –1 is the first negative term of the given AP.

Question 11

Is 51 a term of the A.P. 5, 8, 11, 14, ...?

Sol :
AP = 5, 8, 11, 14, …

Here, a = 5 and d = 8 – 5 = 3

Let 51 be a term, say, nth term of this AP.

We know that

a_n = a + (n – 1)d

So, 51 = 5 + (n – 1)(3)
⇒ 51 = 5 + 3n – 3
⇒ 51 = 2 + 3n
⇒ 51 – 2 = 3n
⇒ 49 = 3n
$\Rightarrow \mathrm{n}=\frac{49}{3}$
But n should be a positive integer because n is the number of terms. So, 51 is not a term of this given AP.

Question 12

Is 56 a term of the A.P. $4,4 \frac{1}{2}, 5,5 \frac{1}{2}, 6, \ldots ?$

Sol :
AP $=4, \frac{9}{2}, 5, \frac{11}{2}, 6, \ldots$

Here, a = 4 and $\mathrm{d}=5-\frac{9}{2}=\frac{10-9}{2}=\frac{1}{2}$

Let 56 be a term, say, nth term of this AP.

We know that

a_n = a + (n – 1)d

So, $56=4+(\mathrm{n}-1) \times \frac{1}{2}$

⇒ 2 × (56 – 4) = n – 1

⇒ 2 × 52 = n – 1

⇒ 104 = n – 1

⇒ 105 = n

Hence, 56 is the 105^th term of this given AP.

Question 13

The 7th term of an A.P. is 20 and its 13th term is 32. Find the A.P.

[CBSE 2004]

Sol :
We have

a₇ = a + (7 – 1)d = a + 6d = 20 …(1)

and a₁₃ = a + (13 – 1)d = a + 12d = 32 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 12d = 20 – 32

⇒ – 6d = –12

⇒ d = 2

Putting the value of d in eq (1), we get

a + 6(2) = 20

⇒ a + 12 = 20

⇒ a = 8

Hence, the required AP is 8, 10, 12, 14,…

Question 14

The 7th term of an A.P. is – 4 and its 13th term is – 16. Find the A.P.

[CBSE 2004]

Sol :
We have

a₇ = a + (7 – 1)d = a + 6d = –4 …(1)

and a₁₃ = a + (13 – 1)d = a + 12d = –16 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 12d = –4 – (–16)

⇒ – 6d = –4 + 16

⇒ – 6d = 12

⇒ d = –2

Putting the value of d in eq (1), we get

a + 6(–2) = –4

⇒ a – 12 = –4

⇒ a = 8

Hence, the required AP is 8, 6, 4, 2,…

Question 15

The 8th term of an A.P. is 37, and its 12th term is 57. Find the A.P.

Sol :
We have

a₈ = a + (8 – 1)d = a + 7d = 37 …(1)

and a₁₂ = a + (12 – 1)d = a + 11d = 57 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 7d – a – 11d = 37 – 57

⇒ – 4d = –20

⇒ d = 5

Putting the value of d in eq (1), we get
a + 7(5) = 37
⇒ a + 35 = 37
⇒ a = 2
Hence, the required AP is 2, 7, 12, 17,…

Question 16

Find the 10th term of the A.P. whose 7th and 12th terms are 34 and 64 respectively.

Sol :
We have

a₇ = a + (7 – 1)d = a + 6d = 34 …(1)

and a₁₂ = a + (12 – 1)d = a + 11d = 64 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 11d = 34 – 64

⇒ – 5d = –30

⇒ d = 6

Putting the value of d in eq (1), we get

a + 6(6) = 34

⇒ a + 36 = 34

⇒ a = –2

Hence, the required AP is –2, 4, 10, 16,…

Now, we to find the 10^th term

So, a_n = a + (n – 1)d

a₁₀ = –2 + (10 – 1)6

a₁₀ = –2 + 9 × 6

a₁₀ = 52

Question 17 A

For what value of n are the nth term of the following two A.P's the same. Also find this term
13, 19, 25, ... and 69, 68, 67, ...

Sol :
1^st AP = 13, 19, 25, …

Here, a = 13, d = 19 – 13 = 6

and 2^nd AP = 69, 68, 67, …

Here, a = 69, d = 68 – 69 = –1

According to the question,

13 + (n – 1)6 = 69 + (n – 1)(–1)

⇒ 13 + 6n – 6 = 69 – n + 1

⇒ 7 + 6n = 70 – n

⇒ 6n + n = 70 – 7

⇒ 7n = 63

⇒ n = 9

9^th term of the given AP’s are same.

Now, we will find the 9^th term

We have,

a_n = a + (n – 1)d

a₉ = 13 + (9 – 1)6

a₉ = 13 + 8 × 6

a₉ = 13 + 48

a₉ = 61

Question 17 B

For what value of n are the nth term of the following two A.P's the same. Also find this term
23, 25, 27, 29, ... and – 17, – 10, – 3, 4, ...

Sol :
1^st AP = 23, 25, 27, 29, ...

Here, a = 23, d = 25 – 23 = 2

and 2^nd AP = – 17, – 10, – 3, 4, ...

Here, a = –17, d = –10 – (–17) = –10 + 17 = 7

According to the question,

23 + (n – 1)2 = –17 + (n – 1)7

⇒ 23 + 2n – 2 = –17 + 7n – 7

⇒ 21 + 2n = –24 + 7n

⇒ 2n – 7n = –24 – 21

⇒ –5n = –45

⇒ n = 9

9^th term of the given AP’s are same.

Now, we will find the 9^th term

We have,

a_n = a + (n – 1)d

a₉ = 23 + (9 – 1)2

a₉ = 23 + 8 × 2

a₉ = 23 + 16

a₉ = 39

Question 17 C

For what value of n are the nth term of the following two A.P's the same. Also find this term
24, 20, 16, 12, ... and – 11, – 8, – 5, – 2, ...

Sol :
1^st AP = 24, 20, 16, 12, ...

Here, a = 24, d = 20 – 24 = –4

and 2^nd AP = – 11, – 8, – 5, – 2, ...

Here, a = –11, d = –8 – (–11) = –8 + 11 = 3

According to the question,

24 + (n – 1)(–4) = –11 + (n – 1)3

⇒ 24 – 4n + 4 = –11 + 3n – 3

⇒ 28 – 4n = –14 + 3n

⇒ 28 + 14 = 3n + 4n

⇒ 7n = 42

⇒ n = 6

6^th term of the given AP’s are same.

Now, we will find the 6^th term

We have,
a_n = a + (n – 1)d
a₆ = 24 + (6 – 1)(–4)
a₆ = 24 + 5 × –4
a₆ = 24 – 20
a₆ = 4

Question 17 D

For what value of n are the nth term of the following two A.P's the same. Also find this term
63, 65, 67, ... and 3, 10, 17, ...

Sol :
1^st AP = 63, 65, 67, ...

Here, a = 63, d = 65 – 63 = 2

and 2^nd AP = 3, 10, 17, ...

Here, a = 3, d = 10 – 3 = 7

According to the question,

63 + (n – 1)2 = 3 + (n – 1)7

⇒ 63 + 2n – 2 = 3 + 7n – 7

⇒ 61 + 2n = 7n – 4

⇒ 65 = 7n – 2n

⇒ 5n = 65

⇒ n = 13

13^th term of the given AP’s are same.

Now, we will find the 13^th term

We have,
a_n = a + (n – 1)d
a₁₃ = 63 + (13 – 1)2
a₁₃ = 63 + 12 × 2
a₁₃ = 63 + 24
a₁₃ = 87

Question 18 A

In the following A.P., find the missing terms:
5, □,□, $9\frac{1}{2}$

Sol :
Here, a = 5 , n = 4 and $1=\frac{19}{2}$

We have,

l = a + (n – 1)d

$\Rightarrow \frac{19}{2}=5+(4-1) \mathrm{d}$

⇒19 = 10 + 6d

⇒9 = 6d

$\Rightarrow \mathrm{d}=\frac{9}{6}=\frac{3}{2}$

So, the missing terms are –

a₂ = a + d $=5+\frac{3}{2}=\frac{10+3}{2}=\frac{13}{2}$

a₃ = a + 2d $=5+2 \times \frac{3}{2}=5+3=8$

Hence, the missing terms are $\frac{13}{2}$ and 8

Question 18 B

In the following A.P., find the missing terms:
54, □,□, 42

Sol :
Here, a = 54 , n = 4 and l = 42

We have,

l = a + (n – 1)d

⇒42 = 54 + (4 – 1)d

⇒42 = 54 + 3d

⇒–12 = 3d

$\Rightarrow \mathrm{d}=\frac{-12}{3}=-4$

So, the missing terms are –
a₂ = a + d = 54 – 4 = 50
a₃ = a + 2d = 54 + 2(–4) = 54 – 8 = 46
Hence, the missing terms are 50 and 46

Question 18 C

In the following A.P., find the missing terms:
– 4, □,□,□,□, 6

Sol :
Here, a = –4, n = 6 and l = 6

We have,

l = a + (n – 1)d

⇒6 = –4 + (6 – 1)d

⇒6 = –4 + 5d

⇒10 = 5d

$\Rightarrow \mathrm{d}=\frac{10}{5}=2$

So, the missing terms are –

a₂ = a + d = –4 + 2 = –2

a₃ = a + 2d = –4 + 2(2) = –4 + 4 = 0

a₄ = a + 3d = –4 + 3(2) = –4 + 6 = 2

a₅ = a + 4d = –4 + 4(2) = –4 + 8 = 4

Hence, the missing terms are –2, 0, 2 and 4

Question 18 D

In the following A.P., find the missing terms:
□, 13, □, 3

Sol :
Given: a₂ = 13 and a₄ = 3

We know that,

a_n = a + (n – 1)d

a₂ = a + (2 – 1)d

13 = a + d …(i)

and a₄ = a +(4 – 1)d

3 = a + 3d …(ii)

Solving linear equations (i) and (ii), we get

a + d – a – 3d = 13 – 3

⇒ –2d = 10

⇒ d = –5

Putting the value of d in eq. (i), we get
a – 5 = 13
⇒ a = 18
Now, a₃ = a + 2d = 18 + 2(–5) = 18 – 10 = 8
Hence, the missing terms are 18 and 8

Question 18 E

In the following A.P., find the missing terms:
7, □,□,□,27

Sol :
Here, a = 7, n = 5 and l = 27

We have,

l = a + (n – 1)d

⇒27 = 7 + (5 – 1)d

⇒27 = 7 + 4d

⇒20 = 4d

$\Rightarrow \mathrm{d}=\frac{20}{4}=5$

So, the missing terms are –
a₂ = a + d = 7 + 5 = 12
a₃ = a + 2d = 7 + 2(5) = 7 + 10 = 17
a₄ = a + 3d = 7 + 3(5) = 7 + 15 = 22
Hence, the missing terms are 12, 17 and 22

Question 18 F

In the following A.P., find the missing terms:
2, □, 26

Sol:
Here, a = 2, n = 3 and l = 26

We have,

l = a + (n – 1)d

⇒26 = 2 + (3 – 1)d

⇒26 = 2 + 2d

⇒24 = 2d

$\Rightarrow \mathrm{d}=\frac{24}{2}=12$

So, the missing terms are –

a₂ = a + d = 2 + 12 = 14

Hence, the missing terms is 14

Question 18 G

In the following A.P., find the missing terms:
□, □, 13, □, □, 22

Sol :
Given: a₃ = 13 and a₆ = 22

We know that,

a_n = a + (n – 1)d

a₃ = a + (3 – 1)d

13 = a + 2d …(i)

and a₆ = a +(6 – 1)d

22 = a + 5d …(ii)

Solving linear equations (i) and (ii), we get

a + 2d – a – 5d = 13 – 22

⇒ –3d = 9

⇒ d = 3

Putting the value of d in eq. (i), we get

a + 2(3) = 13

⇒ a + 6 = 13

⇒ a = 7

Now, a₂ = a + d = 7 + 3 = 10

a₄ = a + 3d = 7 + 3(3) = 7 + 9 = 16

a₅ = a + 4d = 7 + 4(3) = 7 + 12 = 19

Hence, the missing terms are 7, 10, 16 and 19

Question 18 H

In the following A.P., find the missing terms:
– 4, □, □, □, 6

Sol :
Here, a = –4, n = 5 and l = 6

We have,

l = a + (n – 1)d

⇒6 = –4 + (5 – 1)d

⇒6 = –4 + 4d

⇒10 = 4d

$\Rightarrow \mathrm{d}=\frac{10}{4}=\frac{5}{2}$

So, the missing terms are –

a₂ = a + d $=-4+\frac{5}{2}=\frac{-8+5}{2}=\frac{-3}{2}$

a₃ = a + 2d $=-4+2 \times \frac{5}{2}=-4+5=1$

a₄ = a + 3d $=-4+3 \times \frac{3}{2}=\frac{-8+9}{2}=\frac{1}{2}$

Hence, the missing terms are $\frac{-3}{2}, 1$ and $\frac{1}{2}$

Question 18 I

In the following A.P., find the missing terms:
□, 38, □,□,□, – 22

Sol :
Given: a₂ = 38 and a₆ = –22

We know that,

a_n = a + (n – 1)d

a₂ = a + (2 – 1)d

38 = a + d …(i)

and a₆ = a +(6 – 1)d

–22 = a + 5d …(ii)

Solving linear equations (i) and (ii), we get

a + d – a – 5d = 38 – (–22)

⇒ –4d = 60

⇒ d = –15

Putting the value of d in eq. (i), we get

a + (–15) = 38

⇒ a – 15 = 38

⇒ a = 53

Now, a₃ = a + 2d = 53 + 2(–15) = 53 – 30 = 23

a₄ = a + 3d = 53 + 3(–15) = 53 – 45 = 8

a₅ = a + 4d = 53 + 4(–15) = 53 – 60 = –7

Hence, the missing terms are 53, 23, 8 and –7

Question 19 A

If 10th term of an A.P. is 52 and 17th term is 20 more than the 13th term, find the A.P.

Sol :
Given: a₁₀ = 52 and a₁₇ = 20 + a₁₃

Now, a_n = a + (n – 1)d

a₁₀ = a + (10 – 1)d

52 = a + 9d …(i)

and a₁₇ = 20 + a₁₃

a + (17 – 1)d = 20 + a + (13 – 1)d

⇒ a + 16d = 20 + a + 12d

⇒ 16d –12d = 20

⇒ 4d = 20

⇒ d = 5

Putting the value of d in eq. (i), we get

a + 9(5) = 52

⇒ a + 45 = 52

⇒ a = 52 – 45

⇒ a = 7

Therefore, the AP is 7, 12, 17, …

Question 19 B

Which term of the A.P. 3, 15, 27, 39, ... will be 132 more than its 54th term?

Sol :
Given: 3, 15, 27, 39, …

First we need to calculate 54^th term.

We know that

a_n = a + (n – 1)d

Here, a = 3, d = 15 – 3 = 12 and n = 54

So, a₅₄ = 3 + (54 – 1)12

⇒ a₅₄ = 3 + 53 × 12

⇒ a₅₄ = 3 + 636

⇒ a₅₄ = 639

Now, the term is 132 more than a₅₄ is 132 + 639 = 771

Now,

a + (n – 1)d = 771

⇒ 3 + (n – 1)12 = 771

⇒ 3 + 12n – 12 = 771

⇒ 12n = 771 + 12 – 3

⇒ 12n = 780
⇒ n = 65
Hence, the 65^th term is 132 more than the 54^th term.

Question 20

Which term of the A.P. 3, 10, 17, 24, ... will be 84 more than its 13th term ?

Sol :
Given: 3, 10, 17, 24, ...

First we need to calculate 13^th term.

We know that

a_n = a + (n – 1)d

Here, a = 3, d = 10 – 3 = 7 and n = 13

So, a₁₃ = 3 + (13 – 1)7

⇒ a₁₃ = 3 + 12 × 7

⇒ a₁₃ = 3 + 84

⇒ a₁₃ = 87

Now, the term is 84 more than a₁₃ is 84 + 87 = 171

Now,

a + (n – 1)d = 171

⇒ 3 + (n – 1)7 = 171

⇒ 3 + 7n – 7 = 171

⇒ 7n = 171 + 7 – 3

⇒ 7n = 175

⇒ n = 25

Hence, the 25^th term is 84 more than the 13^th term.

Question 21

The 4th term of an A.P. is zero. Prove that its 25th term is triple its 11th term.

Sol :
Given: a₄ = 0

To Prove: a₂₅ = 3 × a₁₁

Now, a₄ = 0

⇒ a + 3d = 0

⇒ a = –3d

We know that,

a_n = a + (n – 1)d

a₁₁ = –3d + (11 – 1)d [from (i)]

a₁₁ = –3d + 10d

a₁₁ = 7d …(ii)

Now,

a₂₅ = a + (25 – 1)d

a₂₅ = –3d + 24d [from(i)]

a₂₅ = 21d

a₂₅ = 3 × 7d

a₂₅ = 3 × a₁₁ [from(ii)]

Hence Proved

Question 22

If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that its 25th term is zero.

Sol :
Given: 10 × a₁₀ = 15 × a₁₅

To Prove: a₂₅ = 0

Now,

10 × (a + 9d) = 15 × (a + 14d)

⇒ 10a + 90d = 15a + 210d

⇒ 10a – 15a = 210d – 90d

⇒ –5a = 120d

⇒ a = –24d …(i)

Now,

a_n = a + (n – 1)d

a₂₅ = –24d + (25 – 1)d [from (i)]

a₂₅ = –24d + 24d

a₂₅ = 0

Hence Proved

Question 23

If (m + 1)th term of an A.P. is twice the (n + 1)th term,

prove that (3m + 1)th term is twice the (m + n + 1)th term.

Sol :
Given: a_m+1 = 2a_n+1
To Prove: a_3m+1 = 2a_m+n+1

Now,
a_n = a + (n – 1)d
⇒ a_m+1 = a + (m + 1 – 1)d
⇒ a_m+1 = a + md
and a_n+1 = a + (n + 1 – 1)d
⇒ a_n+1 = a + nd

Given: a_m+1 = 2a_n+1
a +md = 2(a + nd)
⇒ a + md = 2a + 2nd
⇒ md – 2nd = 2a – a
⇒ d(m – 2n) = a …(i)

Now,
a_m+n+1 = a + (m + n + 1 – 1)d
= a + (m + n)d
= md – 2nd + md + nd [from (i)]
= 2md – nd
a_m+n+1 = d (2m – n) …(ii)
a_3m+1 = a + (3m + 1 – 1)d
= a + 3md
= md – 2nd + 3md [from (i)]
= 4md – 2nd
= 2d( 2m – n)
a_3m+1 = 2a_m+n+1 [from (ii)]
Hence Proved

Question 24

If t_n be the nth term of an A.P. such that $\frac{t_{4}}{t_{7}}=\frac{2}{3}$ find $\frac{t_{8}}{t_{9}}$.

Sol :
Given: $\frac{t_{4}}{t_{7}}=\frac{2}{3}$

To find: $\frac{t_{8}}{t_{9}}$

We know that,

t_n = a + (n – 1)d

So,

$\frac{t_{4}}{t_{7}}=\frac{a+(4-1) d}{a+(7-1) d}=\frac{2}{3}$

$\Rightarrow \frac{a+3 d}{a+6 d}=\frac{2}{3}$

⇒ 3(a+3d) = 2(a+6d)

⇒ 3a + 9d = 2a + 12d

⇒ 3a – 2a = 12d – 9d

⇒ a = 3d …(i)

Now, $\frac{t_{8}}{t_{9}}=\frac{a+(8-1) d}{a+(9-1) d}=\frac{3 d+7 d}{3 d+8 d}=\frac{10 d}{11 d}=\frac{10}{11}$ [from (i)]

Question 25

Find the number of all positive integers of 3 digits which are divisible by 5.

Sol :
The list of 3 digit numbers divisible by 5 is:

100, 105, 110,…,995

Here a = 100, d = 105 – 100 = 5, a_n = 995

We know that

a_n = a + (n – 1)d

995 = 100 + (n – 1)5

⇒ 895 = (n – 1)5

⇒ 179 = n – 1

⇒ 180 = n

So, there are 180 three– digit numbers divisible by 5.

Question 26

How many three digit numbers are divisible by 7.

Sol :
The list of 3 digit numbers divisible by 7 is:

105, 112, 119,…,994

Here a = 105, d = 112 – 105 = 7, a_n = 994

We know that

a_n = a + (n – 1)d

994 = 105 + (n – 1)7

⇒ 889 = (n – 1)7

⇒ 127 = n – 1

⇒ 128 = n

So, there are 128 three– digit numbers divisible by 7.

Question 27

If t_n denotes the nth term of an A.P., show that t_m + t_2n+m = 2 t_m+n.

Sol :
To show: t_m + t_2n+m = 2 t_m+n

Taking LHS

t_m + t_2n+m = a + (m – 1)d + a + (2n + m – 1)d

= 2a + md – d + 2nd + md – d

= 2a + 2md + 2nd – 2d

= 2 {a + (m + n – 1)d}

= 2t_m+n

= RHS

∴LHS = RHS

Hence Proved

Question 28

Find a if 5a + 2, 4a – I, a + 2 are in A.P.

Sol :
Let 5a + 2, 4a – 1, a + 2 are in AP

So, first term a = 5a + 2

d = 4a – 1 – 5a – 2 = – a – 3

n = 3

l = a + 2

So,

l = a + (n – 1)d

⇒ a + 2 =5a + 2 + (3 – 1)(–a – 3)

⇒ a + 2 – 5a – 2 = –3a – 9 + a + 3

⇒ – 4a = –2a – 6

⇒ – 4a + 2a = – 6

⇒ –2a = – 6

⇒ a = 3

Question 29

nth term of a sequence is 2n + 1. Is this sequence an A.P.? If so find its first term and common difference.

Sol :
We know that nth term of an A.P is given by,

a_n = a + (n – 1) d

Now equating it with the expression given we get,

2 n + 1 = a + (n – 1) d

2 n + 1 = a + nd – d

2 n + 1 = nd + (a – d)

Equating both sides we get,

d = 2 and a – d = 1

So we get,

a = 3 and d = 2.

So the first term of this sequence is 3, and the common difference is 2.

Question 30

The sum of the 4th and Sth terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of A.P.

Sol :
Given: a₄ + a₈ = 24

⇒ a +3d + a + 7d = 24

⇒ 2a + 10d = 24 …(i)

and a₆ + a₁₀ = 44

⇒ a +5d + a + 9d = 44

⇒ 2a + 14d = 44 …(ii)

Solving Linear equations (i) and (ii), we get

2a + 10d – 2a – 14d = 24 – 44

⇒ –4d = – 20

⇒ d = 5

Putting the value of d in eq. (i), we get

2a + 10×5 = 24

⇒ 2a + 50 = 24

⇒ 2a =24 –50

⇒ 2a =–26

⇒ a = –13

So, the first three terms are –13, –8, –3.

Question 31

A person was appointed in the pay scale of Rs. 700–40–1500. Find in how many years he will reach maximum of the scale.

Sol :
Let the required number of years = n

Given t_n = 1500, a= 700, d = 40

We know that,

t_n = a +(n – 1)d

⇒ 1500 = 700 + (n – 1)40

⇒ 800 = (n – 1)40

⇒ 20 = n – 1

⇒ n = 21

Hence, in 21years he will reach maximum of the scale.

Question 32

A sum of money kept in a hank amounts to Rs. 600/– in 4 years and Rs. 800/– in 12 years. Find the sum and interest carried every year.

Sol :
Let the required sum = a

and the interest carried every year = d

According to question,

In 4years, a sum of money kept in bank account = Rs. 600

i.e. t₅ = 600 ⇒ a + 4d = 600 …(i)

and in 12 years , sum of money kept = Rs. 800

i.e. t₁₃ = 800 ⇒ a + 12d = 800 …(ii)

Solving linear equations (i) and (ii), we get

a + 4d – a – 12d = 600 – 800

⇒ – 8d = –200

⇒ d = 25

Putting the value of d in eq.(i), we get

a + 4(25) = 600

⇒ a + 100 = 600

⇒ a = 500

Hence, the sum and interest carried every year is Rs 500 and Rs 25 respectively.

Question 33

A man starts repaying, a loan with the first instalment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30th instalment ?

Sol :
The first instalment of the loan = Rs. 100

The 2^nd instalment of the loan = Rs. 105

The 3^rd instalment of the loan = Rs. 110

and so, on

The amount that the man repays every month forms an AP.

Therefore, the series is

100, 105, 110, 115,…

Here, a = 100, d = 105 – 100 = 5

We know that,
⇒a_n = a + (n – 1)d
⇒a₃₀ = 100 + (30 – 1)5
⇒ a₃₀ = 100 + 29 × 5
⇒ a₃₀ = 100 +145
⇒ a₃₀ = 245
Hence, the amount he will pay in the 30^th installment is Rs 245.

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