| Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4 |
Exercise 8.1
Question 1 A
Write the first three terms of the following sequences defined by:
tn = 3n + 1
Sol :tn = 3n + 1
Given: tn = 3n + 1
Taking n = 1, we get
t1 = 3(1) + 1 = 3 + 1 = 4
Taking n = 2, we get
t2 = 3(2) + 1 = 6 + 1 = 7
Taking n = 3, we get
t3 = 3(3) + 1 = 9 + 1 = 10
Hence, the first three terms are 4, 7 and 10.
Question 1 B
Write the first three terms of the following sequences defined by:
tn = 2nSol :
Given: tn = 2n
Taking n = 1, we get
t1 = 21 = 2
Taking n = 2, we get
t2 = 22 = 2 × 2 = 4
Taking n = 3, we get
t3 = 23 = 2 × 2 × 2 = 8
Hence, the first three terms are 2, 4 and 8.
Given: tn = n2 + 1
Taking n = 1, we get
t1 = (1)2 + 1 = 1 + 1 = 2
Taking n = 2, we get
t2 = (2)2 + 1 = 4 + 1 = 5
Taking n = 3, we get
t3 = (3)2 + 1 = 9 + 1 = 10
Hence, the first three terms are 2, 5 and 10.
Given: tn = n(n+2)
Taking n = 1, we get
t1 = (1)(1+2) = (1)(3) = 3
Taking n = 2, we get
t2 = (2)(2+2) = (2)(4) = 8
Taking n = 3, we get
t3 = (3)(3+2) = (3)(5) = 15
Hence, the first three terms are 3, 8 and 15.
Given: tn = 2n + 5
Taking n = 1, we get
t1 = 2(1) + 5 = 2 + 5 = 7
Taking n = 2, we get
t2 = 2(2) + 5 = 4 + 5 = 9
Taking n = 3, we get
t3 = 2(3) + 5 = 6 + 5 = 11
Hence, the first three terms are 7, 9 and 11.
Given: $\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}-3}{4}$
Taking n = 1, we get
$\mathrm{t}_{1}=\frac{1-3}{4}=\frac{-2}{4}=\frac{-1}{2}$
Given: tn = 2n
Taking n = 1, we get
t1 = 21 = 2
Taking n = 2, we get
t2 = 22 = 2 × 2 = 4
Taking n = 3, we get
t3 = 23 = 2 × 2 × 2 = 8
Hence, the first three terms are 2, 4 and 8.
Question 1 C
Write the first three terms of the following sequences defined by:
tn = n2 + 1
Sol :tn = n2 + 1
Given: tn = n2 + 1
Taking n = 1, we get
t1 = (1)2 + 1 = 1 + 1 = 2
Taking n = 2, we get
t2 = (2)2 + 1 = 4 + 1 = 5
Taking n = 3, we get
t3 = (3)2 + 1 = 9 + 1 = 10
Hence, the first three terms are 2, 5 and 10.
Question 1 D
Write the first three terms of the following sequences defined by:
tn = n(n + 2)
Sol :tn = n(n + 2)
Given: tn = n(n+2)
Taking n = 1, we get
t1 = (1)(1+2) = (1)(3) = 3
Taking n = 2, we get
t2 = (2)(2+2) = (2)(4) = 8
Taking n = 3, we get
t3 = (3)(3+2) = (3)(5) = 15
Hence, the first three terms are 3, 8 and 15.
Question 1 E
Write the first three terms of the following sequences defined by:
tn = 2n+5
Sol :tn = 2n+5
Given: tn = 2n + 5
Taking n = 1, we get
t1 = 2(1) + 5 = 2 + 5 = 7
Taking n = 2, we get
t2 = 2(2) + 5 = 4 + 5 = 9
Taking n = 3, we get
t3 = 2(3) + 5 = 6 + 5 = 11
Hence, the first three terms are 7, 9 and 11.
Question 1 F
Write the first three terms of the following sequences defined by:
$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}-3}{4}$
Sol :$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}-3}{4}$
Given: $\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}-3}{4}$
Taking n = 1, we get
$\mathrm{t}_{1}=\frac{1-3}{4}=\frac{-2}{4}=\frac{-1}{2}$
Taking n = 2, we get
$\mathrm{t}_{2}=\frac{2-3}{4}=\frac{-1}{4}$
$\mathrm{t}_{2}=\frac{2-3}{4}=\frac{-1}{4}$
Taking n = 3, we get
$\mathrm{t}_{3}=\frac{3-3}{4}=0$
Hence, the first three terms are $\frac{-1}{2}, \frac{-1}{4}$ and 0
Question 2 A
Find the indicated terms in each of the following sequence whose nth terms are:
$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}^{2}(\mathrm{n}+1)}{3}$
Sol :
Given: $\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}^{2}(\mathrm{n}+1)}{3}$Now, we have to find t1 and t2.
So, in t1, n = 1
$\therefore \mathrm{t}_{1}=\frac{(1)^{2}(1+1)}{3}$
$=\frac{(1)(2)}{3}=\frac{2}{3}$
Now, t2, n = 2
Now, t2, n = 2
$\therefore \mathrm{t}_{2}=\frac{(2)^{2}(2+1)}{3}$
$=\frac{(4)(3)}{3}=4$
Given: $\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-2)}{\mathrm{n}+3}$
So,$\mathrm{t}_{20}=\frac{20(20-2)}{20+3}$
Question 2 B
Find the indicated terms in each of the following sequence whose nth terms are:
$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-2)}{\mathrm{n}+3} ; \mathrm{t}_{20}$
Sol :$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-2)}{\mathrm{n}+3} ; \mathrm{t}_{20}$
Given: $\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}-2)}{\mathrm{n}+3}$
So,$\mathrm{t}_{20}=\frac{20(20-2)}{20+3}$
$=\frac{20 \times 18}{23}=\frac{360}{23}$
Question 2 C
Find the indicated terms in each of the following sequence whose nth terms are:
tn= (n - 1)(2 - n)(3 + n) ; t20
Sol :tn= (n - 1)(2 - n)(3 + n) ; t20
Given: tn = (n – 1)(2 – n)(3+n)
So, t20 = (20 – 1)(2 – 20)(3+20)
= (19)(-18)(23)
= -7866
Question 2 D
Find the indicated terms in each of the following sequence whose nth terms are:
$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{t}_{\mathrm{n}-1}}{\mathrm{n}^{2}}, \mathrm{t}_{1}=3$;
$\mathrm{t}_{2}, \mathrm{t}_{3},(\mathrm{n} \geq 2)$
Sol :Given: $\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{t}_{\mathrm{n}-1}}{\mathrm{n}^{2}}$
So, $\mathrm{t}_{2}=\frac{\mathrm{t}_{2-1}}{(2)^{2}}=\frac{\mathrm{t}_{1}}{4}=\frac{3}{4}$ [given: t1 = 3]
and $\mathrm{t}_{3}=\frac{\mathrm{t}_{3-1}}{(3)^{2}}=\frac{\mathrm{t}_{2}}{9}$
$=\frac{\frac{3}{4}}{9}=\frac{3}{4 \times 9}=\frac{1}{12}$ $\left[\because \mathrm{t}_{2}=\frac{3}{4}\right]$
Question 3 A
Write the next three terms of the following sequences:
$\mathrm{t}_{1}=3, \mathrm{t}_{\mathrm{n}}=3 \mathrm{t}_{\mathrm{n}-1}+2$
Sol :$\mathrm{t}_{1}=3, \mathrm{t}_{\mathrm{n}}=3 \mathrm{t}_{\mathrm{n}-1}+2$
Given: t2 = 2 and tn = tn-1 + 1
Now, we have to find next three terms i.e. t3, t4 and t5
Taking n = 3, we get
t3 = t3-1 + 1
= t2 + 1
= 2 + 1 [given: t2 = 2]
t3 = 3 …(i)
Taking n = 4, we get
t4 = t4-1 + 1
= t3 + 1
= 3 + 1 [from (i)]
t4 = 4 …(ii)
Taking n = 5, we get
t5 = t5-1 + 1
= t4 + 1
= 4 +1
t5 = 5 [from (ii)]
Hence, the next three terms are 3, 4 and 5.
Question 3 B
Write the next three terms of the following sequences:
$\mathrm{t}_{1}=3, \mathrm{t}_{\mathrm{n}}=3 \mathrm{t}_{\mathrm{n}-1}+2$ for all n>1
Sol :$\mathrm{t}_{1}=3, \mathrm{t}_{\mathrm{n}}=3 \mathrm{t}_{\mathrm{n}-1}+2$ for all n>1
Given: t1 = 3 and tn = 3tn-1 + 2
Now, we have to find next three terms i.e. t2, t3 and t4
Taking n = 2, we get
t2 = 3t2-1 + 2
= 3t1 + 2
= 3(3) + 2 [given: t1 = 3]
t2 = 9 + 2
t2 = 11 …(i)
Taking n = 3, we get
t3 = 3t3-1 + 2
= 3t2 + 2
= 3(11) + 2 [from (i)]
= 33 + 2
t3 = 35 …(ii)
Taking n = 4, we get
t4 = 3t4-1 + 2
= 3t3 + 2
= 3(35) +2
t5 = 105 + 2
t5 = 107 [from (ii)]
Hence, the next three terms are 11, 35 and 107.
Question 4 A
Write the first four terms of the A.P. when first term a and common difference d are given as follows:
a = 1, d = 1
Sol :a = 1, d = 1
Given: a = 1 and d = 1
The general form of an A.P is a, a+d, a+2d, a+3d,…
So, the first term a is 1 and the common difference d is 1, then the first four terms of the AP is
1, (1+1), (1+2×1), (1+3×1)
⇒ 1, 2, 3, 4
Hence, the first four terms of the A.P. is 1, 2, 3 and 4.
Question 4 B
Write the first four terms of the A.P. when first term a and common difference d are given as follows:
a= 3, d=0
Sol :a= 3, d=0
Given: a = 3 and d = 0
The general form of an A.P is a, a+d, a+2d, a+3d,…
So, the first term a is 3 and the common difference d is 0, then the first four terms of the AP is
3, (3+0), (3+2×0), (3+3×0)
⇒ 3, 3, 3, 3
Hence, first four terms of the A.P. is 3, 3, 3 and 3.
Question 4 C
Write the first four terms of the A.P. when first term a and common difference d are given as follows:
a = 10, d = 10
Sol :a = 10, d = 10
Given: a = 10 and d = 10
The general form of an A.P is a, a+d, a+2d, a+3d,…
So, the first term a is 10 and the common difference d is 10, then the first four terms of the AP is
10, (10+10), (10+2×10), (10+3×10)
⇒ 10, (20), (10+20), (10+30)
⇒ 10, 20, 30, 40
Hence, first four terms of the A.P. is 10, 20, 30 and 40.
Question 4 D
Write the first four terms of the A.P. when first term a and common difference d are given as follows:
a= -2, d=0
Sol :a= -2, d=0
Given: a = -2 and d = 0
The general form of an A.P is a, a+d, a+2d, a+3d,…
So, the first term a is -2 and the common difference d is 0, then the first four terms of the AP is
-2, (-2+0), (-2+2×0), (-2+3×0)
⇒ -2, -2, -2, -2
Hence, the first four terms of the A.P. is -2, -2, -2 and -2.
Question 4 E
Write the first four terms of the A.P. when first term a and common difference d are given as follows:
a = 100, d = -30
Sol :a = 100, d = -30
Given: a = 100 and d = -30
The general form of an A.P is a, a+d, a+2d, a+3d,…
So, the first term a is 100 and the common difference d is -30, then the first four terms of the AP is
100, (100+(-30)), (100+2×(-30)),(100+3×(-30))
⇒ 100, (100 – 30), (100 – 60), (100 – 90)
⇒ 100, 70, 40, 10
Hence, first four terms of the A.P. is 100, 70, 40 and 10.
Question 4 F
Write the first four terms of the A.P. when first term a and common difference d are given as follows:
a= -1, d= 1/2
Sol :a= -1, d= 1/2
Given: a = -1 and d $=\frac{1}{2}$
The general form of an A.P is a, a+d, a+2d, a+3d,…
So, the first term a is 1 and the common difference d is $\frac{1}{2}$, then the first four terms of the AP is
$-1,\left(-1+\frac{1}{2}\right),\left(-1+2 \times \frac{1}{2}\right),\left(-1+3 \times \frac{1}{2}\right)$
⇒ $-1,\left(\frac{-2+1}{2}\right),(-1+1),\left(\frac{-2+3}{2}\right)$
$-1, \frac{-1}{2}, 0, \frac{1}{2}$
Hence, first four terms of the A.P. is $-1, \frac{-1}{2}, 0, \frac{1}{2}$
Question 4 G
Write the first four terms of the A.P. when first term a and common difference d are given as follows:
a = -7, d = -7
Sol :a = -7, d = -7
Given: a = -7 and d = -7
The general form of an A.P is a, a+d, a+2d, a+3d,…
So, the first term a is -7, and the common difference d is -7, then the first four terms of the AP is
-7, (-7+(-7)), (-7+2×(-7)), (-7+3×(-7))
⇒ -7, (-7 – 7), (-7 – 14), (-7 – 21)
⇒ -7, -14, -21, -28
Hence, the first four terms of the A.P. is -7, -14, -21 and -28.
Question 4 H
Write the first four terms of the A.P. when first term a and common difference d are given as follows:
a = 1, d = 0.1
Sol :a = 1, d = 0.1
Given: a = 1 and d = 0.1
The general form of an A.P is a, a+d, a+2d, a+3d,…
So, the first term a is 1, and the common difference d is 0.1, then the first four terms of the AP is
1, (1+0.1), (1+2×(0.1)), (1+3×(0.1))
⇒ 1, 1.1, 1.2, 1.3
Hence, the first four terms of the A.P. is 1, 1.1, 1.2 and 1.3.
Question 5 A
For the following A.P's write the first term and common difference:
6, 3, 0, - 3, ...
Sol :6, 3, 0, - 3, ...
In general, for an AP a1, a2, . . . .,an, we have
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
For the list of numbers: 6, 3, 0, -3, . . .
a2 – a1 = 3 – 6 = -3
a3 – a2 = 0 – 3 = -3
a4 – a3 = -3 – 0 = -3
Here, the difference of any two consecutive terms in each case is -3.
So, the given list is an AP whose first term a is 6, and common difference d is -3.
Question 5 B
For the following A.P's write the first term and common difference:
- 3.1, - 3.0, - 2.9, - 2.8, ...
Sol :- 3.1, - 3.0, - 2.9, - 2.8, ...
In general, for an AP a1, a2, . . . .,an, we have
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
For the list of numbers: - 3.1, - 3.0, - 2.9, - 2.8, ...
a2 – a1 = -3.0 – (-3.1) = -3.0 + 3.1 = 0.1
a3 – a2 = -2.9 – (-3.0) = -2.9 + 3.0 = 0.1
a4 – a3 = -2.8 – (-2.9) = -2.8 + 2.9 = 0.1
Here, the difference of any two consecutive terms in each case is 0.1. So, the given list is an AP whose first term a is -3.1 and common difference d is 0.1.
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
For the list of numbers: - 3.1, - 3.0, - 2.9, - 2.8, ...
a2 – a1 = -3.0 – (-3.1) = -3.0 + 3.1 = 0.1
a3 – a2 = -2.9 – (-3.0) = -2.9 + 3.0 = 0.1
a4 – a3 = -2.8 – (-2.9) = -2.8 + 2.9 = 0.1
Here, the difference of any two consecutive terms in each case is 0.1. So, the given list is an AP whose first term a is -3.1 and common difference d is 0.1.
Question 5 C
For the following A.P's write the first term and common difference:
147, 148, 149, 150, ...
Sol :147, 148, 149, 150, ...
In general, for an AP a1, a2, . . . .,an, we have
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
For the list of numbers: 147, 148, 149, 150, ...
a2 – a1 = 148 – 147 = 1
a3 – a2 = 149 – 148 = 1
a4 – a3 = 150 – 149 = 1
Here, the difference of any two consecutive terms in each case is -1. So, the given list is an AP whose first term a is 147 and common difference d is 1.
Question 5 D
For the following A.P's write the first term and common difference:
- 5, - 1, 3, 7, ...
Sol :- 5, - 1, 3, 7, ...
In general, for an AP a1, a2, . . . .,an, we have
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
For the list of numbers: - 5, - 1, 3, 7, ...
a2 – a1 = -1 – (-5) = -1 + 5 = 4
a3 – a2 = 3 – (-1) = 3 + 1 = 4
a4 – a3 = 7 – 3 = 4
Here, the difference of any two consecutive terms in each case is -4. So, the given list is an AP whose first term a is -5 and common difference d is 4.
Question 5 E
For the following A.P's write the first term and common difference:
3, 1, - 1, - 3, ...
Sol :3, 1, - 1, - 3, ...
In general, for an AP a1, a2, . . . .,an, we have
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
For the list of numbers: 3, 1, - 1, - 3, ...
a2 – a1 = 1 – 3 = -2
a3 – a2 =-1 – 1 = -1 - 1 = -2
a4 – a3 =-3 – (-1) = -3 + 1= -2
Here, the difference of any two consecutive terms in each case is --2. So, the given list is an AP whose first term a is 3 and common difference d is -2.
Question 5 F
For the following A.P's write the first term and common difference:
$2,2 \frac{1}{3}, 2 \frac{2}{3},-3$
Sol :$2,2 \frac{1}{3}, 2 \frac{2}{3},-3$
In general, for an AP a1, a2, . . . .,an, we have
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
For the list of numbers: $2, \frac{7}{3}, \frac{8}{3}, 3, \ldots$
$a_{2}-a_{1}=\frac{7}{3}-2=\frac{7-6}{3}=\frac{1}{3}$
$a_{3}-a_{2}=\frac{8}{3}-\frac{7}{3}=\frac{1}{3}$
$a_{4}-a_{3}=3-\frac{8}{3}=\frac{9-8}{3}=\frac{1}{3}$
Here, the difference of any two consecutive terms in each case is $\frac{1}{3}$. So, the given list is an AP whose first term a is 2 and common difference d is $\frac{1}{3}$.
In general, for an AP a1, a2, . . . .,an, we have
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
$a_{2}-a_{1}=\frac{7}{3}-2=\frac{7-6}{3}=\frac{1}{3}$
$a_{3}-a_{2}=\frac{8}{3}-\frac{7}{3}=\frac{1}{3}$
$a_{4}-a_{3}=3-\frac{8}{3}=\frac{9-8}{3}=\frac{1}{3}$
Here, the difference of any two consecutive terms in each case is $\frac{1}{3}$. So, the given list is an AP whose first term a is 2 and common difference d is $\frac{1}{3}$.
Question 5 G
For the following A.P's write the first term and common difference:
$\frac{3}{2}, \frac{1}{2},-\frac{1}{2},-\frac{3}{2}, \ldots$
Sol :$\frac{3}{2}, \frac{1}{2},-\frac{1}{2},-\frac{3}{2}, \ldots$
In general, for an AP a1, a2, . . . .,an, we have
d = ak+1 - ak
where ak+1 and ak are the (k+1)th and kth terms respectively.
For the list of numbers: $\frac{3}{2}, \frac{1}{2}, \frac{-1}{2}, \frac{-3}{2}, \ldots$
$a_{2}-a_{1}=\frac{1}{2}-\frac{3}{2}=\frac{-2}{2}=-1$
$a_{3}-a_{2}=\frac{-1}{2}-\frac{1}{2}=\frac{-2}{2}=-1$
$a_{4}-a_{3}=-\frac{3}{2}-\left(\frac{-1}{2}\right)=\frac{-3+1}{2}=\frac{-2}{2}=-1$
$a_{2}-a_{1}=\frac{1}{2}-\frac{3}{2}=\frac{-2}{2}=-1$
$a_{3}-a_{2}=\frac{-1}{2}-\frac{1}{2}=\frac{-2}{2}=-1$
$a_{4}-a_{3}=-\frac{3}{2}-\left(\frac{-1}{2}\right)=\frac{-3+1}{2}=\frac{-2}{2}=-1$
Here, the difference of any two consecutive terms in each case is -1. So, the given list is an AP whose first term a is $\frac{3}{2}$ and common difference d is -1.
Question 6 A
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
1, - 1, - 3, - 5,
Sol :1, - 1, - 3, - 5,
We have,
a2 – a1 = -1 – 1 = -2
a3 – a2 =-3 – (-1) = -3 + 1 = -2
a4 – a3 =-5 – (-3) = -5 + 3= -2
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = -2
Now, we have to find the next three terms.
We have a1 = 1, a2 = -1, a3 = -3 and a4 = -5
Now, we will find a5, a6 and a7
So, a5 = -5 + (-2) = -5 – 2 = -7
a6 = -7 + (-2) = -7 – 2 = -9
and a7 = -9 + (-2) = -9 – 2 = -11
Hence, the next three terms are -7, -9 and -11
Question 6 B
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
2, 4, 8, 16, ...
Sol :2, 4, 8, 16, ...
We have,
a2 – a1 = 4 – 2 = 2
a3 – a2 = 8 – 4 = 4
a4 – a3 = 16 – 8 = 8
i.e. ak+1 – ak is not same every time.
So, the given list of numbers do not form an AP.
Question 6 C
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
- 2, 2, - 2, 2, - 2, ...
Sol :- 2, 2, - 2, 2, - 2, ...
We have,
a2 – a1 = 2 – (-2) = 2 + 2 = 4
a3 – a2 = -2 – 2 = -4
a4 – a3 = 2 – (-2) = 2 + 2 = 4
i.e. ak+1 – ak is not same every time.
So, the given list of numbers do not form an AP.
Question 6 D
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$
Sol :$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$
We have,
$a_{2}-a_{1}=-\frac{1}{2}-\left(-\frac{1}{2}\right)=\frac{-1+1}{2}=0$
$a_{3}-a_{2}=-\frac{1}{2}-\left(-\frac{1}{2}\right)=\frac{-1+1}{2}=0$
$a_{4}-a_{3}=-\frac{1}{2}-\left(-\frac{1}{2}\right)=\frac{-1+1}{2}=0$
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = 0
Now, we have to find the next three terms.
We have $a_{1}=-\frac{1}{2}$,$ a_{2}=-\frac{1}{2}$,$ a_{3}=-\frac{1}{2}$,$ a_{4}=-\frac{1}{2}$
Now, we will find a5, a6 and a7
So, $a_{5}=-\frac{1}{2}+0=-\frac{1}{2}$
$a_{6}=-\frac{1}{2}+0=-\frac{1}{2}$
and $a_{7}=-\frac{1}{2}+0=-\frac{1}{2}$
Hence, the next three terms are $-\frac{1}{2},-\frac{1}{2}$ and $-\frac{1}{2}$
Question 6 E
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
$2, \frac{5}{2}, 3 \frac{7}{2}, \ldots \ldots$
Sol :$2, \frac{5}{2}, 3 \frac{7}{2}, \ldots \ldots$
We have,
$\mathrm{a}_{2}-\mathrm{a}_{1}=\frac{5}{2}-2$ $=\frac{5-4}{2}=\frac{1}{2}$
$\mathrm{a}_{3}-\mathrm{a}_{2}=3-\left(\frac{5}{2}\right)$ $=\frac{6-5}{2}=\frac{1}{2}$
$a_{4}-a_{3}=\frac{7}{2}-3$ $=\frac{7-6}{2}=\frac{1}{2}$
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference $\mathrm{d}=\frac{1}{2}$
Now, we have to find the next three terms.
We have $a_{1}=2$, $a_{2}=\frac{5}{2}$,$ a_{3}=3$,$ a_{4}=\frac{7}{2}$
Now, we will find a5, a6 and a7
So, $a_{5}=\frac{7}{2}+\frac{1}{2}$ $=\frac{8}{2}=4$
$a_{6}=4+\frac{1}{2}$ $=\frac{8+1}{2}=\frac{9}{2}$
and $a_{7}=\frac{9}{2}+\frac{1}{2}$ $=\frac{10}{2}=5$
Hence, the next three terms are $4, \frac{9}{2}$ and 5
We have,
a2 – a1 = -4 – 0 = -4
a3 – a2 =-8 – (-4) = -8 + 4 = -4
a4 – a3 =-12 – (-8) = -12 + 8= -4
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = -4
Now, we have to find the next three terms.
We have a1 = 0, a2 = -4, a3 = -8 and a4 = -12
Now, we will find a5, a6 and a7
So, a5 = -12 + (-4) = -12 – 4 = -16
a6 = -16 + (-4) = -16 – 4 = -20
and a7 = -20 + (-4) = -20 – 4 = -24
Hence, the next three terms are -16, -20 and -24
We have,
a2 – a1 = 10 – 4 = 6
a3 – a2 = 16 – 10 = 6
a4 – a3 = 22 – 16 = 6
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = 6
Now, we have to find the next three terms.
We have a1 = 4, a2 = 10, a3 = 16 and a4 = 22
Now, we will find a5, a6 and a7
So, a5 = 22 + 6 = 28
a6 = 28 + 6 = 34
and a7 = 34 + 6 = 40
Hence, the next three terms are 28, 34 and 40
We have,
a2 – a1 = 2a – a = a
a3 – a2 = 3a – 2a = a
a4 – a3 = 4a – 3a = a
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = a
Now, we have to find the next three terms.
We have a1 = a, a2 = 2a, a3 = 3a and a4 = 4a
Now, we will find a5, a6 and a7
So, a5 = 4a + a = 5a
a6 = 5a + a = 6a
and a7 = 6a + a = 7a
Hence, the next three terms are 5a, 6a and 7a
We have,
a2 – a1 = -3.2 – (-1.2) =-3.2 + 1.2 = -2.0
a3 – a2 = -5.2 – (-3.2) = -5.2 + 3.2 = -2.0
a4 – a3 = -7.2 – (-5.2) = -7.2 + 5.2 = -2.0
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = -2
Now, we have to find the next three terms.
We have a1 = -1.2, a2 = -3.2, a3 = -5.2 and a4 = -7.2
Now, we will find a5, a6 and a7
So, a5 = -7.2 + (-2) = -7.2 – 2.0 = -9.2
a6 = -9.2 + (-2) = -9.2 – 2.0 = -11.2
and a7 = -11.2 + (-2) = -11.2 – 2.0 = -13.2
Hence, the next three terms are -9.2, -11.2 and -13.2
We have,
a2 – a1 = √12 - √3 = 2√3 - √3 = √3
a3 – a2 = √48 - √12 = 4√3 - 2√3 = 2√3
a4 – a3 = √192 - √48 = 8√3 - 4√3 = 4√3
i.e. ak+1 – ak is not same every time.
So, the given list of numbers do not form an AP.
We have,
a2 – a1 = a2 – a = a (a – 1)
a3 – a2 = a3 – a2 = a2(a – 1)
a4 – a3 = a4 – a3 = a3(a – 1)
i.e. ak+1 – ak is not same every time.
So, the given list of numbers does not form an AP.
We have,
a2 – a1 = 3 – 1 = 2
a3 – a2 = 9 – 3 = 6
a4 – a3 = 27 – 9 = 18
i.e. ak+1 – ak is not same every time.
So, the given list of numbers does not form an AP.
We have,
a2 – a1 = 22 – (1)2 = 4 – 1 = 3
a3 – a2 = 32 – (2)2 = 9 – 4 = 5
a4 – a3 = 42 – (3)2 = 16 – 9 =7
i.e. ak+1 – ak is not same every time.
So, the given list of numbers does not form an AP.
We have,
a2 – a1 = 52 – (1)2 = 25 – 1 = 24
a3 – a2 = 72 – (5)2 = 49 – 25 = 24
a4 – a3 = 72 – (7)2 = 0
i.e. ak+1 – ak is not same every time.
So, the given list of numbers do not form an AP.
We have,
a2 – a1 = 32 – (1)2 = 9 – 1 = 8
a3 – a2 = 52 – (3)2 = 25 – 9 = 16
a4 – a3 = 72 – (5)2 = 49 – 25 =24
i.e. ak+1 – ak is not same every time.
So, the given list of numbers does not form an AP.
Salary for the 1st year = Rs. 8000
and according to the question,
There is an annual increment of Rs. 500
⇒ The salary for the 2nd year = Rs. 8000 + 500 = Rs.8500
Now, again there is an increment of Rs. 500
⇒ The salary for the 3rd year = Rs. 8500 +500 = Rs. 9000
Therefore, the series is
8000 , 8500 , 9000 , …
Difference between 2nd term and 1st term = 8500 – 8000 = 500
Difference between 3rd term and 2nd term = 9000 – 8500 = 500
Since, the difference is same.
Hence, salary in successive years are in AP with common difference d = 500 and first term a is 8000.
Taxi fare for 1km = 15
According to question, Rs. 8 for each additional km
⇒ Taxi fare for 2km = 15 + 8 =23
and Taxi fare for 3km = 23 + 8 =31
Therefore, series is
15, 23, 31 ,…
Difference between 2nd and 1st term = 23 – 15 = 8
Difference between 3rd and 2nd term = 31 – 23 = 8
Since, difference is same.
Hence, the taxi fare after each km form an AP with the first term, a = Rs. 15 and common difference, d = Rs. 8
The length of the bottom rung = 45cm
According to the question,
Length of rungs decreases by 2cm from bottom to top. The lengths (in cm) of the 1st, 2nd, 3rd, … from the bottom to top respectively are 45, 43, 41, …
Difference between 2nd and 1st term = 43 – 45 = -2
Difference between 3rd and 2nd term = 41 – 43 = -2
Since, the difference is same.
Hence, the length of the rungs form an AP with a = 45cm and d = -2 cm.
Original Amount = Rs. 10,000
Interest earned in first year = 10,000 × 8%
$=10,000 \times \frac{8}{100}$
= Rs 800
Total amount outstanding after one year = Rs 10000 + 800
= Rs 10800
Now, interest earned in 2nd year
Question 6 F
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
0, - 4, - 8, - 12,
Sol :0, - 4, - 8, - 12,
We have,
a2 – a1 = -4 – 0 = -4
a3 – a2 =-8 – (-4) = -8 + 4 = -4
a4 – a3 =-12 – (-8) = -12 + 8= -4
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = -4
Now, we have to find the next three terms.
We have a1 = 0, a2 = -4, a3 = -8 and a4 = -12
Now, we will find a5, a6 and a7
So, a5 = -12 + (-4) = -12 – 4 = -16
a6 = -16 + (-4) = -16 – 4 = -20
and a7 = -20 + (-4) = -20 – 4 = -24
Hence, the next three terms are -16, -20 and -24
Question 6 G
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
4, 10, 16, 22, ...
Sol :4, 10, 16, 22, ...
We have,
a2 – a1 = 10 – 4 = 6
a3 – a2 = 16 – 10 = 6
a4 – a3 = 22 – 16 = 6
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = 6
Now, we have to find the next three terms.
We have a1 = 4, a2 = 10, a3 = 16 and a4 = 22
Now, we will find a5, a6 and a7
So, a5 = 22 + 6 = 28
a6 = 28 + 6 = 34
and a7 = 34 + 6 = 40
Hence, the next three terms are 28, 34 and 40
Question 6 H
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
a, 2a, 3a, 4a, ...
Sol :a, 2a, 3a, 4a, ...
We have,
a2 – a1 = 2a – a = a
a3 – a2 = 3a – 2a = a
a4 – a3 = 4a – 3a = a
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = a
Now, we have to find the next three terms.
We have a1 = a, a2 = 2a, a3 = 3a and a4 = 4a
Now, we will find a5, a6 and a7
So, a5 = 4a + a = 5a
a6 = 5a + a = 6a
and a7 = 6a + a = 7a
Hence, the next three terms are 5a, 6a and 7a
Question 6 I
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
- 1.2, - 3.2, - 5.2, - 7.2, ...
Sol :- 1.2, - 3.2, - 5.2, - 7.2, ...
We have,
a2 – a1 = -3.2 – (-1.2) =-3.2 + 1.2 = -2.0
a3 – a2 = -5.2 – (-3.2) = -5.2 + 3.2 = -2.0
a4 – a3 = -7.2 – (-5.2) = -7.2 + 5.2 = -2.0
i.e. ak+1 – ak is the same every time.
So, the given list of numbers forms an AP with the common difference d = -2
Now, we have to find the next three terms.
We have a1 = -1.2, a2 = -3.2, a3 = -5.2 and a4 = -7.2
Now, we will find a5, a6 and a7
So, a5 = -7.2 + (-2) = -7.2 – 2.0 = -9.2
a6 = -9.2 + (-2) = -9.2 – 2.0 = -11.2
and a7 = -11.2 + (-2) = -11.2 – 2.0 = -13.2
Hence, the next three terms are -9.2, -11.2 and -13.2
Question 6 J
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
$\sqrt{3}, \sqrt{12}, \sqrt{48}, \sqrt{192}$
Sol :$\sqrt{3}, \sqrt{12}, \sqrt{48}, \sqrt{192}$
We have,
a2 – a1 = √12 - √3 = 2√3 - √3 = √3
a3 – a2 = √48 - √12 = 4√3 - 2√3 = 2√3
a4 – a3 = √192 - √48 = 8√3 - 4√3 = 4√3
i.e. ak+1 – ak is not same every time.
So, the given list of numbers do not form an AP.
Question 6 K
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
a, a2, a3, a4, …..
Sol :a, a2, a3, a4, …..
We have,
a2 – a1 = a2 – a = a (a – 1)
a3 – a2 = a3 – a2 = a2(a – 1)
a4 – a3 = a4 – a3 = a3(a – 1)
i.e. ak+1 – ak is not same every time.
So, the given list of numbers does not form an AP.
Question 6 L
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
1, 3, 9, 27, ..
Sol :1, 3, 9, 27, ..
We have,
a2 – a1 = 3 – 1 = 2
a3 – a2 = 9 – 3 = 6
a4 – a3 = 27 – 9 = 18
i.e. ak+1 – ak is not same every time.
So, the given list of numbers does not form an AP.
Question 6 M
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
12, 22, 32, 42, …
Sol :12, 22, 32, 42, …
We have,
a2 – a1 = 22 – (1)2 = 4 – 1 = 3
a3 – a2 = 32 – (2)2 = 9 – 4 = 5
a4 – a3 = 42 – (3)2 = 16 – 9 =7
i.e. ak+1 – ak is not same every time.
So, the given list of numbers does not form an AP.
Question 6 N
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
12, 52, 72, 72, …
Sol :12, 52, 72, 72, …
We have,
a2 – a1 = 52 – (1)2 = 25 – 1 = 24
a3 – a2 = 72 – (5)2 = 49 – 25 = 24
a4 – a3 = 72 – (7)2 = 0
i.e. ak+1 – ak is not same every time.
So, the given list of numbers do not form an AP.
Question 6 O
Which of the following list of numbers form A.R's? If they form an A.P., find the common difference d and also write its next three terms.
12, 32, 52, 72, …
Sol :12, 32, 52, 72, …
We have,
a2 – a1 = 32 – (1)2 = 9 – 1 = 8
a3 – a2 = 52 – (3)2 = 25 – 9 = 16
a4 – a3 = 72 – (5)2 = 49 – 25 =24
i.e. ak+1 – ak is not same every time.
So, the given list of numbers does not form an AP.
Question 7 A
In which of the following situations does the list of numbers involved arithmetic progression, and why?
The salary of a teacher in successive years when starting salary is Rs. 8000, with an annual increment of Rs. 500.
Sol :The salary of a teacher in successive years when starting salary is Rs. 8000, with an annual increment of Rs. 500.
Salary for the 1st year = Rs. 8000
and according to the question,
There is an annual increment of Rs. 500
⇒ The salary for the 2nd year = Rs. 8000 + 500 = Rs.8500
Now, again there is an increment of Rs. 500
⇒ The salary for the 3rd year = Rs. 8500 +500 = Rs. 9000
Therefore, the series is
8000 , 8500 , 9000 , …
Difference between 2nd term and 1st term = 8500 – 8000 = 500
Difference between 3rd term and 2nd term = 9000 – 8500 = 500
Since, the difference is same.
Hence, salary in successive years are in AP with common difference d = 500 and first term a is 8000.
Question 7 B
In which of the following situations does the list of numbers involved arithmetic progression, and why?
The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
Sol :The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
Taxi fare for 1km = 15
According to question, Rs. 8 for each additional km
⇒ Taxi fare for 2km = 15 + 8 =23
and Taxi fare for 3km = 23 + 8 =31
Therefore, series is
15, 23, 31 ,…
Difference between 2nd and 1st term = 23 – 15 = 8
Difference between 3rd and 2nd term = 31 – 23 = 8
Since, difference is same.
Hence, the taxi fare after each km form an AP with the first term, a = Rs. 15 and common difference, d = Rs. 8
Question 7 C
In which of the following situations does the list of numbers involved arithmetic progression, and why?
The lengths of the rungs of a ladder when the bottom rung is 45cm, and length of rungs decrease by 2 cm from bottom to top.
Sol :The lengths of the rungs of a ladder when the bottom rung is 45cm, and length of rungs decrease by 2 cm from bottom to top.
The length of the bottom rung = 45cm
According to the question,
Length of rungs decreases by 2cm from bottom to top. The lengths (in cm) of the 1st, 2nd, 3rd, … from the bottom to top respectively are 45, 43, 41, …
Difference between 2nd and 1st term = 43 – 45 = -2
Difference between 3rd and 2nd term = 41 – 43 = -2
Since, the difference is same.
Hence, the length of the rungs form an AP with a = 45cm and d = -2 cm.
Question 7 D
In which of the following situations does the list of numbers involved arithmetic progression, and why?
The amount of money in the account every year when Rs. 10000 is deposited at compound interest 8% per annum.
Sol :The amount of money in the account every year when Rs. 10000 is deposited at compound interest 8% per annum.
Original Amount = Rs. 10,000
Interest earned in first year = 10,000 × 8%
$=10,000 \times \frac{8}{100}$
= Rs 800
Total amount outstanding after one year = Rs 10000 + 800
= Rs 10800
Now, interest earned in 2nd year
$=10800 \times \frac{8}{100}=\mathrm{Rs} .864$
Total amount outstanding after 2nd year = Rs10800 + 864
= Rs 11664
Interest earned in 3rd year
Total amount outstanding after 2nd year = Rs10800 + 864
= Rs 11664
Interest earned in 3rd year
$=11664 \times \frac{8}{100}$
= Rs 933.12
Total amount outstanding after 3rd year = Rs 11664 + 933.12
= Rs 12597.12
Therefore, the series is
10800, 11664, 12597.12,…
Difference between second and first term = 11664 – 10800
= 864
Difference between third and second term = 12597.12 – 11664
= 933.12
Since the difference is not same
Therefore, it doesn’t form an AP.
The money saved by Sudha in the first year = Rs. 100
According to the question,
Sudha increased the amount by Rs. 50 every year
⇒ The money saved by Sudha in a 2nd year = Rs. 100 +50
= Rs. 150
The money saved by Sudha in a 3rd year = Rs. 150 + 50
= Rs. 200
Therefore, the series is
100, 150, 200, 250,…
Difference in the 2nd term and 1st term = 150 – 100 = 50
Difference in the 3rd term and 2nd term = 200 – 150 = 50
Since the difference is the same.
Therefore, the money saved by Sudha in successive years form an AP with a = Rs 100 and d =Rs 50
Assuming that no rabbit dies,
the number of pairs of rabbits at the start of the 1st month = 1
the number of pairs of rabbits at the start of the 2nd month = 1
the number of pairs of rabbits at the start of the 3rd month = 2
the number of pairs of rabbits at the start of the 4th month = 3
the number of pairs of rabbits at the start of the 5th month = 5
Therefore, the series is
1, 1, 2, 3, 5, 8,…
Difference between 2nd and 1st term = 1 – 1 = 0
Difference between 3rd and 2nd term = 2 – 1 = 1
Since, the difference is not same.
Therefore, the number of pair of rabbits in successive months are 1,1,2,3,5,8,… and they don’t form an AP.
Let the initial investment be I,
After one year it increases by $\frac{5}{4}$ I,
So the investment becomes, I + $\frac{5}{4}$ I
=$\frac{9}{4}$ I
At the end of 2nd year it again increase to $\frac{5}{4}$ I,
So the investment becomes, $\left(1+\frac{5}{4} I+\frac{5}{4} I\right)$
$=\left(\frac{9}{4} I+\frac{5}{4} I\right)$
$=\frac{14}{4} I$
At the end of the 3rd year it again increases to $\frac{5}{4} I$
So the investment becomes
Total amount outstanding after 3rd year = Rs 11664 + 933.12
= Rs 12597.12
Therefore, the series is
10800, 11664, 12597.12,…
Difference between second and first term = 11664 – 10800
= 864
Difference between third and second term = 12597.12 – 11664
= 933.12
Since the difference is not same
Therefore, it doesn’t form an AP.
Question 7 E
In which of the following situations does the list of numbers involved arithmetic progression, and why?
The money saved by Sudha in successive years when she saves Rs. 100 the first year and increased the amount by Rs. 50 every year.
Sol :The money saved by Sudha in successive years when she saves Rs. 100 the first year and increased the amount by Rs. 50 every year.
The money saved by Sudha in the first year = Rs. 100
According to the question,
Sudha increased the amount by Rs. 50 every year
⇒ The money saved by Sudha in a 2nd year = Rs. 100 +50
= Rs. 150
The money saved by Sudha in a 3rd year = Rs. 150 + 50
= Rs. 200
Therefore, the series is
100, 150, 200, 250,…
Difference in the 2nd term and 1st term = 150 – 100 = 50
Difference in the 3rd term and 2nd term = 200 – 150 = 50
Since the difference is the same.
Therefore, the money saved by Sudha in successive years form an AP with a = Rs 100 and d =Rs 50
Question 7 F
In which of the following situations does the list of numbers involved arithmetic progression, and why?
Number of pairs of rabbits in successive months when the pair of rabbits is too young to produce in their first month. In the second month and every subsequent month, they produce a new pair. Each new pair of rabbits pr a new pair in their second months and every subsequent month (see Fig.) (assume that no rabbit dies).
Sol :Number of pairs of rabbits in successive months when the pair of rabbits is too young to produce in their first month. In the second month and every subsequent month, they produce a new pair. Each new pair of rabbits pr a new pair in their second months and every subsequent month (see Fig.) (assume that no rabbit dies).
Assuming that no rabbit dies,
the number of pairs of rabbits at the start of the 1st month = 1
the number of pairs of rabbits at the start of the 2nd month = 1
the number of pairs of rabbits at the start of the 3rd month = 2
the number of pairs of rabbits at the start of the 4th month = 3
the number of pairs of rabbits at the start of the 5th month = 5
Therefore, the series is
1, 1, 2, 3, 5, 8,…
Difference between 2nd and 1st term = 1 – 1 = 0
Difference between 3rd and 2nd term = 2 – 1 = 1
Since, the difference is not same.
Therefore, the number of pair of rabbits in successive months are 1,1,2,3,5,8,… and they don’t form an AP.
Question 7 G
In which of the following situations does the list of numbers involved arithmetic progression, and why?
The values of an investment after 1, 2, 3, 4, ... years if after each subsequent year it increases by 5/4 times the initial investment.
Sol :The values of an investment after 1, 2, 3, 4, ... years if after each subsequent year it increases by 5/4 times the initial investment.
Let the initial investment be I,
After one year it increases by $\frac{5}{4}$ I,
So the investment becomes, I + $\frac{5}{4}$ I
=$\frac{9}{4}$ I
At the end of 2nd year it again increase to $\frac{5}{4}$ I,
So the investment becomes, $\left(1+\frac{5}{4} I+\frac{5}{4} I\right)$
$=\left(\frac{9}{4} I+\frac{5}{4} I\right)$
$=\frac{14}{4} I$
At the end of the 3rd year it again increases to $\frac{5}{4} I$
So the investment becomes
$\left(1+\frac{5}{4} I+\frac{5}{4} I+\frac{5}{4} I\right)$
$=\frac{9}{4} I+\frac{5}{4} I+\frac{5}{4} I$
$=\frac{14}{4} \mathrm{I}+\frac{5}{4} \mathrm{I}$
$=\frac{19}{4} \mathrm{I}$
Therefore the series is:
$I, \frac{9}{4} I, \frac{14}{4} I, \frac{19}{4} I\dots$
Now difference between 2nd and 1st term is
$=\frac{9}{4} I+\frac{5}{4} I+\frac{5}{4} I$
$=\frac{14}{4} \mathrm{I}+\frac{5}{4} \mathrm{I}$
$=\frac{19}{4} \mathrm{I}$
Therefore the series is:
$I, \frac{9}{4} I, \frac{14}{4} I, \frac{19}{4} I\dots$
Now difference between 2nd and 1st term is
$\frac{9}{4} I-I=\frac{5}{4} I$
difference between 3rd and 2nd term is
$\frac{14}{4} \mathrm{I}-\frac{9}{4} \mathrm{I}=\frac{5}{4} \mathrm{I}$
Since the difference is same,
Hence the obtained series is an A.P.
Since the difference is same,
Hence the obtained series is an A.P.
| 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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