KC Sinha Mathematics Solution Class 10 Chapter 8 Arithmetic Progressions Exercise 8.1

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:
t_n = 3n + 1

Sol :
Given: t_n = 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:

t_n = 2ⁿ

Sol :
Given: t_n = 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:
t_n = n² + 1

Sol :
Given: t_n = 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:
t_n = n(n + 2)

Sol :
Given: t_n = 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:
t_n = 2n+5

Sol :
Given: t_n = 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 :
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
$\mathrm{t}_{2}=\frac{2-3}{4}=\frac{-1}{4}$

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

$\therefore \mathrm{t}_{2}=\frac{(2)^{2}(2+1)}{3}$

$=\frac{(4)(3)}{3}=4$

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 :
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:
t_n= (n - 1)(2 - n)(3 + n) ; t₂₀

Sol :
Given: t_n = (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 :
Given: t₂ = 2 and t_n = t_n-1 + 1
Now, we have to find next three terms i.e. t₃, t₄ and t₅
Taking n = 3, we get
t₃ = t_3-1 + 1
= t₂ + 1
= 2 + 1 [given: t₂ = 2]
t₃ = 3 …(i)
Taking n = 4, we get
t₄ = t_4-1 + 1
= t₃ + 1
= 3 + 1 [from (i)]
t₄ = 4 …(ii)
Taking n = 5, we get
t₅ = t_5-1 + 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 :
Given: t₁ = 3 and t_n = 3t_n-1 + 2
Now, we have to find next three terms i.e. t₂, t₃ and t₄
Taking n = 2, we get
t₂ = 3t_2-1 + 2
= 3t₁ + 2
= 3(3) + 2 [given: t₁ = 3]
t₂ = 9 + 2
t₂ = 11 …(i)

Taking n = 3, we get
t₃ = 3t_3-1 + 2
= 3t₂ + 2
= 3(11) + 2 [from (i)]
= 33 + 2
t₃ = 35 …(ii)
Taking n = 4, we get
t₄ = 3t_4-1 + 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 :
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 :
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 :
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 :
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 :
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 :
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 :
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 :
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 :
In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th 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 :

In general, for an AP a₁, a₂, . . . .,a_n, we have
d = a_k+1 - a_k
where a_k+1 and a_k are the (k+1)^th and k^th 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 :
In general, for an AP a₁, a₂, . . . .,a_n, we have
d = a_k+1 - a_k
where a_k+1 and a_k are the (k+1)^th and k^th 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 :
In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th 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 :
In general, for an AP a₁, a₂, . . . .,a_n, we have
d = a_k+1 - a_k
where a_k+1 and a_k are the (k+1)^th and k^th 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 :
In general, for an AP a₁, a₂, . . . .,a_n, we have
d = a_k+1 - a_k
where a_k+1 and a_k are the (k+1)^th and k^th 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 $\frac{1}{3}$.

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 :
In general, for an AP a₁, a₂, . . . .,a_n, we have
d = a_k+1 - a_k
where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers: $\frac{3}{2}, \frac{1}{2}, \frac{-1}{2}, \frac{-3}{2}, \ldots$
$a_{2}-a_{1}=\frac{1}{2}-\frac{3}{2}=\frac{-2}{2}=-1$
$a_{3}-a_{2}=\frac{-1}{2}-\frac{1}{2}=\frac{-2}{2}=-1$
$a_{4}-a_{3}=-\frac{3}{2}-\left(\frac{-1}{2}\right)=\frac{-3+1}{2}=\frac{-2}{2}=-1$

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 :
We have,
a₂ – a₁ = -1 – 1 = -2
a₃ – a₂ =-3 – (-1) = -3 + 1 = -2
a₄ – a₃ =-5 – (-3) = -5 + 3= -2
i.e. a_k+1 – a_k 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 :
We have,

a₂ – a₁ = 4 – 2 = 2

a₃ – a₂ = 8 – 4 = 4

a₄ – a₃ = 16 – 8 = 8

i.e. a_k+1 – a_k 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 :
We have,
a₂ – a₁ = 2 – (-2) = 2 + 2 = 4
a₃ – a₂ = -2 – 2 = -4
a₄ – a₃ = 2 – (-2) = 2 + 2 = 4
i.e. a_k+1 – a_k 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 :
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. a_k+1 – a_k 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 :
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. a_k+1 – a_k 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

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 :
We have,

a₂ – a₁ = -4 – 0 = -4

a₃ – a₂ =-8 – (-4) = -8 + 4 = -4

a₄ – a₃ =-12 – (-8) = -12 + 8= -4

i.e. a_k+1 – a_k 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 :
We have,

a₂ – a₁ = 10 – 4 = 6

a₃ – a₂ = 16 – 10 = 6

a₄ – a₃ = 22 – 16 = 6

i.e. a_k+1 – a_k 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 :
We have,

a₂ – a₁ = 2a – a = a

a₃ – a₂ = 3a – 2a = a

a₄ – a₃ = 4a – 3a = a

i.e. a_k+1 – a_k 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 :
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. a_k+1 – a_k 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 :
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. a_k+1 – a_k 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 :
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. a_k+1 – a_k 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 :
We have,
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
a₄ – a₃ = 27 – 9 = 18
i.e. a_k+1 – a_k 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 :
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. a_k+1 – a_k 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 :
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. a_k+1 – a_k 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 :
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. a_k+1 – a_k 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 :
Salary for the 1^st year = Rs. 8000
and according to the question,
There is an annual increment of Rs. 500
⇒ The salary for the 2^nd year = Rs. 8000 + 500 = Rs.8500
Now, again there is an increment of Rs. 500
⇒ The salary for the 3^rd year = Rs. 8500 +500 = Rs. 9000
Therefore, the series is
8000 , 8500 , 9000 , …
Difference between 2^nd term and 1^st term = 8500 – 8000 = 500
Difference between 3^rd term and 2^nd 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 :
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 2^nd and 1^st term = 23 – 15 = 8
Difference between 3^rd and 2^nd 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 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 1^st, 2^nd, 3^rd, … from the bottom to top respectively are 45, 43, 41, …
Difference between 2^nd and 1^st term = 43 – 45 = -2
Difference between 3^rd and 2^nd 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 :
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 2^nd year

$=10800 \times \frac{8}{100}=\mathrm{Rs} .864$

Total amount outstanding after 2^nd year = Rs10800 + 864
= Rs 11664

Interest earned in 3^rd year

$=11664 \times \frac{8}{100}$

= Rs 933.12

Total amount outstanding after 3^rd 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 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 2^nd year = Rs. 100 +50

= Rs. 150

The money saved by Sudha in a 3^rd year = Rs. 150 + 50

= Rs. 200

Therefore, the series is

100, 150, 200, 250,…

Difference in the 2^nd term and 1^st term = 150 – 100 = 50

Difference in the 3^rd term and 2^nd 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 :
Assuming that no rabbit dies,
the number of pairs of rabbits at the start of the 1^st month = 1
the number of pairs of rabbits at the start of the 2^nd month = 1
the number of pairs of rabbits at the start of the 3^rd month = 2
the number of pairs of rabbits at the start of the 4^th month = 3
the number of pairs of rabbits at the start of the 5^th month = 5

Therefore, the series is
1, 1, 2, 3, 5, 8,…
Difference between 2^nd and 1^st term = 1 – 1 = 0
Difference between 3^rd and 2^nd 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 :
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 2^nd 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 2^nd and 1^st term is

$\frac{9}{4} I-I=\frac{5}{4} I$

difference between 3^rd and 2^nd 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.

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