# Class 10 SELINA Solutions Maths Chapter 10 - Arithmetic Progression

## Arithmetic Progression Exercise TEST YOURSELF

### Solution 1(a)

Correct option: (iv) All are A.P.

A + B = (2 + 3), (4 + 6), (6 + 9), (8 + 12), …… = 5, 10, 15, 20, ….. is an A.P.

A – C = (2 – 0), (4 – 4), (6 – 8), (8 – 12), …. = 2, 0, –2, –4, …. is an A.P.

C – B = (0 – 3), (4 – 6), (8 – 9), (12 – 12), ….. = –3, –2, –1, 0, …. is an A.P.

B – A = (3 – 2), (6 – 4), (9 – 6), (12 – 8), ….. = 1, 2, 3, 4, …. is an A.P.

### Solution 1(b)

Correct option: (iv) 5

2k – 7, k + 5 and 3k + 2 are in A.P.

(k + 5) – (2k – 7) = (3k + 2) – (k + 5)

12 – k = 2k – 3

3k = 15

k = 5

### Solution 1(c)

Correct option: (ii) 5

l = a + (n – 1)d

36 = –36 + (n – 1)18

72 = (n – 1)18

n – 1 = 4

n = 5

### Solution 1(d)

Correct option: (iii) Yes, 97

1^{2}, 5^{2}, 7^{2}, 73 …. = 1, 25, 49, 73 ….

This forms an A.P. with first term 1 and common difference 24.

Therefore, next term = 73 + 24 = 97

### Solution 1(e)

Correct option: (ii) 110

First 10 even natural numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

First term = 2

Common difference = 2

Options (i) and (iv) are same in textbook.

### Solution 1(f)

Correct option: (i)

The statement given in assertion is correct.

Hence, assertion A is true.

The difference between the consecutive terms is same, i.e. .

Hence, reason (R) is true.

### Solution 2

Let 'a' be the first term and 'd' be the common difference of the given A.P.

Now, t_{6} = 16 (given)

⇒ a + 5d = 16 ….(i)

And,

t_{14} = 32 (given)

⇒ a + 13d = 32 ….(ii)

Subtracting (i) from (ii), we get

8d = 16

⇒ d = 2

⇒ a + 5(2) = 16

⇒ a = 6

Hence, 36^{th} term = t_{36} = a + 35d = 6 + 35(2) = 76

### Solution 3

### Solution 4

For a given A.P.,

Number of terms, n = 50

3^{rd} term, t_{3} = 12

⇒ a + 2d = 12 ….(i)

Last term, l = 106

⇒ t_{50} = 106

⇒ a + 49d = 106 ….(ii)

Subtracting (i) from (ii), we get

47d = 94

⇒ d = 2

⇒ a + 2(2) = 12

⇒ a = 8

Hence, t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64

### Solution 5

### Solution 6

Here,

First term, a = 3

Last term, l = 57

n = 20

### Solution 7

Here, we find that

15 - 18 = 12 - 15 = -3

Thus, the given series is an A.P. with first term 18 and common difference -3.

Let the number of term to be added be 'n'.

⇒ 90 = n[36 - 3n + 3]

⇒ 90 = n[39 - 3n]

⇒ 90 = 3n[13 - n]

⇒ 30 = 13n - n^{2}

⇒ n^{2} - 13n + 30 = 0

⇒ n^{2} - 10n - 3n + 30 = 0

⇒ n(n - 10) - 3(n - 10) = 0

⇒ (n - 10)(n - 3) = 0

⇒ n - 10 = 0 or n - 3 = 0

⇒ n = 10 or n = 3

Thus, required number of term to be added is 3 or 10.

### Solution 8

t_{n} = 8 - 5n

Replacing n by (n + 1), we get

t_{n+1} = 8 - 5(n + 1) = 8 - 5n - 5 = 3 - 5n

Now,

t_{n+1} - t_{n} = (3 - 5n) - (8 - 5n) = -5

Since, (t_{n+1} - t_{n}) is independent of n and is therefore a constant.

Hence, the given sequence is an A.P.

### Solution 9

The given sequence is 1, -1, -3, …..

Now,

1 - 3 = -1 - 1 = -3 - (-1) = -2

Hence, the given sequence is an A.P. with first term a = 3 and common difference d = -2.

The general term (n^{th} term) of an A.P. is given by

t_{n} = a + (n - 1)d

= 3 + (n - 1)(-2)

= 3 - 2n + 2

= 5 - 2n

Hence, 23^{rd} term = t_{23} = 5 - 2(23) = 5 - 46 = -41

### Solution 10

The given sequence is 11, 8, 5, 2, …..

Now,

8 - 11 = 5 - 8 = 2 - 5 = -3

Hence, the given sequence is an A.P. with first term a = 11 and common difference d = -3.

The general term of an A.P. is given by

t_{n} = a + (n - 1)d

⇒ -150 = 11 + (n - 1)(-5)

⇒ -161 = -5n + 5

⇒ 5n = 166

The number of terms cannot be a fraction.

So, clearly, -150 is not a term of the given sequence.

### Solution 11

### Solution 12

### Solution 13

Let 'a' be the first term and 'd' be the common difference of the given A.P.

Given,

S_{14} = 1050

⇒ 7[2a + 13d] = 1050

⇒ 2a + 13d = 150

⇒ a + 6.5d = 75 ….(i)

And, t_{14} = 140

⇒ a + 13d = 140 ….(ii)

Subtracting (i) from (ii), we get

6.5d = 65

⇒ d = 10

⇒ a + 13(10) = 140

⇒ a = 10

Thus, 20^{th} term = t_{20} = 10 + 19d = 10 + 19(10) = 200

### Solution 14

n^{th} term of an A.P. is given by t_{n}= a + (n - 1) d.

⇒ t_{25 } = a + (25 - 1)d = a + 24d and

t_{9} = a + (9 - 1)d = a + 8d

According to the condition in the question, we get

t_{25 }= t_{9} + 16

⇒ a + 24d = a + 8d + 16

⇒ 16d = 16

⇒ d = 1

### Solution 15

Let a and d be the first term and common difference respectively.

⇒(m + n)^{th} term = a + (m + n - 1)d …. (i) and

(m - n)^{th} term = a + (m - n - 1)d …. (ii)

From (i) + (ii), we get

(m + n)^{th} term + (m - n)^{th} term

= a + (m + n - 1)d + a + (m - n - 1)d

= a + md + nd - d + a + md - nd - d

= 2a + 2md - 2d

= 2a + (m - 1)2d

= 2[ a + (m - 1)d]

= 2 × m^{th}term

Hence proved.

### Solution 16

In the first A.P. 58, 60, 62,....

a = 58 and d = 2

t_{n} = a + (n - 1)d

⇒ t_{n} = 58 + (n - 1)2 …. (i)

In the first A.P. -2, 5, 12, ….

a = -2 and d = 7

t_{n} = a + (n - 1)d

⇒ t_{n}= -2 + (n - 1)7 …. (ii)

Given that the n^{th} term of first A.P is equal to the n^{th} term of the second A.P.

⇒58 + (n - 1)2 = -2 + (n - 1)7 … from (i) and (ii)

⇒58 + 2n - 2 = -2 + 7n - 7

⇒ 65 = 5n

⇒ n = 15

### Solution 17

Here a = 105 and d = 101 - 105 = -4

Let a_{n} be the first negative term.

⇒ a_{n} < 0

⇒ a + (n - 1)d < 0

⇒ 105 + (n - 1)(-4)<0

⇒ 105 - 4n + 4 <0

⇒ 109 - 4n < 0

⇒ 109 <4n

⇒ 27.25 < n

The value of n = 28.

Therefore 28^{th} term is the first negative term of the given A.P.

### Solution 18

Let the three parts of 216 in A.P be (a - d), a, (a + d).

⇒a - d + a + a + d = 216

⇒ 3a = 216

⇒ a = 72

Given that the product of the two smaller parts is 5040.

⇒a(a - d ) = 5040

⇒ 72(72 - d) = 5040

⇒ 72 - d = 70

⇒ d = 2

∴ a - d = 72 - 2 = 70, a = 72 and a + d = 72 + 2 = 74

Therefore the three parts of 216 are 70, 72 and 74.

### Solution 19

We have 2n^{2} - 7,

Substitute n = 1, 2, 3, … , we get

2(1)^{2} - 7, 2(2)^{2} - 7, 2(3)^{2} - 7, 2(4)^{2} - 7, ….

-5, 1, 11, ….

Difference between the first and second term = 1 - (-5) = 6

And Difference between the second and third term = 11 - 1 = 10

Here, the common difference is not same.

Therefore the n^{th} term of an A.P can't be 2n^{2} - 7.

### Solution 20

Here a = 14 , d = 7 and t_{n} = 168

t_{n} = a + (n - 1)d

⇒ 168 = 14 + (n - 1)7

⇒ 154 = 7n - 7

⇒ 154 = 7n - 7

⇒ 161 = 7n

⇒ n = 23

We know that,

Therefore the sum of the A.P., 14, 21, 28, …, 168 is 2093.

### Solution 21

Here a = 20 and S_{7} = 2100

We know that,

To find: t_{31} =?

t_{n} = a + (n - 1)d

Therefore the 31^{st} term of the given A.P. is 2820.** **

### Solution 22

First we will reverse the given A.P. as we have to find the sum of last 8 terms of the A.P.

58, …., -8, -10, -12.

Here a = 58 , d = -2

Therefore the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58 is 408.

## Arithmetic Progression Exercise Ex. 10(A)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

The given A.P. is 1, 4, 7, 10, ……….

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

For a given A.P.,

Number of terms, n = 60

First term, a = 7

Last term, l = 125

⇒ t_{60}
= 125

⇒ a + 59d = 125

⇒ 7 + 59d = 125

⇒ 59d = 118

⇒ d = 2

Hence,
t_{31} = a + 30d = 7 + 30(2) = 7 + 60 = 67

### Solution 19

Let 'a' be the first term and 'd' be the common difference of the given A.P.

t_{4}
+ t_{8} = 24 (given)

⇒ (a + 3d) + (a + 7d) = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 ….(i)

And,

t_{6}
+ t_{10} = 34 (given)

⇒ (a + 5d) + (a + 9d) = 34

⇒ 2a + 14d = 34

⇒ a + 7d = 17 ….(ii)

Subtracting (i) from (ii), we get

2d = 5

### Solution 20

Let 'a' be the first term and 'd' be the common difference of the given A.P.

Now,
t_{3} = 5 (given)

⇒ a + 2d = 5 ….(i)

And,

t_{7}
= 9 (given)

⇒ a + 6d = 9 ….(ii)

Subtracting (i) from (ii), we get

4d = 4

⇒ d = 1

⇒ a + 2(1) = 5

⇒ a = 3

Hence,
17^{th} term = t_{17} = a + 16d = 3 + 16(1) = 19

### Solution 1(a)

Correct option: (ii) 8, 3, –2, –7, …..

First term = a = 8

Common difference = d = –5

Then, the A.P. is 8, 8 + (–5), 8 + 2(–5), 8 + 3(–5), ……

That is, 8, 3, –2, –7, …..

### Solution 1(b)

Correct option: (ii) yes

The given sequence –8, –8, –8, –8, ….. is an A.P. with the first term –8 and common difference zero.

### Solution 1(c)

Correct option: (iv) –39

Given A.P. is 3, 0, –3, –6, …..

First term, a = 3

Common difference, d = 0 – 3 = –3

15^{th} term, t_{15} = a + 14d

= 3 + 14(–3)

= 3 – 42

= –39

### Solution 1(d)

Correct option: (ii) 2

t_{24} – t_{19} = 10

(a + 23d) – (a + 18d) = 10

5d = 10

d = 2

### Solution 1(e)

Correct option: (ii) 16

Given A.P. is 8, 13, 18, …..

First term, a = 8

Common difference, d = 13 – 8 = 5

n^{th} term, t_{n} = 83

a + (n – 1)d = 83

8 + (n – 1)5 = 83

(n – 1)5 = 75

n – 1 = 15

n = 16

## Arithmetic Progression Exercise Ex. 10(B)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 1(a)

Correct option: (iii) 8

Common difference of two A.P.,s = d

Let the first and nth terms of the first A.P. be a and t_{n,} respectively.

Let the first and nth terms of the second A.P. be A and T_{n,} respectively.

Then, t_{25} – T_{25} = 8

a + 24d – A – 24d = 8

a – A = 8

Now, t_{50} – T_{50} = a + 49d – A – 49d = a – A = 8

### Solution 1(b)

Correct option: (i) 0

Let the first term and the common difference of an A.P. be a and d, respectively.

Then, 10 × t_{10} = 20 × (t_{20})

10 × (a + 9d) = 20 × (a + 19d)

a + 9d = 2(a + 19d)

a + 9d = 2a + 38d

a + 29d = 0

t_{30} = 0

### Solution 1(c)

Correct option: (iv) 7

t_{n} = 7n – 5

Then, t_{1} = 7(1) – 5 = 2 and t_{2} = 7(2) – 5 = 9

Therefore, common difference = t_{2} – t_{1} = 9 – 2 = 7

### Solution 1(d)

Correct option: (iv) 3

t_{40} – t_{16} = 72

(a + 39d) – (a + 15d) = 72

24d = 72

d = 3

### Solution 1(e)

Correct option: (i) 17

First term = a = 6

Common difference = d = 11 – 6 = 5

t_{n} = 106

a + (n – 1)d = 106

6 + (n – 1)(5) = 106

(n – 1)(5) = 100

n – 1 = 20

n – 1 – 3 = 17

n – 4 = 17

## Arithmetic Progression Exercise Ex. 10(C)

### Solution 2

### Solution 3

### Solution 4(i)

### Solution 4(ii)

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 1(a)

Correct option: (ii) 1640

Total terms = 41

Then, middle term = 21^{st} term = 40

That is, a + 20d = 40

### Solution 1(b)

Correct option: (iv) 4905

Two digit numbers are 10, 11, 12, ……99

This forms an A.P. with first term, a = 10, last term, l = 99 and common difference, d = 1

Hence, n = 100 – 10 = 90

### Solution 1(c)

Correct option: (i) 650

First term, a = 4

Common difference, d = 7 – 4 = 3

n = 20

### Solution 1(d)

Correct option: (ii) 2620

First term, a = 7

Common difference, d = 10 – 7 = 3

n = 40

### Solution 1(e)

Correct option: (iii) 26

t_{n} = 6n + 4

Then, t_{1} = 6(1) + 4 = 10 and t_{2} = 6(2) + 4 = 16

Sum of first two terms = t_{1} + t_{2} = 10 + 16 = 26

## Arithmetic Progression Exercise Ex. 10(D)

### Solution 2

### Solution 3

### Solution 4

### Solution 5

### Solution 6

We know that,

Sum of n terms of an A.P =

Let the first term be 2x and the last term be 3x.

∴ Sum of 5 terms of an A.P =

First term = 2x =2 × 1 = 2 and the last term = 3x = 3 × 1 = 3

n^{th} term of an A.P. is given by

t_{n} = a + (n - 1)d

⇒ a_{5} = 2 + (5 - 1)d

⇒ 3 = 2 + 4d

⇒ 1 = 4d

⇒ d = = 0.25

Therefore the five numbers in an A.P are 2, 2.25, 2.50, 2.75 and 3.

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

### Solution 12

### Solution 1(a)

t_{1} = k + 2, t_{2} = 2k + 7 and t_{3} = 4k + 12

t_{2} – t_{1} = t_{3} – t_{2}

(2k + 7) – (k + 2) = (4k + 12) – (2k + 7)

k + 5 = 2k + 5

k = 0

Therefore, first term = 0 + 2 = 2

### Solution 1(b)

Correct option: (iii) 9

S_{n} = 3n^{2}

S_{1} = 3(1)^{2} = 3

S_{2} = 3(2)^{2} = 12

Second term = S_{2} – S_{1}^{}= 12 – 3 = 9

### Solution 1(c)

Correct option: (ii) 5 × 7, 7 × 7 and 9 × 7

5, 7 and 9 are in A.P.

If each term of an A.P. is multiplied by a non-zero fixed number, the resulting sequence is an A.P.