Class 10 SELINA Solutions Maths Chapter 10 - Arithmetic Progression

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.

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

Correct option: (ii) 5

l = a + (n – 1)d

36 = –36 + (n – 1)18

72 = (n – 1)18

n – 1 = 4

n = 5

Correct option: (iii) Yes, 97

1², 5², 7², 73 …. = 1, 25, 49, 73 ….

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

Therefore, next term = 73 + 24 = 97

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.

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.

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

Now, t₆ = 16 (given)

⇒ a + 5d = 16 ….(i)

And,

t₁₄ = 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 = t₃₆ = a + 35d = 6 + 35(2) = 76

For a given A.P.,

Number of terms, n = 50

3^rd term, t₃ = 12

⇒ a + 2d = 12 ….(i)

Last term, l = 106

⇒ t₅₀ = 106

⇒ a + 49d = 106 ….(ii)

Subtracting (i) from (ii), we get

47d = 94

⇒ d = 2

⇒ a + 2(2) = 12

⇒ a = 8

Hence, t₂₉ = a + 28d = 8 + 28(2) = 8 + 56 = 64

Here,

First term, a = 3

Last term, l = 57

n = 20

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 - n²

⇒ n² - 13n + 30 = 0

⇒ n² - 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.

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.

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 = t₂₃ = 5 - 2(23) = 5 - 46 = -41

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. 

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

Given,

S₁₄ = 1050

⇒ 7[2a + 13d] = 1050

⇒ 2a + 13d = 150 

⇒ a + 6.5d = 75 ….(i)

And, t₁₄ = 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 = t₂₀ = 10 + 19d = 10 + 19(10) = 200

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

⇒ t₂₅  = a + (25 - 1)d = a + 24d and

 t₉ = a + (9 - 1)d = a + 8d

According to the condition in the question, we get

t₂₅ = t₉ + 16

⇒ a + 24d = a + 8d + 16

⇒ 16d = 16

⇒ d = 1

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^thterm

Hence proved.

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

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.

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.

We have 2n² - 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 2n² - 7.

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. 

Here a = 20 and S₇ = 2100

We know that,

To find: t₃₁ =?

t_n = a + (n - 1)d

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

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.

The given A.P. is 1, 4, 7, 10, ………. 

For a given A.P.,

Number of terms, n = 60

First term, a = 7

Last term, l = 125

⇒ t₆₀ = 125

⇒ a + 59d = 125

⇒ 7 + 59d = 125

⇒ 59d = 118

⇒ d = 2

Hence, t₃₁ = a + 30d = 7 + 30(2) = 7 + 60 = 67

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

t₄ + t₈ = 24 (given)

⇒ (a + 3d) + (a + 7d) = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 ….(i)

And,

t₆ + t₁₀ = 34 (given)

⇒ (a + 5d) + (a + 9d) = 34

⇒ 2a + 14d = 34

⇒ a + 7d = 17 ….(ii)

Subtracting (i) from (ii), we get

2d = 5

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

Now, t₃ = 5 (given)

⇒ a + 2d = 5 ….(i)

And,

t₇ = 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 = t₁₇ = a + 16d = 3 + 16(1) = 19

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, …..

Correct option: (ii) yes

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

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, t₁₅ = a + 14d

= 3 + 14(–3)

= 3 – 42

= –39

Correct option: (ii) 2

t₂₄ – t₁₉ = 10

(a + 23d) – (a + 18d) = 10

5d = 10

d = 2

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

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, t₂₅ – T₂₅ = 8

a + 24d – A – 24d = 8

a – A = 8

Now, t₅₀ – T₅₀ = a + 49d – A – 49d = a – A = 8

Correct option: (i) 0

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

Then, 10 × t₁₀ = 20 × (t₂₀)

10 × (a + 9d) = 20 × (a + 19d)

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

a + 9d = 2a + 38d

a + 29d = 0

t₃₀ = 0

Correct option: (iv) 7

t_n = 7n – 5

Then, t₁ = 7(1) – 5 = 2 and t₂ = 7(2) – 5 = 9

Therefore, common difference = t₂ – t₁ = 9 – 2 = 7

Correct option: (iv) 3

t₄₀ – t₁₆ = 72

(a + 39d) – (a + 15d) = 72

24d = 72

d = 3

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

Correct option: (ii) 1640

Total terms = 41

Then, middle term = 21^st term = 40

That is, a + 20d = 40

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

Correct option: (i) 650

First term, a = 4

Common difference, d = 7 – 4 = 3

n = 20

Correct option: (ii) 2620

First term, a = 7

Common difference, d = 10 – 7 = 3

n = 40

Correct option: (iii) 26

t_n = 6n + 4

Then, t₁ = 6(1) + 4 = 10 and t₂ = 6(2) + 4 = 16

Sum of first two terms = t₁ + t₂ = 10 + 16 = 26

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

⇒ a₅ = 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.

Correct option: (iii[RK1] ) 2

t₁ = k + 2, t₂ = 2k + 7 and t₃ = 4k + 12

t₂ – t₁ = t₃ – t₂

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

k + 5 = 2k + 5

k = 0

Therefore, first term = 0 + 2 = 2

Correct option: (iii) 9

S_n = 3n²

S₁ = 3(1)² = 3

S₂ = 3(2)² = 12

Second term = S₂ – S₁= 12 – 3 = 9

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.