Arithmetic Progression — Exercise 10(B)

Two A.P.'s have same common difference. If the difference between their 25^th terms is 8, the difference between their 50^th terms is :

1. 16

2. 5

3. 8

4. 25

Answer

Let two A.P.'s have first term a₁ and a₂.

Their common difference be d.

By formula,

⇒ a_n = a + (n - 1)d

25th term of first A.P. = a₁ + (25 - 1)d = a₁ + 24d.

25th term of second A.P. = a₂ + (25 - 1)d = a₂ + 24d.

Given, Difference between 25th terms is 8.

∴ (a₁ + 24d) - (a₂ + 24d) = 8

a₁ - a₂ + 24d - 24d = 8

a₁ - a₂ = 8.

50th term of first A.P. = a₁ + (50 - 1)d = a₁ + 49d.

50th term of second A.P. = a₂ + (50 - 1)d = a₂ + 49d.

Difference = (a₁ + 49d) - (a₂ + 49d) = a₁ - a₂ = 8.

Hence, Option 3 is the correct option.

Ten times the 10^th term of an A.P. is equal to twenty times the 20^th term of the same A.P. The 30^th term of this A.P. is :

1. 0

2. 40

3. 20

4. 2 × (30 + 10)

Answer

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

By formula,

a_n = a + (n - 1)d

According to question,

⇒ 10a₁₀ = 20a₂₀

⇒ 10[a + (10 - 1)d] = 20[a + (20 - 1)d]

⇒ 10[a + 9d] = 20[a + 19d]

⇒ 10a + 90d = 20a + 380d

⇒ 20a - 10a = 90d - 380d

⇒ 10a = -290d

⇒ a = -29d.

a₃₀ = a + (30 - 1)d

= a + 29d

= -29d + 29d

= 0.

Hence, Option 1 is the correct option.

The n^th term of an A.P. is 7n - 5. Its common difference is :

1. 2

2. 9

3. 16

4. 7

Answer

Given,

n^th term = 7n - 5

(n - 1)^th term = 7(n - 1) - 5 = 7n - 7 - 5

= 7n - 12.

By formula,

d = n^th term - (n - 1)^th term

= (7n - 5) - (7n - 12)

= 7n - 7n - 5 + 12

= 7.

Hence, Option 4 is the correct option.

The 40^th term of an A.P. exceeds its 16^th term by 72. Then its common difference is :

1. 40

2. 40 - 16

3. 72

4. 3

Answer

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

By formula,

a_n = a + (n - 1)d

Given,

40^th term of an A.P. exceeds its 16^th term by 72.

∴ a₄₀ - a₁₆ = 72

⇒ [a + (40 - 1)d] - [a + (16 - 1)d] = 72

⇒ a + 39d - [a + 15d] = 72

⇒ a - a + 39d - 15d = 72

⇒ 24d = 72

⇒ d = $\dfrac{72}{24}$ = 3.

Hence, Option 4 is the correct option.

The n^th term of the A.P. 6, 11, 16, 21, ....... is 106, the the value of n - 4 is :

1. 17

2. 15

3. 16

4. 20

Answer

In A.P.,

6, 11, 16, 21, ........

First term (a) = 6,

Common difference (d) = 11 - 6 = 5.

nth term (a_n) = 106

By formula,

⇒ a_n = a + (n - 1)d

⇒ 106 = 6 + (n - 1)5

⇒ 106 = 6 + 5n - 5

⇒ 106 = 5n + 1

⇒ 106 - 1 = 5n

⇒ 5n = 105

⇒ n = $\dfrac{105}{5}$ = 21.

⇒ n - 4 = 21 - 4 = 17.

Hence, Option 1 is the correct option.

In an A.P. ten times of its tenth term is equal to thirty times of its 30^th term. Find its 40^th term.

Answer

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

According to question,

⇒ 10a₁₀ = 30a₃₀

⇒ 10[a + (10 - 1)d] = 30[a + (30 - 1)d]

⇒ a + 9d = 3(a + 29d)

⇒ a + 9d = 3a + 87d

⇒ 3a - a = 9d - 87d

⇒ 2a = -78d

⇒ a = -39d.

∴ a₄₀ = a + (40 - 1)d

= -39d + 39d

= 0.

Hence, a₄₀ = 0.

How many two-digit numbers are divisible by 3 ?

Answer

Two digits no. divisible by 3 are,

12, 15, 18 ........., 99.

The above series is an A.P. with, a = 12 and d = 15 - 12 = 3 and last term = 99.

Let total no. of terms be n,

∴ a_n = a + (n - 1)d

⇒ 99 = 12 + (n - 1)3

⇒ 99 = 12 + 3n - 3

⇒ 99 = 9 + 3n

⇒ 99 - 9 = 3n

⇒ 90 = 3n

⇒ n = 30.

Hence, two-digit numbers divisible by 3 are 30.

Which term of A.P. 5, 15, 25, ....... will be 130 more than its 31^st term ?

Answer

In the A.P. 5, 15, 25, ......., a = 5 and d = 15 - 5 = 10.

Let n^th term be 130 more than 31^st term.

∴ a_n = 130 + a₃₁

⇒ a + (n - 1)d = 130 + a + (31 - 1)d

⇒ 5 + (n - 1)10 = 130 + 5 + 30(10)

⇒ 5 + 10n - 10 = 135 + 300

⇒ 10n - 5 = 435

⇒ 10n = 440

⇒ n = 44.

Hence, 44^th term of the A.P. will be 130 more than its 31^st term .

Find the value of p, if x, 2x + p and 3x + 6 are in A.P.

Answer

Since, x, 2x + p and 3x + 6 are in A.P.

Hence, difference between consecutive terms will be equal.

∴ 2x + p - x = 3x + 6 - (2x + p)

⇒ x + p = x + 6 - p

⇒ p + p = x + 6 - x

⇒ 2p = 6

⇒ p = 3.

Hence, p = 3.

If the 3^rd and the 9^th terms of an arithmetic progression are 4 and -8 respectively, which term of it is zero ?

Answer

We know that,

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

a_n = a + (n - 1)d

Given, 3^rd term is 4

∴ a₃ = a + (3 - 1)d

⇒ 4 = a + 2d

⇒ a + 2d = 4 ........(i)

Given, 9^th term is -8

∴ a₉ = a + (9 - 1)d

⇒ -8 = a + 8d

⇒ a + 8d = -8 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 8d - (a + 2d) = -8 - 4

⇒ 6d = -12

⇒ d = -2.

Substituting value of d in (i) we get,

⇒ a + 2(-2) = 4

⇒ a - 4 = 4

⇒ a = 8.

Let n^th term be zero.

∴ a_n = a + (n - 1)d = 0

⇒ 8 + (n - 1)(-2) = 0

⇒ 8 - 2n + 2 = 0

⇒ 2n = 10

⇒ n = 5.

Hence, 5^th term of the A.P. is zero.

How many three digit numbers are divisible by 87?

Answer

Three digit numbers divisible by 87 are,

174, 261, ........., 957.

The above series is an A.P. with a = 174 and d = 261 - 174 = 87 and last term = 957.

Let no. of terms in the A.P. be n.

∴ a_n = 957

⇒ a + (n - 1)d = 957

⇒ 174 + (n - 1)(87) = 957

⇒ 174 + 87n - 87 = 957

⇒ 87n + 87 = 957

⇒ 87n = 957 - 87

⇒ 87n = 870

⇒ n = 10.

Hence, there are 10 three digit numbers which are divisible by 87.

For what value of n, the n^th term of A.P. 63, 65, 67, ....... and n^th term of A.P. 3, 10, 17, ....., are equal to each other?

Answer

In the A.P. 63, 65, 67, ....... first term = a = 63 and common difference = d = 65 - 63 = 2.

In the A.P. 3, 10, 17, ....... first term = a₁ = 3 and common difference = d₁ = 10 - 3 = 7.

Given, n^th term of both A.P. are equal,

∴ a + (n - 1)d = a₁ + (n - 1)d₁

⇒ 63 + (n - 1)2 = 3 + (n - 1)7

⇒ 63 + 2n - 2 = 3 + 7n - 7

⇒ 61 + 2n = 7n - 4

⇒ 7n - 2n = 61 + 4

⇒ 5n = 65

⇒ n = 13.

Hence, n = 13.

Determine the A.P. whose 3^rd term is 16 and the 7^th term exceeds the 5^th term by 12.

Answer

Given,

⇒ a₃ = 16

⇒ a + (3 - 1)d = 16

⇒ a + 2d = 16 .........(i)

Given,

⇒ a₇ - a₅ = 12

⇒ [a + (7 - 1)d] - [a + (5 - 1)d] = 12

⇒ (a + 6d) - (a + 4d) = 12

⇒ a - a + 6d - 4d = 12

⇒ 2d = 12

⇒ d = 6.

Substituting value of d in (i) we get,

⇒ a + 2(6) = 16

⇒ a + 12 = 16

⇒ a = 4.

A.P. = a, (a + d), (a + 2d), ..........

= 4, (4 + 6), (4 + 2(6)), .........

= 4, 10, 16, .........

Hence, A.P. = 4, 10, 16, .........

If numbers n - 2, 4n - 1 and 5n + 2 are in A.P., find the value of n and its next two terms.

Answer

Since, n - 2, 4n - 1 and 5n + 2 are in A.P.

Hence, difference between consecutive terms are equal.

∴ 4n - 1 - (n - 2) = (5n + 2) - (4n - 1)

⇒ 4n - n - 1 + 2 = 5n - 4n + 2 - (-1)

⇒ 3n + 1 = n + 3

⇒ 3n - n = 3 - 1

⇒ 2n = 2

⇒ n = 1.

Substituting n in n - 2, 4n - 1 and 5n + 2 we get,

= 1 - 2, 4(1) - 1, 5(1) + 2 .........

= -1, 3, 7, .........

The above A.P. has first term = -1 and common term = 3 - (-1) = 4.

Next two terms = 7 + 4 = 11 and 7 + 2(4) = 15.

Hence, n = 1 and next two terms of the A.P. are 11 and 15.

Determine the value of k for which k² + 4k + 8, 2k² + 3k + 6 and 3k² + 4k + 4 are in A.P.

Answer

Since, k² + 4k + 8, 2k² + 3k + 6 and 3k² + 4k + 4 are in A.P.

Hence, difference between consecutive terms are equal.

∴ 2k² + 3k + 6 - (k² + 4k + 8) = 3k² + 4k + 4 - (2k² + 3k + 6)

⇒ 2k² - k² + 3k - 4k + 6 - 8 = 3k² - 2k² + 4k - 3k + 4 - 6

⇒ k² - k - 2 = k² + k - 2

⇒ k² - k² + k + k = -2 + 2

⇒ 2k = 0

⇒ k = 0.

Hence, k = 0.

State, true or false : if a, b and c are in A.P. then :

(i) 4a, 4b and 4c are in A.P.

(ii) a + 4, b + 4 and c + 4 are in A.P.

Answer

(i) Given,

a, b and c are in A.P.

We know that,

If each term of an A.P. is multiplied by a constant, then the resulting sequence is also an A.P.

Thus, 4a, 4b and 4c are also in A.P.

Hence, yes the terms 4a, 4b and 4c are in A.P.

(ii) To prove,

a + 4, b + 4 and c + 4 are in A.P., difference between consecutive terms should be same.

⇒ (b + 4) - (a + 4) = (c + 4) - (b + 4)

⇒ b - a + 4 - 4 = c - b + 4 - 4

⇒ b - a = c - b [As, a, b and c are in A.P., thus this equation is correct]

Hence, yes the terms a + 4, b + 4 and c + 4 are in A.P.

An A.P. consists of 57 terms of which 7^th term is 13 and the last term is 108. Find the 45^th term of this A.P.

Answer

Let A.P. has first term = a and common difference = d.

According to question,

⇒ a₅₇ = 108

⇒ a + (57 - 1)d = 108

⇒ a + 56d = 108 .........(i)

Also,

⇒ a₇ = 13

⇒ a + (7 - 1)d = 13

⇒ a + 6d = 13 ........(ii)

Subtracting (ii) from (i) we get,

⇒ (a + 56d) - (a + 6d) = 108 - 13

⇒ a - a + 56d - 6d = 95

⇒ 50d = 95

⇒ d = 1.9

Substituting value of d in (ii) we get,

⇒ a + 6(1.9) = 13

⇒ a + 11.4 = 13

⇒ a = 13 - 11.4 = 1.6

45^th term of A.P. = a₄₅

= a + (45 - 1)d

= 1.6 + 44(1.9)

= 1.6 + 83.6

= 85.2

Hence, 45^th term of A.P. = 85.2

4^th term of an A.P. is equal to 3 times its first term and 7^th term exceeds twice the 3^rd term by 1. Find the first term and the common difference.

Answer

Let first term be a and common difference be d.

According to question,

⇒ a₄ = 3a

⇒ a + (4 - 1)d = 3(a)

⇒ a + 3d = 3a

⇒ 2a = 3d

⇒ a = $\dfrac{3d}{2}$ ........(i)

Also,

⇒ a₇ - 2a₃ = 1

⇒ a + (7 - 1)d - 2[a + (3 - 1)d] = 1

⇒ a + 6d - 2(a + 2d) = 1

⇒ a + 6d - 2a - 4d = 1

⇒ a - 2a + 2d = 1

⇒ -a + 2d = 1

Substituting value of a from (i) in above equation,

$\Rightarrow -\dfrac{3d}{2} + 2d = 1 \\[1em] \Rightarrow \dfrac{-3d + 4d}{2} = 1 \\[1em] \Rightarrow \dfrac{d}{2} = 1 \\[1em] \Rightarrow d = 2.$

Substituting value of d in (i) we get,

a = $\dfrac{3 \times 2}{2}$ = 3.

Hence, first term = 3 and common difference = 2.

The sum of the 2^nd term and the 7^th term of an A.P. is 30. If its 15^th term is 1 less than twice of its 8^th term, find the A.P.

Answer

Let first term be a and common difference be d.

According to question,

⇒ a₂ + a₇ = 30

⇒ a + (2 - 1)d + a + (7 - 1)d = 30

⇒ a + d + a + 6d = 30

⇒ 2a + 7d = 30 .........(i)

Also,

⇒ 2a₈ - 1 = a₁₅

⇒ 2[a + (8 - 1)d] - 1 = a + (15 - 1)d

⇒ 2[a + 7d] - 1 = a + 14d

⇒ 2a + 14d - 1 = a + 14d

⇒ 2a - a - 1 = 14d - 14d

⇒ a - 1 = 0

⇒ a = 1.

Substituting value of a in (i) we get,

⇒ 2(1) + 7d = 30

⇒ 2 + 7d = 30

⇒ 7d = 28

⇒ d = 4.

A.P. = a, (a + d), (a + 2d),..........

= 1, (1 + 4), (1 + 2.4),.........

= 1, 5, 9,.......

Hence, A.P. = 1, 5, 9,........

In an A.P. if m^th term is n and n^th term is m, show that its r^th term is (m + n - r).

Answer

Let first term be a and common difference be d.

Given,

⇒ a_m = n

⇒ a + (m - 1)d = n

⇒ a + md - d = n

⇒ a = n - md + d .......(i)

Also,

⇒ a_n = m

⇒ a + (n - 1)d = m

⇒ a + nd - d = m

Substituting value of a from (i) in above equation,

⇒ n - md + d + nd - d = m

⇒ nd - md = m - n

⇒ d(n - m) = m - n

⇒ d = $\dfrac{m - n}{n - m} = \dfrac{m - n}{-(m - n)}$ = -1.

Substituting value of d in (i) we get,

⇒ a = n - m(-1) + (-1) = n + m - 1.

a_r = a + (r - 1)d

= n + m - 1 + (r - 1)(-1)

= n + m - 1 - r + 1

= n + m - r.

Hence, proved that r^th term is (m + n - r).

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

Answer

In the above A.P.,

a = 3 and d = 10 - 3 = 7.

⇒ a₁₃ = a + (13 - 1)d = 3 + 12(7) = 3 + 84 = 87.

Let n^th term be 84 more than its 13^th term.

⇒ a_n = 84 + a₁₃

⇒ a + (n - 1)d = 84 + 87

⇒ 3 + (n - 1)7 = 171

⇒ (n - 1)7 = 168

⇒ n - 1 = 24

⇒ n = 25.

Hence, 25^th term will be 84 more than its 13^th term.