Arithmetic Progression — Exercise 10(A)

The first term and the common difference of an A.P. are 8 and -5 respectively. The A.P. is :

1. 8, 13, 18, 23, 28, .........

2. 8, 3, -2, -7, .......

3. -5, 3, 11, 19, 27, ........

4. -5, -13, -21, -29, .......

Answer

Given,

First term (a) = 8

Common difference (d) = -5

A.P. is given by a, a + d, a + 2d, a + 3d, ..........

⇒ 8, 8 + (-5), 8 + 2(-5), 8 + 3(-5), .........

⇒ 8, 8 - 5, 8 - 10, 8 - 15, ........

⇒ 8, 3, -2, -7, ........

Hence, Option 2 is the correct option.

Is -8, -8, -8, -8, ..... an A.P.?

1. no

2. yes

3. may be

4. none of the above

Answer

The above list is an A.P. with common difference = -8 - (-8) = -8 + 8 = 0.

Hence, Option 2 is the correct option.

The 15^th term of the A.P. 3, 0, -3, -6, ...... is :

1. -42

2. 39

3. 42

4. -39

Answer

In A.P.,

3, 0, -3, -6, ......

First term (a) = 3

Common difference (d) = 0 - 3 = -3.

By formula,

a_n = a + (n - 1)d

a₁₅ = 3 + (15 - 1)(-3)

= 3 + 14 × -3

= 3 - 42

= -39.

Hence, Option 4 is the correct option.

The 24th term of an A.P. exceeds its 19^th term by 10, its common difference is :

1. 5

2. 2

3. 10

4. 1

Answer

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

By formula,

⇒ a_n = a + (n - 1)d

⇒ a₂₄ = a + (24 - 1)d = a + 23d.

⇒ a₁₉ = a + (19 - 1)d = a + 18d.

Given,

24th term exceeds 19^th term by 10.

⇒ a₂₄ - a₁₉ = 10

⇒ a + 23d - (a + 18d) = 10

⇒ a - a + 23d - 18d = 10

⇒ 5d = 10

⇒ d = $\dfrac{10}{5}$ = 2.

Hence, Option 2 is the correct option.

In the A.P. 8, 13, 18, ......., the n^th term is 83, then n is equal to :

1. 14

2. 16

3. 13

4. 15

Answer

In the A.P.,

First term (a) = 8

Common difference (d) = 13 - 8 = 5.

By formula,

⇒ a_n = a + (n - 1)d

Substituting values we get :

⇒ 83 = 8 + 5(n - 1)

⇒ 83 = 8 + 5n - 5

⇒ 83 = 5n + 3

⇒ 83 - 3 = 5n

⇒ 80 = 5n

⇒ n = $\dfrac{80}{5}$ = 16.

Hence, Option 2 is the correct option.

The n^th term of a sequence is (2n - 3), find the fifteenth term.

Answer

Given,

a_n = 2n - 3

So, a₁₅ = 2(15) - 3 = 30 - 3 = 27.

Hence, fifteenth term = 27.

If the p^th term of an A.P. is (2p + 3); find the A.P.

Answer

Given,

⇒ a_p = 2p + 3

So,

⇒ a₁ = 2(1) + 3 = 5

⇒ a₂ = 2(2) + 3 = 7

⇒ a₃ = 2(3) + 3 = 9

Hence, the A.P. is 5, 7, 9 .......

Find the 24^th term of the sequence :

12, 10, 8, 6 ........

Answer

Since, 10 - 12 = -2, 8 - 10 = -2 and 6 - 8 = -2.

Hence, the series is an A.P. with common difference = -2.

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

⇒ a_n = a + (n - 1)d, where a is the first term.

So, 24^th term is,

⇒ a₂₄ = 12 + (24 - 1)(-2) = 12 + 23(-2) = 12 - 46 = -34.

Hence, 24^th term of the sequence = -34.

Find the 30^th term of the sequence :

$\dfrac{1}{2}, 1, \dfrac{3}{2}, .........$

Answer

Since, $1 - \dfrac{1}{2} = \dfrac{1}{2}, \dfrac{3}{2} - 1 = \dfrac{1}{2}$.

Hence, the series is an A.P. with common difference = $\dfrac{1}{2}$.

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

⇒ a_n = a + (n - 1)d, where a is the first term.

So, 30^th term is,

$\Rightarrow a_{30} = \dfrac{1}{2} + (30 - 1) \times \dfrac{1}{2} \\[1em] = \dfrac{1}{2} + \dfrac{29}{2} \\[1em] = \dfrac{30}{2} \\[1em] = 15.$

Hence, 30^th term of the sequence = 15.

Find the 100^th term of the sequence :

$\sqrt{5}, 2\sqrt{5}, 3\sqrt{5}, .......$

Answer

In the sequence :

$3\sqrt{5} - 2\sqrt{5} = 2\sqrt{5} - \sqrt{5} = \sqrt{5}$.

Since, the difference between consecutive terms are equal, thus the sequence is an A.P.

First term (a) = $\sqrt{5}$

Common difference (d) = $\sqrt{5}$

We know that,

$\Rightarrow a_n = a + (n - 1)d \\[1em] \Rightarrow a_{100} = \sqrt{5} + (100 - 1) \times \sqrt{5} \\[1em] \Rightarrow a_{100} = \sqrt{5} + 99\sqrt{5} \\[1em] \Rightarrow a_{100} = 100\sqrt{5}.$

Hence, 100^th term of the sequence = $100\sqrt{5}$.

Find the 50^th term of the sequence :

$\dfrac{1}{n}, \dfrac{n + 1}{n}, \dfrac{2n + 1}{n}, ......$

Answer

Since, $\dfrac{n + 1}{n} - \dfrac{1}{n} = \dfrac{n}{n} = 1 \text{ and } \dfrac{2n + 1}{n} - \dfrac{n + 1}{n} = \dfrac{n}{n}$ = 1.

Hence, the series is an A.P. with common difference = 1

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

⇒ a_n = a + (n - 1)d, where a is the first term.

So, 50^th term is,

$\Rightarrow a_{50} = \dfrac{1}{n} + (50 - 1) \times 1 \\[1em] = \dfrac{1}{n} + 49.$

Hence, 50^th term of the sequence = $\dfrac{1}{n} + 49.$

Is 402 a term of the sequence :

8, 13, 18, 23, ........?

Answer

Since, 13 - 8 = 5, 18 - 13 = 5 and 23 - 18 = 5.

Hence, the series is an A.P. with common difference = 5.

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

⇒ a_n = a + (n - 1)d, where a is the first term.

Let 402 be n^th term,

⇒ 402 = a + (n - 1)d

⇒ 402 = 8 + (n - 1)5

⇒ 402 = 8 + 5n - 5

⇒ 402 = 5n + 3

⇒ 399 = 5n

⇒ n = $\dfrac{399}{5}$

Since, the no. of term cannot be in fraction.

Hence, 402 is not a term of the sequence.

Find the common difference and 99^th term of the arithmetic progression :

$7\dfrac{3}{4}, 9\dfrac{1}{2}, 11\dfrac{1}{4}, .......$

Answer

The sequence = $\dfrac{31}{4}, \dfrac{19}{2}, \dfrac{45}{4}$ .......

Common difference = $\dfrac{19}{2} - \dfrac{31}{4}$

$\phantom{Common difference}= \dfrac{38 - 31}{4} \\[1em] \phantom{Common difference}= \dfrac{7}{4} \\[1em] \phantom{Common difference}= 1\dfrac{3}{4}$

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

⇒ a_n = a + (n - 1)d, where a is the first term.

So, 99^th term,

$a_{99} = \dfrac{31}{4} + (99 - 1) \times \dfrac{7}{4} \\[1em] = \dfrac{31}{4} + \dfrac{98 \times 7}{4} \\[1em] = \dfrac{31}{4} + \dfrac{686}{4} \\[1em] = \dfrac{717}{4} \\[1em] = 179\dfrac{1}{4}.$

Hence, 99^th term of the sequence = $179\dfrac{1}{4}$ and common difference = $1\dfrac{3}{4}$.

How many terms are there in the series :

4, 7, 10, 13, ..........., 148 ?

Answer

Since, 7 - 4 = 3, 10 - 7 = 3 and 13 - 10 = 3.

Hence, the series is an A.P. with common difference = 3 and last term = 148.

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

a_n = a + (n - 1)d

⇒ 148 = 4 + (n - 1)3

⇒ 148 = 4 + 3n - 3

⇒ 148 = 3n + 1

⇒ 148 - 1 = 3n

⇒ 3n = 147

⇒ n = 49.

Hence, no. terms in the series = 49.

How many terms are there in the series :

0.5, 0.53, 0.56, ........, 1.1 ?

Answer

Since, 0.53 - 0.5 = 0.03, 0.56 - 0.53 = 0.03.

Hence, the series is an A.P. with common difference = 0.03 and last term = 1.1.

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

a_n = a + (n - 1)d

⇒ 1.1 = 0.5 + (n - 1) × 0.03

⇒ 1.1 = 0.5 + 0.03n - 0.03

⇒ 1.1 = 0.47 + 0.03n

⇒ 1.1 - 0.47 = 0.03n

⇒ 0.63 = 0.03n

⇒ n = 21.

Hence, no. terms in the series = 21.

How many terms are there in the series :

$\dfrac{3}{4}, 1, 1\dfrac{1}{4}, ........., 3$ ?

Answer

Since, $1 - \dfrac{3}{4} = \dfrac{1}{4}, \dfrac{5}{4} - 1 = \dfrac{1}{4}.$

Hence, the series is an A.P. with common difference = $\dfrac{1}{4}$ and last term = 3.

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

a_n = a + (n - 1)d

$\Rightarrow 3 = \dfrac{3}{4} + (n - 1) \times \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{3 + n - 1}{4} = 3 \\[1em] \Rightarrow n + 2 = 12 \\[1em] \Rightarrow n = 10.$

Hence, no. terms in the series = 10.

Which term of the A.P. 1, 4, 7, 10, ....... is 52 ?

Answer

In the above A.P,

Common difference = 4 - 1 = 3,

First term = 1.

Let 52 be n^th term,

∴ 52 = 1 + (n - 1)(3)

⇒ 52 = 1 + 3n - 3

⇒ 52 = 3n - 2

⇒ 54 = 3n

⇒ n = 18.

Hence, 52 is 18th term of the A.P.

If 5^th and 6^th terms of an A.P. are respectively 6 and 5, find the 11^th term of the A.P.

Answer

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

a_n = a + (n - 1)d

Given,

5^th term = 6

⇒ a₅ = a + (5 - 1)d

⇒ 6 = a + 4d .........(i)

Also,

6^th term = 5

⇒ a₆ = a + (6 - 1)d

⇒ 5 = a + 5d .........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 5d - (a + 4d) = 5 - 6

⇒ a - a + 5d - 4d = -1

⇒ d = -1.

Substituting value of d in (i) we get,

⇒ 6 = a + 4(-1)

⇒ 6 = a - 4

⇒ a = 10.

11^th term of the sequence,

⇒ a₁₁ = 10 + (11 - 1)(-1)

= 10 + 10(-1)

= 10 - 10 = 0.

Hence, 11^th term of the sequence = 0.

If t_n represents n^th term of an A.P., t₂ + t₅ - t₃ = 10 and t₂ + t₉ = 17, find its first term and its common difference.

Answer

t_n = a + (n - 1)d .......(1)

Substituting above value in t₂ + t₅ - t₃ = 10,

⇒ [a + (2 - 1)d] + [a + (5 - 1)d] - [a + (3 - 1)d] = 10

⇒ (a + d) + (a + 4d) - (a + 2d) = 10

⇒ a + 3d = 10 .......(i)

Substituting value from 1 in t₂ + t₉ = 17,

⇒ [a + (2 - 1)d] + [a + (9 - 1)d] = 17

⇒ a + d + a + 8d = 17

⇒ 2a + 9d = 17 ........(ii)

Multiplying (i) by 2 and then subtracting from (ii) we get,

⇒ 2a + 9d - 2(a + 3d) = 17 - 2(10)

⇒ 2a + 9d - 2a - 6d = 17 - 20

⇒ 3d = -3

⇒ d = -1.

Substituting value of d in (i) we get,

⇒ a + 3(-1) = 10

⇒ a - 3 = 10

⇒ a = 10 + 3 = 13.

Hence, first term = 13 and common difference = -1.

Find the 10^th term from the end of the A.P. 4, 9, 14, ......., 254.

Answer

We know that r^th term from the end is (n - r + 1)^th term from the beginning, where n is no. of terms.

Let the A.P. 4, 9, 14, ......., 254 has n terms.

The above A.P. has common difference = 9 - 4 = 5 and first term = 4.

Substituting values in a_n = a + (n - 1)d,

⇒ 254 = 4 + (n - 1)5

⇒ 254 = 4 + 5n - 5

⇒ 254 = 5n - 1

⇒ 255 = 5n

⇒ n = 51.

So, 10^th term from the end = (51 - 10 + 1)^th = 42^nd term from beginning.

Substituting values in a_n = a + (n - 1)d,

⇒ a_n = 4 + (42 - 1)5

= 4 + 41(5)

= 4 + 205

= 209.

Hence, 10^th term from the end = 209.

Determine the arithmetic progression whose 3^rd term is 5 and 7^th term is 9.

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 5

∴ a₃ = a + (3 - 1)d

⇒ 5 = a + 2d

⇒ a + 2d = 5 ........(i)

Given, 7^th term is 9

∴ a₇ = a + (7 - 1)d

⇒ 9 = a + 6d

⇒ a + 6d = 9 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 6d - (a + 2d) = 9 - 5

⇒ 4d = 4

⇒ d = 1.

Substituting value of d in (i) we get,

⇒ a + 2(1) = 5

⇒ a + 2 = 5

⇒ a = 3.

A.P. = a, (a + d), (a + 2d), (a + 3d) ..........

= 3, 4, 5, 6, 7, ...........

Hence, A.P. = 3, 4, 5, 6, 7, ...........

Find the 31^st term of an A.P. whose 10^th term is 38 and 16^th term is 74.

Answer

We know that,

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

a_n = a + (n - 1)d

Given, 10^th term is 38

∴ a₁₀ = a + (10 - 1)d

⇒ 38 = a + 9d

⇒ a + 9d = 38 ........(i)

Given, 16^th term is 74

∴ a₁₆ = a + (16 - 1)d

⇒ 74 = a + 15d

⇒ a + 15d = 74 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 15d - (a + 9d) = 74 - 38

⇒ 6d = 36

⇒ d = 6.

Substituting value of d in (i) we get,

⇒ a + 9(6) = 38

⇒ a + 54 = 38

⇒ a = -16.

31^st term = a₃₁

a₃₁ = a + (31 - 1)d

= -16 + 30(6)

= -16 + 180

= 164.

Hence, 31^st term of A.P. = 164.

Which term of the series :

21, 18, 15, ........ is -81 ?

Can any term of this series be zero ?

If yes, find the number of terms.

Answer

In A.P. 21, 18, 15, ........,

d = 18 - 21 = -3 and a = 21.

Let n^th term be -81.

⇒ a_n = -81

⇒ a + (n - 1)d = -81

⇒ 21 + (n - 1)(-3) = -81

⇒ 21 - 3n + 3 = -81

⇒ 24 - 3n = -81

⇒ 3n = 24 + 81

⇒ 3n = 105

⇒ n = 35.

Let r^th term be 0.

⇒ a_r = 0

⇒ a + (r - 1)d = 0

⇒ 21 + (r - 1)(-3) = 0

⇒ 21 - 3r + 3 = 0

⇒ 24 - 3r = 0

⇒ 3r = 24

⇒ r = 8.

Hence, -81 is 35^th and 0 is 8^th term of an A.P.

An A.P. consists of 60 terms. If the first and the last term be 7 and 125 respectively, find the 31^st term.

Answer

Given,

First term = a = 7,

Last term = a₆₀ = a + (60 - 1)d

⇒ 125 = 7 + (59)d

⇒ 125 - 7 = 59d

⇒ 59d = 118

⇒ d = 2.

a₃₁ = a + (31 - 1)d

= 7 + 30(2)

= 7 + 60

= 67.

Hence, 31^st term = 67.

The sum of the 4^th term and the 8^th terms of an A.P. is 24 and the sum of 6^th and the 10^th terms of the same A.P. is 34. Find the first three terms of the A.P.

Answer

According to question,

a₄ + a₈ = 24 .......(i)

a₆ + a₁₀ = 34 .......(ii)

Solving (i) we get,

⇒ a + (4 - 1)d + a + (8 - 1)d = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ a + 5d = 12 .........(iii)

Solving (ii) we get,

⇒ a + (6 - 1)d + a + (10 - 1)d = 34

⇒ a + 5d + a + 9d = 34

⇒ 2a + 14d = 34

⇒ a + 7d = 17 .........(iv)

Subtracting (iii) from (iv) we get,

⇒ a + 7d - (a + 5d) = 17 - 12

⇒ 2d = 5

⇒ d = $\dfrac{5}{2}$.

Substituting d in (iv) we get,

⇒ a + $7 \times \dfrac{5}{2}$ = 17

⇒ a + $\dfrac{35}{2}$ = 17

⇒ a = $17 - \dfrac{35}{2} = \dfrac{34 - 35}{2} = -\dfrac{1}{2}$.

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

$= -\dfrac{1}{2}, -\dfrac{1}{2} + \dfrac{5}{2}, -\dfrac{1}{2} + 2 \times \dfrac{5}{2} ......... \\[1em] = -\dfrac{1}{2}, \dfrac{-1 + 5}{2}, -\dfrac{1}{2} + 5.......... \\[1em] = -\dfrac{1}{2}, 2, \dfrac{9}{2} ........$

Hence, first three terms of the A.P. = $-\dfrac{1}{2}, 2, \dfrac{9}{2}$

If the third term of an A.P. is 5 and the seventh term is 9, find the 17^th term.

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 5

∴ a₃ = a + (3 - 1)d

⇒ 5 = a + 2d

⇒ a + 2d = 5 ........(i)

Given, 7^th term is 9

∴ a₇ = a + (7 - 1)d

⇒ 9 = a + 6d

⇒ a + 6d = 9 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 6d - (a + 2d) = 9 - 5

⇒ 4d = 4

⇒ d = 1.

Substituting value of d in (i) we get,

⇒ a + 2(1) = 5

⇒ a + 2 = 5

⇒ a = 3.

17^th term = a₁₇

a₁₇ = a + (17 - 1)d

= 3 + 16(1)

= 3 + 16

= 19.

Hence, 17^th term of A.P. = 19.