Arithmetic Progression

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,

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

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, 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.

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.

If a, b and c are in A.P. show that :

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

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

Answer

Given,

a, b and c are in A.P.

∴ b - a = c - b

⇒ b + b = c + a

⇒ 2b = a + c ..........(1)

(i) To prove,

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

⇒ 4b - 4a = 4c - 4b

⇒ 4(b - a) = 4(c - b)

⇒ b - a = c - b

⇒ 2b = c + a, which is equal to equation 1.

Hence, proved that 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

⇒ 2b = c + a, which is equal to equation 1.

Hence, proved that 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.

The sum of 41 terms of an A.P. with middle term 40 is :

1. 820

2. 1640

3. 2460

4. none of these

Answer

No. of terms = 41

Middle term = $\dfrac{41 + 1}{2} = \dfrac{42}{2}$ = 21st term

Given,

Middle term = 40

⇒ a₂₁ = 40

⇒ a + (21 - 1)d = 40

⇒ a + 20d = 40 ............(1)

By formula,

Sum of A.P. = $\dfrac{n}{2}$[2a + (n - 1)d]

= $\dfrac{41}{2}[2a + (41- 1)d]$

= $\dfrac{41}{2}[2a + 40d]$

= $\dfrac{41}{2} \times 2 \times [a + 20d]$

= 41 × 40 ..........[From (1)]

= 1640.

Hence, Option 2 is the correct option.

The sum of all two digit numbers is :

1. 9810

2. 9045

3. 4509

4. 4905

Answer

Two digit numbers : 10, 11, ......., 99.

The above list is an A.P. with first term (a) = 10 and common difference (d) = 11 - 10 = 1.

Last term = 99.

Let n^th term be the last term.

∴ a_n = a + (n - 1)d

⇒ 99 = 10 + (n - 1)1

⇒ 99 = 10 + n - 1

⇒ 99 = n + 9

⇒ n = 99 - 9 = 90.

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

Sum of the above A.P.

$= \dfrac{90}{2}[2 \times 10 + (90 - 1)(1)] \\[1em] = 45[20 + 89] \\[1em] = 45 \times 109 \\[1em] = 4905.$

Hence, Option 4 is the correct option.

The sum of A.P. 4, 7, 10, 13, ........ upto 20 terms is :

1. 650

2. 10 × 27

3. 510

4. 1300

Answer

In A.P. 4, 7, 10, 13, ........ upto 20 terms

First term (a) = 4

Common difference (d) = 7 - 4 = 3

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

Sum of above A.P. (upto 20 terms)

$= \dfrac{20}{2}[2 \times 4 + (20 - 1) \times 3] \\[1em] = 10 \times [8 + 19 \times 3] \\[1em] = 10 \times [8 + 57] \\[1em] = 10 \times 65 \\[1em] = 650.$

Hence, Option 1 is the correct option.

The sum of 40 terms of the A.P. 7 + 10 + 13 + 16 + ..... is :

1. 5240

2. 2620

3. 1310

4. 2680

Answer

In the A.P., 7 + 10 + 13 + 16 + .........

First term (a) = 7

Common difference (d) = 10 - 7 = 3.

By formula,

Sum of n terms = $\dfrac{n}{2}[2a + (n - 1)d]$

Sum of above A.P. (upto 20 terms)

$= \dfrac{40}{2}[2 \times 7 + (40 - 1) \times 3] \\[1em] = 20 \times [14 + 39 \times 3] \\[1em] = 20 \times [14 + 117] \\[1em] = 20 \times 131 \\[1em] = 2620.$

Hence, Option 2 is the correct option.

The n^th term of an A.P. is 6n + 4. The sum of its first two terms is :

1. 16

2. 20

3. 26

4. none of these

Answer

Given,

n^th term = 6n + 4

First term = 6(1) + 4 = 10

Second term = 6(2) + 4 = 12 + 4 = 16.

Sum of first two terms = 10 + 16 = 26.

Hence, Option 3 is the correct option.

How many terms of the A.P. :

24, 21, 18, ........... must be taken so that their sum is 78 ?

Answer

In above A.P. a = 24 and d = -3.

We know that,

S = $\dfrac{n}{2}(2a + (n - 1)d)$

Let sum of n terms be 78.

$\Rightarrow 78 = \dfrac{n}{2}(2 \times 24 + (n - 1)(-3)) \\[1em] \Rightarrow 78 = \dfrac{n}{2}(48 - 3n + 3) \\[1em] \Rightarrow 78 \times 2 = n(51 - 3n) \\[1em] \Rightarrow 156 = 51n - 3n^2 \\[1em] \Rightarrow 3n^2 - 51n + 156 = 0 \\[1em] \Rightarrow 3(n^2 - 17n + 52) = 0 \\[1em] \Rightarrow n^2 - 17n + 52 = 0 \\[1em] \Rightarrow n^2 - 13n - 4n + 52 = 0 \\[1em] \Rightarrow n(n - 13) - 4(n - 13) = 0 \\[1em] \Rightarrow (n - 4)(n - 13) = 0 \\[1em] \Rightarrow n - 4 = 0 \text{ or } n - 13 = 0 \\[1em] \Rightarrow n = 4, 13.$

Hence, no. of terms = 4 or 13.

Find the sum of 28 terms of an A.P. whose n^th term is 8n - 5.

Answer

Given,

a_n = 8n - 5

So,

a₁ = 8(1) - 5 = 3,

a₂₈ = 8(28) - 5 = 219.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{28}{2}(3 + 219) \\[1em] = 14 \times 222 \\[1em] = 3108.$

Hence, sum = 3108.

Find the sum of all odd natural numbers less than 50.

Answer

Odd numbers less than 50 = 1, 3, 5, ........., 49.

The above series is an A.P. with a = 1, d = 2 and l = 49.

Let there be n terms in series,

⇒ a_n = 49

⇒ a + (n - 1)d = 49

⇒ 1 + 2(n - 1) = 49

⇒ 1 + 2n - 2 = 49

⇒ 2n - 1 = 49

⇒ 2n = 50

⇒ n = 25.

We know that,

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{25}{2}(1 + 49) \\[1em] = \dfrac{25}{2} \times 50 \\[1em] = 625.$

Hence, sum of odd natural numbers less than 50 = 625.

Find the sum of first 12 natural numbers each of which is a multiple of 7.

Answer

First 12 natural numbers that are a multiple of 7 are,

7, 14, 21, ..........., 12^th term.

The above sequence is an A.P. with a = 7 and common difference = 7.

$S = \dfrac{n}{2}(2a + (n - 1)d) \\[1em] = \dfrac{12}{2}(2 \times 7 + (12 - 1) \times 7) \\[1em] = 6(14 + 77) \\[1em] = 6 \times 91 \\[1em] = 546.$

Hence, sum of first 12 natural numbers that are divisible by 7 is 546.

Find the sum of first 51 terms of an A.P. whose 2^nd and 3^rd terms are 14 and 18 respectively.

Answer

Given,

⇒ a₂ = a + (2 - 1)d

⇒ 14 = a + d

⇒ a = 14 - d .......(i)

Also,

⇒ a₃ = a + (3 - 1)d

⇒ 18 = a + 2d

⇒ a = 18 - 2d .......(ii)

From (i) and (ii) we get,

⇒ 14 - d = 18 - 2d

⇒ -d + 2d = 18 - 14

⇒ d = 4.

Substituting value of d in (i) we get,

⇒ a = 14 - 4 = 10.

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{51}{2}[2 \times 10 + (51 - 1) \times 4] \\[1em] = \dfrac{51}{2} \times (20 + 200) \\[1em] = \dfrac{51}{2} \times 220 \\[1em] = 51 \times 110 \\[1em] = 5610.$

Hence, sum of first 51 terms = 5610.

The sum of first 7 terms of an A.P. is 49 and that of first 17 terms of it is 289. Find the sum of first n terms.

Answer

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

Given, sum of first 7 terms of an A.P. is 49,

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 49 = \dfrac{7}{2}[2 \times a + (7 - 1)d] \\[1em] \Rightarrow 49 = \dfrac{7}{2}[2a + 6d] \\[1em] \Rightarrow 49 = 7(a + 3d) \\[1em] \Rightarrow a + 3d = 7 ........(i)$

Given, sum of first 17 terms of an A.P. is 289,

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 289 = \dfrac{17}{2}[2 \times a + (17 - 1)d] \\[1em] \Rightarrow 289 = \dfrac{17}{2}[2a + 16d] \\[1em] \Rightarrow 289 = 17(a + 8d) \\[1em] \Rightarrow a + 8d = 17 ........(ii)$

Subtracting (i) from (ii) we get,

⇒ a + 8d - (a + 3d) = 17 - 7

⇒ a - a + 8d - 3d = 10

⇒ 5d = 10

⇒ d = 2.

Substituting value of d in (i) we get,

⇒ a + 3(2) = 7

⇒ a + 6 = 7

⇒ a = 1.

$\text{Sum of n terms } = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{n}{2}[2(1) + (n - 1)2] \\[1em] = \dfrac{n}{2}[2 + 2n - 2] \\[1em] = \dfrac{n}{2} \times 2n \\[1em] = n^2.$

Hence, sum of n terms = n².

The first term of an A.P. is 5, the last term is 45 and the sum of its term is 1000. Find the number of terms and the common difference of the A.P.

Answer

Given, a = 5, l = 45 and S = 1000.

$\Rightarrow S = \dfrac{n}{2}(a + l) \\[1em] \Rightarrow 1000 = \dfrac{n}{2}(5 + 45) \\[1em] \Rightarrow 1000 = \dfrac{n}{2} \times 50 \\[1em] \Rightarrow n = \dfrac{1000 \times 2}{50} \\[1em] \Rightarrow n = 40.$

Given, l = a₄₀ = 45

⇒ a + (40 - 1)d = 45
⇒ 5 + 39d = 45
⇒ 39d = 40
⇒ d = $\dfrac{40}{39}$.

Hence, n = 40 and d = $\dfrac{40}{39}$.

Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.

Answer

$\dfrac{250}{9} = 27\dfrac{7}{9} \text{ and } \dfrac{1000}{9} = 111\dfrac{1}{9}$.

The numbers which are divisible by 9 between 250 and 1000 are,

= 28 × 9, 29 × 9, 30 × 9, ............, 111 × 9.

= 252, 261, 270, .........., 999.

The above sequence is an A.P. with common difference = 9 and first term = 252 and last term = 999.

Let n be no. of terms,

∴ a_n = a + (n - 1)d

⇒ 999 = 252 + (n - 1)9

⇒ 999 = 252 + 9n - 9

⇒ 999 = 9n + 243

⇒ 999 - 243 = 9n

⇒ 9n = 756

⇒ n = 84.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{84}{2} \times (252 + 999) \\[1em] = 42 \times 1251 \\[1em] = 52542.$

Hence, sum = 52542.

The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?

Answer

Let n be no. of terms,

∴ a_n = a + (n - 1)d

⇒ 700 = 34 + (n - 1)18

⇒ 700 = 34 + 18n - 18

⇒ 700 = 18n + 16

⇒ 700 - 16 = 18n

⇒ 18n = 684

⇒ n = 38.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{38}{2} \times (34 + 700) \\[1em] = 19 \times 734 \\[1em] = 13946.$

Hence, no. of terms = 38 and sum = 13946.

In an A.P. the first term is 25, n^th term is -17 and the sum of n terms is 132. Find n and the common difference.

Answer

Given,

a = 25, a_n = -17 and S = 132

⇒ a + (n - 1)d = -17

⇒ 25 + (n - 1)d = -17

⇒ (n - 1)d = -42 ........(i)

We know that,

$\Rightarrow S = \dfrac{n}{2}(2a + (n - 1)d) \\[1em] \Rightarrow 132 = \dfrac{n}{2}(2 \times 25 + (-42)) \\[1em] \Rightarrow 132 = \dfrac{n}{2}(50 - 42) \\[1em] 132 = \dfrac{8n}{2} \\[1em] 4n = 132 \\[1em] n = 33.$

Substituting value of n in (i),

⇒ (33 - 1)d = -42

⇒ 32d = -42

⇒ d = -$\dfrac{42}{32} = -\dfrac{21}{16}$

Hence, n = 33 and d = $-\dfrac{21}{16}$.

If the 8^th term of an A.P. is 37 and the 15^th term is 15 more than the 12^th term, find the A.P.

Also, find the sum of first 20 terms of this A.P.

Answer

Given,

⇒ a₈ = a + (8 - 1)d

∴ a + 7d = 37 ........(i)

Also,

⇒ a₁₅ = a₁₂ + 15

⇒ a + (15 - 1)d = a + (12 - 1)d + 15

⇒ a + 14d = a + 11d + 15

⇒ a - a + 14d - 11d = 15

⇒ 3d = 15

⇒ d = 5.

Substituting value of d in (i) we get,

⇒ a + 7(5) = 37

⇒ a + 35 = 37

⇒ a = 2.

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

= 2, 7, 12, ...........

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{20}{2}[2 \times 2 + (20 - 1) \times 5] \\[1em] = 10(4 + 19 \times 5) \\[1em] = 10 \times 99 \\[1em] = 990.$

Hence, A.P. = 2, 7, 12, ........... and sum = 990.

The sum of n natural numbers is 5n² + 4n. Find its 8^th term.

Answer

Here T stands for term.

S_n = 5n² + 4n

S₁ = Sum of first natural number = T₁

= 5(1)² + 4(1) = 5 + 4 = 9.

S₂ = 5(2)² + 4(2) = 20 + 8 = 28.

T₂ = S₂ - S₁ = 28 - 9 = 19.

We know that,

⇒ T₂ = a + d

⇒ 19 = 9 + d

⇒ d = 10.

T₈ = a + (8 - 1)d = 9 + 7(10) = 79.

Hence, 8^th term = 79.

The fourth term of an A.P. is 11 and the eight term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.

Answer

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

Given,

⇒ a₄ = 11

⇒ a + (4 - 1)d = 11

⇒ a + 3d = 11 .......(i)

Also,

⇒ a₈ = 2a₄ + 5

⇒ a + (8 - 1)d = 2[a + (4 - 1)d] + 5

⇒ a + 7d = 2a + 6d + 5

⇒ a - 2a + 7d - 6d = 5

⇒ -a + d = 5 ........(ii)

Adding (i) and (ii) we get,

⇒ a + 3d + (-a + d) = 11 + 5

⇒ 4d = 16

⇒ d = 4.

Substituting value of d in (i) we get,

⇒ a + 3(4) = 11

⇒ a + 12 = 11

⇒ a = -1.

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

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

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{50}{2}[2 \times (-1) + (50 - 1) \times 4] \\[1em] = 25[-2 + 196] \\[1em] = 25 \times 194 \\[1em] = 4850.$

Hence, A.P. = -1, 3, 7, ........... and sum of first 50 terms = 4850.

k + 2, 2k + 7 and 4k + 12 are the first three terms of an A.P. The first term of this A.P. is :

1. -2

2. 0

3. 2

4. 3

Answer

Since,

k + 2, 2k + 7 and 4k + 12 are the first three terms of an A.P.

∴ 2k + 7 - (k + 2) = 4k + 12 - (2k + 7)

⇒ 2k - k + 7 - 2 = 4k - 2k + 12 - 7

⇒ k + 5 = 2k + 5

⇒ 2k - k = 5 - 5

⇒ k = 0.

⇒ k + 2 = 0 + 2 = 2.

Hence, Option 3 is the correct option.

The sum of n terms of an A.P. is 3n². The second term of this A.P. is :

1. 8

2. 3

3. 9

4. 12

Answer

Given,

Sum of n terms = 3n²

Sum upto 2 terms = 3(2)² = 3 × 4 = 12.

Sum upto 1 term = 3(1)² = 3 × 1 = 3.

2nd term = Sum upto 2 terms - Sum upto 1 terms = 12 - 3 = 9.

Hence, Option 3 is the correct option.

If 5, 7 and 9 are in A.P. then which of the following is in A.P.?

1. 5 × 7, 7 × 9 and 9 × 5

2. 5 × 7, 7 × 7 and 9 × 7

3. 2 × 5, 2 × 7 and 5 × 9

4. 5 - 7, 7 - 9 and 9 - 5

Answer

First list :

⇒ 5 × 7, 7 × 9 and 9 × 5

⇒ 35, 63, 45

63 - 35 = 28

45 - 63 = -18

Since, common difference between consecutive terms is not equal, so it is not an A.P.

Second list :

⇒ 5 × 7, 7 × 7 and 9 × 7

⇒ 35, 49, 63

49 - 35 = 14

63 - 49 = 14

Since, common difference between consecutive terms are equal, it is an A.P.

Hence, Option 2 is the correct option.

Find three numbers in A.P. whose sum is 24 and whose product is 440.

Answer

Let three numbers in A.P. be (a - d), a, (a + d).

Sum = 24

∴ a - d + a + a + d = 24

⇒ 3a = 24

⇒ a = 8.

Product = 440

⇒ (a - d)(a)(a + d) = 440

⇒ (8 - d)(8)(8 + d) = 440

⇒ (8 - d)(8 + d) = $\dfrac{440}{8}$

⇒ 64 - d² = 55

⇒ d² = 64 - 55 = 9

⇒ d = ± 3

Let d = 3,

A.P. = (8 - 3), 8, (8 + 3) = 5, 8, 11.

Let d = -3,

A.P. = (8 - (-3)), 8, (8 + (-3)) = 11, 8, 5.

Hence, A.P. = 5, 8, 11 or 11, 8, 5.

The sum of three consecutive terms of an A.P. is 21 and the sum of their squares is 165. Find these terms.

Answer

Let three consecutive terms be (a - d), a, (a + d)

S = 21

⇒ a - d + a + a + d = 21

⇒ 3a = 21

⇒ a = 7.

Given, sum of their squares is 165

⇒ (a - d)² + a² + (a + d)² = 165

⇒ (7 - d)² + 7² + (7 + d)² = 165

⇒ 49 + d² - 14d + 49 + 49 + d² + 14d = 165

⇒ 147 + 2d² = 165

⇒ 2d² = 165 - 47 = 18

⇒ d² = 9

⇒ d = ± 3

Let d = 3,

A.P. = (7 - 3), 7, (7 + 3) = 4, 7, 10.

Let d = -3,

A.P. = (7 - (-3)), 7, (7 + (-3)) = 10, 7, 4.

Hence, the three consecutive terms of A.P. = 4, 7, 10 or 10, 7, 4.

The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.

Answer

Let the angles of quadrilateral are,

a, a + d, a + 2d, a + 3d

∴ a + (a + d) + (a + 2d) + (a + 3d) = 360°

⇒ 4a + 6d = 360°

⇒ 2(2a + 3d) = 360°

⇒ 2a + 3d = 180°

Putting value of d = 20° in above equation we get,

⇒ 2a + 3(20) = 180°

⇒ 2a + 60 = 180°

⇒ 2a = 180° - 60 = 120°

⇒ a = 60°.

Hence, angles = 60°, 80°, 100°, 120°.

Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.

Answer

Let the four terms of A.P. are

(a - 3d), (a - d), (a + d), (a + 3d).

According to question,

⇒ a - 3d + a - d + a + d + a + 3d = 96

⇒ 4a = 96

⇒ a = 24.

So, terms are,

24 - 3d, 24 - d, 24 + d, 24 + 3d.

Given, ratio between product of their means to product of their extremes is 15 : 7.

$\therefore \dfrac{(24 - d)(24 + d)}{(24 - 3d)(24 + 3d)} = \dfrac{15}{7} \\[1em] \Rightarrow \dfrac{576 - d^2}{576 - 9d^2} = \dfrac{15}{7} \\[1em] \Rightarrow 7(576 - d^2) = 15(576 - 9d^2) \\[1em] \Rightarrow 4032 - 7d^2 = 8640 - 135d^2 \\[1em] \Rightarrow 135d^2 - 7d^2 = 8640 - 4032 \\[1em] \Rightarrow 128d^2 = 4608 \\[1em] \Rightarrow d^2 = 36 \\[1em] \Rightarrow d = \pm 6.$

Let d = 6,

Terms = (a - 3d), (a - d), (a + d), (a + 3d) = (24 - 3 × 6), (24 - 6), (24 + 6), (24 + 3 × 6)

= 6, 18, 30, 42.

Let d = -6,

Terms = (a - 3d), (a - d), (a + d), (a + 3d) = (24 - 3 × -6), (24 - (-6)), (24 + (-6)), (24 + 3 × -6)

= 42, 30, 18, 6.

Hence, four parts of 96 are 6, 18, 30, 42 or 42, 30, 18, 6.

Find five numbers in A.P. whose sum is $12\dfrac{1}{2}$ and the ratio of the first to the last term is 2 : 3.

Answer

Let the numbers be (a - 2d), (a - d), a, (a + d), (a + 2d).

Given, sum = $12\dfrac{1}{2}$.

$\therefore a - 2d + a - d + a + a + d + a + 2d = 12\dfrac{1}{2} \\[1em] \Rightarrow 5a = \dfrac{25}{2} \\[1em] \Rightarrow a = \dfrac{5}{2} = 2.5.$

Given, ratio of the first to the last term is 2 : 3.

$\therefore \dfrac{a - 2d}{a + 2d} = \dfrac{2}{3} \\[1em] \Rightarrow 3(a - 2d) = 2(a + 2d) \\[1em] \Rightarrow 3a - 6d = 2a + 4d \\[1em] \Rightarrow 3a - 2a = 4d + 6d \\[1em] \Rightarrow a = 10d \\[1em] \Rightarrow \dfrac{5}{2} = 10d \\[1em] \Rightarrow d = \dfrac{5}{2 \times 10} \\[1em] \Rightarrow d = \dfrac{1}{4} = 0.25.$

Numbers = 2.5 - 2 × 0.25, 2.5 - 0.25, 2.5, 2.5 + 0.25, 2.5 + 2 × 0.25

= 2, 2.25, 2.5, 2.75, 3.

Hence, the numbers are 2, 2.25, 2.5, 2.75, 3.

Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.

Answer

Let the numbers be a - d, a, a + d.

According to question,

⇒ a - d + a + a + d = 207

⇒ 3a = 207

⇒ a = 69.

Given,

Product of the two smaller parts is 4623

⇒ (a - d)(a) = 4623

⇒ 69(69 - d) = 4623

⇒ 4761 - 69d = 4623

⇒ 69d = 4761 - 4623

⇒ 69d = 138

⇒ d = 2.

Numbers = (69 - 2), 69, (69 + 2).

Hence, numbers are 67, 69, 71.

The sum of three numbers in A.P. is 15 and the sum of the squares of the extreme terms is 58. Find the numbers.

Answer

Let the numbers be a - d, a, a + d.

According to question,

⇒ a - d + a + a + d = 15

⇒ 3a = 15

⇒ a = 5.

Given,

Sum of the squares of the extreme terms is 58.

⇒ (a - d)² + (a + d)² = 58

⇒ (5 - d)² + (5 + d)² = 58

⇒ 25 + d² - 10d + 25 + d² + 10d = 58

⇒ 50 + 2d² = 58

⇒ 2d² = 8

⇒ d² = 4

⇒ d = ±2

Let d = 2,

Numbers = (5 - 2), 5, (5 + 2) = 3, 5, 7.

Let d = -2,

Numbers = (5 - (-2)), 5, (5 + (-2)) = 7, 5, 3.

Hence, numbers = 3, 5, 7 or 7, 5, 3.

Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.

Answer

Let numbers be a - 3d, a - d, a + d, a + 3d.

Given, sum = 20

⇒ a - 3d + a - d + a + d + a + 3d = 20

⇒ 4a = 20

⇒ a = 5.

Given, sum of squares is 120.

⇒ (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 120

⇒ (5 - 3d)² + (5 - d)² + (5 + d)² + (5 + 3d)² = 120

⇒ 25 + 9d² - 30d + 25 + d² - 10d + 25 + d² + 10d + 25 + 9d² + 30d = 120

⇒ 100 + 20d² = 120

⇒ 20d² = 20

⇒ d² = 1.

⇒ d = ±1

Let d = 1,

Numbers = (5 - 3(1)), (5 - 1), (5 + 1), (5 + 3(1))

= 2, 4, 6, 8.

Let d = -1,

Numbers = (5 - 3(-1)), (5 - (-1)), (5 + (-1)), (5 + 3(-1))

= 8, 6, 4, 2.

Hence, numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

Insert one arithmetic mean between 3 and 13.

Answer

Arithmetic mean between 3 and 13 = $\dfrac{3 + 13}{2} = \dfrac{16}{2}$ = 8.

Hence, arithmetic mean between 3 and 13 = 8.

The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.

Answer

Given, angles of polygon are in A.P.,

a = 120° and d = 5°.

Let no. of sides be n and so sum of angles = (2n - 4) × 90°

Sum of A.P. of angles = $\dfrac{n}{2}[2 \times 120° + (n - 1) \times 5°]$

$\therefore \dfrac{n}{2}[2 \times 120° + (n - 1) \times 5°] = (2n - 4) × 90° \\[1em] \Rightarrow \dfrac{n}{2}[240° + 5°n - 5°] = 180°n - 360° \\[1em] \Rightarrow 240°n + 5°n^2 - 5°n = 360°n - 720° \\[1em] \Rightarrow 5°n^2 + 235°n = 360°n - 720° \\[1em] \Rightarrow 5°n^2 - 125°n + 720° = 0 \\[1em] \Rightarrow 5°(n^2 - 25n + 144) = 0 \\[1em] \Rightarrow n^2 - 25n + 144 = 0 \\[1em] \Rightarrow n^2 - 16n - 9n + 144 = 0 \\[1em] \Rightarrow n(n - 16) - 9(n - 16) = 0 \\[1em] \Rightarrow (n - 9)(n - 16) = 0 \\[1em] \Rightarrow n = 9, 16.$

Hence, number of sides = 9 or 16.

$\dfrac{1}{a}, \dfrac{1}{b} \text{ and } \dfrac{1}{c}$ are in A.P.

Show that : bc, ca and ab are also in A.P.

Answer

Given,

$\dfrac{1}{a}, \dfrac{1}{b} \text{ and } \dfrac{1}{c} \text{ are in A.P.}$ ........(i)

We know that,

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

Multiplying each term of (i) by abc we get,

bc, ca, ab, ........ are also in A.P.

Hence, proved that bc, ca and ab are also in A.P.

If A, B and C are three arithmetic progressions (APs) as given below :

A = 2, 4, 6, 8, ........ upto n terms

B = 3, 6, 9, 12, ...... upto n terms

C = 0, 4, 8, 12, ...... upto n terms, then

out of A + B, A - C, C - B and B - A which is/are A.P. ?

1. A + B

2. A - C

3. C - B

4. All are A.P.

Answer

Given,

A = 2, 4, 6, 8, ........ upto n terms

B = 3, 6, 9, 12, ...... upto n terms

C = 0, 4, 8, 12, ...... upto n terms, then

A + B = 2 + 3, 4 + 6, 6 + 9, 8 + 12, ........

= 5, 10, 15, 20, ..........

The above list is an A.P. with common difference = 5.

A - C = 2 - 0, 4 - 4, 6 - 8, 8 - 12, ......

= 2, 0, -2, -4, .........

The above list is an A.P. with common difference = -2.

C - B = 0 - 3, 4 - 6, 8 - 9, 12 - 12,.........

= -3, -2, -1, 0, .........

The above list is an A.P. with common difference = 1.

Hence, Option 4 is the correct option.

2k - 7, k + 5 and 3k + 2 are in A.P.; the value of k is :

1. -9

2. 6

3. -7

4. 5

Answer

Given,

2k - 7, k + 5 and 3k + 2 are in A.P.

∴ (k + 5) - (2k - 7) = (3k + 2) - (k + 5)

⇒ k - 2k + 5 + 7 = 3k - k + 2 - 5

⇒ -k + 12 = 2k - 3

⇒ 2k + k = 12 + 3

⇒ 3k = 15

⇒ k = $\dfrac{15}{3}$ = 5.

Hence, Option 4 is the correct option.

In an A.P. a = -36, d = 18 and l = 36, then n is :

1. 10

2. 5

3. 15

4. 20

Answer

Let nth be the last term.

By formula,

a_n = a + (n - 1)d

Substituting values we get :

⇒ 36 = -36 + (n - 1)18

⇒ 36 = -36 + 18n - 18

⇒ 36 + 36 + 18 = 18n

⇒ 18n = 90

⇒ n = $\dfrac{90}{18}$ = 5.

Hence, Option 2 is the correct option.

Do the numbers 1², 5², 7², 73 ..... form an A.P. ? If yes, its next term will be :

1. Yes, 11²

2. No

3. Yes, 97

4. Yes, 24

Answer

Given,

List = 1², 5², 7², 73 .....

= 1, 25, 49, 73, ........

Here,

a₂ - a₁ = 25 - 1 = 24

a₃ - a₂ = 49 - 25 = 24.

Since, common difference between consecutive terms are equal.

Hence, it forms an A.P.

Next term = 73 + d = 73 + 24 = 97.

Hence, Option 3 is the correct option.

The sum of first 10 even natural numbers is :

1. 120

2. 110

3. 65

4. 120

Answer

First 10 even natural numbers are : 2, 4, 6, 8, 10, .... upto 10 terms.

The above list is an A.P. with,

First term (a) = 2

Common difference (d) = 2

By formula,

Sum of n terms = $\dfrac{n}{2}$[2a + (n - 1)d]

Substituting values we get :

$= \dfrac{10}{2}[2 \times 2 + (10 - 1) \times 2] \\[1em] = 5 \times [4 + 9 \times 2] \\[1em] = 5 \times [4 + 18] \\[1em] = 5 \times 22 \\[1em] = 110.$

Hence, Option 2 is the correct option.

For the given numbers $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}............$

Assertion (A):

To find whether these terms form an A.P. or not. Express each term as the product of a natural number and $\sqrt{3 }$ i.e.,

$\sqrt{3} = 1 \times \sqrt{3}, \sqrt{12} = 2\sqrt{3}, \sqrt{27} = 3\sqrt{3}, \sqrt{48} = 4\sqrt{3}$, etc.

Reason (R): Since, for the given number difference between the consecutive term is same. It is an A.P.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, sequence

$\Rightarrow \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}............\\[1em] \Rightarrow \sqrt{3}, \sqrt{4 \times 3}, \sqrt{9 \times 3}, \sqrt{16 \times 3}............\\[1em] \Rightarrow \sqrt{3}, \sqrt{4} \times \sqrt{3}, \sqrt{9} \times \sqrt{3}, \sqrt{16} \times \sqrt{3}............\\[1em] \Rightarrow \sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}............$

Difference between first term and second term = $2\sqrt{3} - \sqrt{3} = \sqrt{3}$

Difference between second term and third term = $3\sqrt{3} - 2\sqrt{3} = \sqrt{3}$

So, the first term = $\sqrt{3}$, common difference = $\sqrt{3}$

So, assertion is true.

Since, for the given number difference between the consecutive term is same. Its an A.P., means reason is true and it clearly explain assertion.

Hence, option 3 is the correct option.

5, 8, 11, 14, ............... are in AP.

Assertion (A): $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,$ ............... are also in AP.

Reason (R): If each term of a given A.P. is divided by the same non zero number, the resulting sequence is an A.P..

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, 5, 8, 11, 14, ............... are in AP.

Here, first term = 5, common difference = 8 - 5 = 11 - 8 = 3

Now, new sequence : $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,.................$

The above sequence is found by dividing 5, 8, 11, 14, ............... the sequence by 2.

In new sequence,

Here,

Difference between second and first term = $4 - \dfrac{5}{2} = \dfrac{8 - 5}{2} = \dfrac{3}{2}$

Difference between third and second term = $\dfrac{11}{2} - 4 = \dfrac{11 - 8}{2} = \dfrac{3}{2}$

Difference between fourth and third term = $7 - \dfrac{11}{2} = \dfrac{14 - 11}{2} = \dfrac{3}{2}$

So, the common difference is same, means the given sequence is also in A.P..

So, Assertion is true.

The sequence $\dfrac{5}{2}, 4, \dfrac{11}{2}, 7,.................$ this sequence is found by each term of the A.P. 5, 8, 11, 14, ............... is divided by 2.

If each term of a given A.P. is divided by the same non-zero number, the resulting sequence is an A.P.

So, Reason is true.

Hence, option 3 is the correct option.

An A.P. with 3^rd term = -8 and 9^th term = 4.

Assertion (A): Common difference = -2.

Reason (R): If first term of the A.P. is a, then (a + 8d) - (a + 2d) = -8 - 4.

1. A is true, R is false.

2. Both A and R are false.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Given, in an A.P. 3^rd term = -8 and 9^th term = 4.

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

By formula :

⇒ a_n = a + (n - 1)d

⇒ a₃ = a + (3 - 1)d

⇒ -8 = a + 2d .................(1)

⇒ a₉ = a + (9 - 1)d

⇒ 4 = a + 8d .................(2)

Subtracting equation (1) from (2), we get :

⇒ (a + 8d) - (a + 2d) = 4 - (-8)

⇒ a + 8d - a - 2d = 4 + 8

⇒ 6d = 12

⇒ d = $\dfrac{12}{6}$

⇒ d = 2.

According to assertion d = -2, which is incorrect.

So, assertion (A) is false.

From above calculation, we get :

⇒ (a + 8d) - (a + 2d) = 4 - (-8)

⇒ (a + 8d) - (a + 2d) = 12

According to reason,

⇒ (a + 8d) - (a + 2d) = -8 - 4

⇒ (a + 8d) - (a + 2d) = -12, which is incorrect.

So, reason (R) is false.

Hence, option 2 is the correct option.

The n^th term of a sequence = 5n² - 3.

Statement (1): The sequence is an A.P.

Statement (2): If the n^th term of a sequence is not linear, the sequence does not form an A.P.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, a_n = 5n² - 3

⇒ a₁ = 5(1)² - 3 = 5 - 3 = 2

⇒ a₂ = 5(2)² - 3 = 5 × 4 - 3 = 20 - 3 = 17

⇒ a₃ = 5(3)² - 3 = 5 × 9 - 3 = 45 - 3 = 42.

Difference between terms :

⇒ a₃ - a₂ = 42 - 17 = 25

⇒ a₂ - a₁ = 17 - 2 = 15

Since, the difference between consecutive terms is not equal. Thus, the sequence is not in an A.P.

So, statement 1 is false.

The general form of an A.P. is a_n = a + (n - 1)d, which is a linear expression in n.

So, if T_n is not linear in n, then the sequence cannot be an A.P.

So, statement 2 is true.

Hence, option 4 is the correct option.

The sum of first 10 term of an A.P. = 3 and the sum of its first 15 term = 16.

Statement (1): The sum of first five terms of the given AP equals to 16 - 3 = 13.

Statement (2): The sum of last 5 terms of the given AP equals to sum of first 15 term minus sum of first 10 terms.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

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

Using the formula; S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Given, the sum of first 10 term of an A.P. = 3

$\Rightarrow S_{10} = 3 \\[1em] \Rightarrow \dfrac{10}{2}[2a + (10 - 1)d] = 3\\[1em] \Rightarrow 5[2a + 9d] = 3\\[1em] \Rightarrow 10a + 45d = 3 .............................(1)$

Given, the sum of its first 15 term = 16

$\Rightarrow \dfrac{15}{2}[2a + (15 - 1)d] = 16\\[1em] \Rightarrow \dfrac{15}{2}[2a + 14d] = 16\\[1em] \Rightarrow \dfrac{15}{2} \times 2[a + 7d] = 16 \\[1em] \Rightarrow 15[a + 7d] = 16\\[1em] \Rightarrow 15a + 105d = 16 .............................(2)$

Subtract equation (1) from (2),

⇒ (15a + 105d) - (10a + 45d) = 16 - 3

⇒ 15a + 105d - 10a - 45d = 13

⇒ 5a + 60d = 13.

Sum of first 5 terms = S₅

$S_5 = \dfrac{5}{2}[2a + (5 - 1)d]\\[1em] = \dfrac{5}{2}[2a + 4d]\\[1em] = \dfrac{5}{2}[2(a + 2d)]\\[1em] = 5[a + 2d]\\[1em] = 5a + 10d.$

Since, value of 5a + 10d cannot be equal to 13.

So, statement 1 is false.

The sum of last 5 terms of the given AP = a₁₁ + a₁₂ + a₁₃ + a₁₄ + a₁₅

= [a + (11 - 1)d] + [a + (12 - 1)d] + [a + (13 - 1)d] + [a + (14 - 1)d] + [a + (15 - 1)d]

= (a + 10d) + (a + 11d) + (a + 12d) + (a + 13d) + (a + 14d)

= 5a + 60d.

We have calculated earlier that,

Sum of first 15 term - Sum of first 10 terms = 5a + 60d

Thus, we can say that the sum of last 5 terms of the given AP equals to sum of first 15 term - sum of first 10 terms.

So, statement 2 is true.

Hence, option 4 is the correct option.

The 6^th term of an A.P. is 16 and the 14^th term is 32. Determine the 36^th term.

Answer

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

Given,

⇒ a₆ = a + (6 - 1)d

⇒ a + 5d = 16 ........(i)

⇒ a₁₄ = a + (14 - 1)d

⇒ a + 13d = 32 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 13d - (a + 5d) = 32 - 16

⇒ 8d = 16

⇒ d = 2.

Substituting value of d in (i) we get,

⇒ a + 5(2) = 16

⇒ a + 10 = 16

⇒ a = 6.

a₃₆ = a + (36 - 1)d = 6 + (35)(2) = 76.

Hence, 36^th term = 76.

If the third and the 9^th terms of an A.P. be 4 and -8 respectively, find which term is zero?

Answer

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

Given,

⇒ a₃ = a + (3 - 1)d

⇒ a + 2d = 4 ........(i)

⇒ a₉ = a + (9 - 1)d

⇒ a + 8d = -8 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 8d - (a + 2d) = -8 - 4

⇒ 8d - 2d = -12

⇒ 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 = 0

⇒ a + (n - 1)d = 0

⇒ 8 + (n - 1)(-2) = 0

⇒ 8 - 2n + 2 = 0

⇒ 2n = 10

⇒ n = 5.

Hence, 5^th term = 0.

An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term of the A.P.

Answer

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

Given,

⇒ a₃ = a + (3 - 1)d

⇒ a + 2d = 12 ........(i)

⇒ a₅₀ = a + (50 - 1)d

⇒ a + 49d = 106 ........(ii)

Subtracting (i) from (ii) we get,

⇒ a + 49d - (a + 2d) = 106 - 12

⇒ 49d - 2d = 94

⇒ 47d = 94

⇒ d = 2.

Substituting value of d in (i) we get,

⇒ a + 2(2) = 12

⇒ a + 4 = 12

⇒ a = 8.

a₂₉ = a + (29 - 1)d

a + 28d = 8 + 28(2) = 8 + 56 = 64.

Hence, 29^th term = 64.

Find the arithmetic mean of :

(i) -5 and 41

(ii) 3x - 2y and 3x + 2y

(iii) (m + n)² and (m - n)²

Answer

(i) Arithmetic mean = $\dfrac{-5 + 41}{2} = \dfrac{36}{2}$ = 18.

Hence, arithmetic mean between -5 and 41 = 18.

(ii) Arithmetic mean = $\dfrac{3x - 2y + 3x + 2y}{2} = \dfrac{6x}{2}$ = 3x.

Hence, arithmetic mean between 3x - 2y and 3x + 2y = 3x.

(iii)

Arithmetic mean = $\dfrac{(m + n)^2 + (m - n)^2}{2}$

$\phantom{Arithmetic mean }= \dfrac{m^2+ n^2 + 2mn + m^2 + n^2 - 2mn}{2} \\[1em] \phantom{Arithmetic mean }= \dfrac{2(m^2 + n^2)}{2} \\[1em] \phantom{Arithmetic mean }= m^2 + n^2.$.

Hence, arithmetic mean between (m + n)² and (m - n)² = m² + n².

Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 57.

Answer

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{20}{2}(3 + 57) \\[1em] = 10 \times 60 \\[1em] = 600.$

Hence, sum of first 20 terms = 600.

How many terms of the series 18 + 15 + 12 + ....... when added together will give 45?

Answer

Let n terms be added to get sum of 45.

In above A.P., a = 18 and d = 15 - 18 = -3.

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 45 = \dfrac{n}{2}(2 \times 18 + (n - 1) \times (-3)) \\[1em] \Rightarrow 45 = \dfrac{n}{2}(36 - 3n + 3) \\[1em] \Rightarrow 45 = \dfrac{n}{2}(39 - 3n)\\[1em] \Rightarrow 90 = n(39 - 3n) \\[1em] \Rightarrow 3n(13 - n) = 90 \\[1em] \Rightarrow n(13 - n) = 30 \\[1em] \Rightarrow 13n - n^2 = 30 \\[1em] \Rightarrow n^2 - 13n + 30 = 0 \\[1em] \Rightarrow n^2 - 10n - 3n + 30 = 0 \\[1em] \Rightarrow n(n - 10) - 3(n - 10) = 0 \\[1em] \Rightarrow (n - 3)(n - 10) = 0 \\[1em] \Rightarrow n = 3, 10.$

Hence, n = 3 or 10.

The n^th term of sequence is 8 - 5n. Show that the sequence is an A.P.

Answer

a_n = 8 - 5n

a₁ = 8 - 5(1) = 8 - 5 = 3

a₂ = 8 - 5(2) = 8 - 10 = -2

a₃ = 8 - 5(3) = 8 - 15 = -7

a₄ = 8 - 5(4) = 8 - 20 = -12

Sequence = 3, -2, -7, -12...........

In above sequence,

a₂ - a₁ = -2 - 3 = -5

a₃ - a₂ = -7 - (-2) = -5

Hence, the above sequence is an A.P. with common difference = -5.

Find the general term (n^th term) and 23^rd term of the sequence, 3, 1, -1, -3, .........

Answer

The above A.P. has common difference = 1 - 3 = -2 and a = 3.

a_n = a + (n - 1)d

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

= 3 - 2n + 2

= 5 - 2n.

a₂₃ = 5 - 2(23) = 5 - 46 = -41.

Hence, a_n = 5 - 2n and a₂₃ = -41.

Is -150 a term of 11, 8, 5, 2, .........?

Answer

In above sequence,

8 - 11 = -3 and 5 - 8 = -3.

Hence, above sequence is an A.P. with common difference = -3.

Let -150 be n^th term of the A.P.

∴ a_n = -150

⇒ a + (n - 1)d = -150

⇒ 11 + (-3)(n - 1) = -150

⇒ 11 - 3n + 3 = -150

⇒ 14 - 3n = -150

⇒ -3n = -164

⇒ n = $\dfrac{164}{3} = 54\dfrac{2}{3}$.

Since, no. of terms cannot be in fraction.

Hence, -150 is not a term of 11, 8, 5, 2, .........

How many multiples of 4 lie between 10 and 250?

Answer

Multiples of 4 lying between 10 and 250 are,

12, 16, 20, 24, .........., 248.

The above sequence is an A.P. with d = 4.

Let 248 be n^th term.

∴ a_n = 248

⇒ 12 + 4(n - 1) = 248

⇒ 12 + 4n - 4 = 248

⇒ 4n + 8 = 248

⇒ 4n = 240

⇒ n = 60.

Hence 60, multiples of 4 lie between 10 and 250.

The sum of the 4^th term and the 8^th term of an A.P. is 24 and the sum of the 6^th and 10^th term is 44. Find the first three terms of the A.P.

Answer

Given,

⇒ a₄ + a₈ = 24

⇒ a + (4 - 1)d + a + (8 - 1)d = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24 .........(i)

Also,

⇒ a₆ + a₁₀ = 44

⇒ a + (6 - 1)d + a + (10 - 1)d = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44 .........(ii)

Subtracting (i) from (ii) we get,

⇒ 2a + 14d - (2a + 10d) = 44 - 24

⇒ 4d = 20

⇒ d = 5.

Substituting value of d in (i) we get,

⇒ 2a + 10(5) = 24

⇒ 2a = 24 - 50

⇒ 2a = -26

⇒ a = -13

First three terms of A.P. = a, a + d, a + 2d.

= -13, -13 + 5, -13 + 2(5)

Hence, the first three terms of A.P. are -13, -8, -3.

The sum of first 14 terms of an A.P. is 1050 and its 14^th term is 140. Find the 20^th term.

Answer

Given,

⇒ a₁₄ = 140

⇒ a + (14 - 1)d = 140

⇒ a + 13d = 140

⇒ a = 140 - 13d .......(i)

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 1050 = \dfrac{14}{2}[2 \times (140 - 13d) + (14 - 1)d] \\[1em] \Rightarrow 1050 = 7[280 - 26d + 13d] \\[1em] \Rightarrow 1050 = 7[280 - 13d] \\[1em] \Rightarrow 1050 = 1960 - 91d \\[1em] \Rightarrow 91d = 1960 - 1050 \\[1em] \Rightarrow 91d = 910 \\[1em] \Rightarrow d = 10.$

Substituting value of d in (i) we get,

⇒ a = 140 - 13(10) = 140 - 130 = 10.

a₂₀ = a + 19d = 10 + 19(10) = 200.

Hence, 20^th term = 200.

The 25^th term of an A.P. exceeds its 9^th term by 16. Find its common difference.

Answer

Given,

⇒ a₂₅ - a₉ = 16

⇒ a + (25 - 1)d - (a + (9 - 1)d) = 16

⇒ a - a + 24d - 8d = 16

⇒ 16d = 16

⇒ d = 1.

Hence, common difference = 1.

For an A.P., show that :

(m + n)^th term + (m - n)^th term = 2 × m^th term.

Answer

L.H.S. = a_{m + n} + a_{m - n}

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

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

= 2a + 2md - 2d

= 2(a + md - d)

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

= 2a_m = R.H.S.

Hence, proved that (m + n)^th term + (m - n)^th term = 2 × m^th term.

If the n^th term of the A.P. 58, 60, 62, ...... is equal to the n^th term of the A.P. -2, 5, 12, ......., find the value of n.

Answer

n^th term of the A.P. 58, 60, 62, ......

In above sequence,

60 - 58 = 2 and 62 - 60 = 2.

Hence, above sequence is an A.P. with common difference = 2.

a_n = a + (n - 1)d

= 58 + (n - 1)2

= 58 + 2n - 2

= 2n + 56.

n^th term of the A.P. -2, 5, 12, .......

In above sequence,

5 - (-2) = 7 and 12 - 5 = 7.

Hence, above sequence is an A.P. with common difference = 7.

a_n = a + (n - 1)d

= -2 + (n - 1)7

= -2 + 7n - 7

= 7n - 9.

Since, n^th term of the A.P. 58, 60, 62, ...... is equal to the n^th term of the A.P. -2, 5, 12, .......

∴ 2n + 56 = 7n - 9

⇒ 7n - 2n = 56 + 9

⇒ 5n = 65

⇒ n = 13.

Hence, n = 13.

Which term of the A.P. 105, 101, 97, ........., is the first negative term?

Answer

Let the first negative term be n.

⇒ a_n < 0

⇒ a + (n - 1)d < 0

⇒ 105 + (n - 1)(-4) < 0

⇒ 105 - 4n + 4 < 0

⇒ 109 - 4n < 0

⇒ 4n > 109

⇒ n > $\dfrac{109}{4} = 27\dfrac{1}{4}.$

Given, n > $27\dfrac{1}{4}$.

So after 27^th term next terms will be negative so the first term to be negative will be 28^th.

Hence, 28^th term is the first negative term.

Divide 216 into three parts which are in A.P. and the product of two smaller parts is 5040.

Answer

Let numbers be a - d, a, a + d

Sum = 216

⇒ a - d + a + a + d = 216

⇒ 3a = 216

⇒ a = 72.

According to question,

⇒ (a - d)(a) = 5040

⇒ (72 - d)72 = 5040

⇒ 5184 - 72d = 5040

⇒ 72d = 5184 - 5040

⇒ 72d = 144

⇒ d = 2.

Numbers = (72 - 2), 72, (72 + 2)

Hence, numbers are 70, 72, 74.

Can 2n² - 7 be the n^th term of an A.P. Explain.

Answer

Given,

a_n = 2n² - 7

a₁ = 2(1)² - 7 = 2 - 7 = -5

a₂ = 2(2)² - 7 = 8 - 7 = 1

a³ = 2(3)² - 7 = 18 - 7 = 11

Here, a₂ - a₁ = 1 - (-5) = 6 and a₃ - a₂ = 11 - 1 = 10.

Since, difference is not common.

Hence, 2n² - 7 cannot be the n^th term of an A.P.

Find the sum of the A.P.; 14, 21, 28, ......., 168.

Answer

Let 168 be n^th term.

⇒ a_n = a + (n - 1)d

⇒ 168 = 14 + (n - 1)7

⇒ 168 = 14 + 7n - 7

⇒ 168 = 7n + 7

⇒ 161 = 7n

⇒ n = 23.

$S = \dfrac{n}{2}(a + l) \\[1em] = \dfrac{23}{2}(14 + 168) \\[1em] = \dfrac{23}{2} \times 182 \\[1em] = 23 \times 91 \\[1em] = 2093.$

Hence, n = 23 and sum = 2093.

The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^st term of this A.P.

Answer

$\Rightarrow S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 2100 = \dfrac{7}{2}[2 \times 20 + (7 - 1)d] \\[1em] \Rightarrow 2100 = \dfrac{7}{2}[40 + 6d] \\[1em] \Rightarrow 2100 = 7(20 + 3d) \\[1em] \Rightarrow 300 = 20 + 3d \\[1em] \Rightarrow 3d = 280 \\[1em] \Rightarrow d = \dfrac{280}{3}.$

We know that,

a_n = a + (n - 1)d

⇒ 31^st term = a₃₁

= a + (31 - 1)d

= $20 + 30 \times \dfrac{280}{3}$

= 20 + 2800

= 2820.

Hence, 31^st term of A.P. = 2820.

Find the sum of last 8 terms of the A.P.

-12, -10, -8, ........, 58.

Answer

Sum of last 8 terms of A.P. -12, -10, -8, ........, 58 = Sum of first 8 terms of A.P. 58, 56, 54, ........., -10, -12.

In A.P. 58, 56, 54, ........., -10, -12,

$S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] = \dfrac{8}{2}[2 \times 58 + (8 - 1) \times (-2)] \\[1em] = 4[116 - 14] \\[1em] = 4 \times 102 \\[1em] = 408.$

Hence, sum = 408.