Arithmetic Progression

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.

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

Hence, Option 2 is the correct option.

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.

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 = $\frac{10}{5}$ = 2.

Hence, Option 2 is the correct option.

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 = $\frac{80}{5}$ = 16.

Hence, Option 2 is the correct option.

Given,

a_n = 2n - 3

So, a₁₅ = 2(15) - 3 = 30 - 3 = 27.

Hence, fifteenth term = 27.

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 .......

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.

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

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

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,

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

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

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,

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

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

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

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

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

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 = $\frac{399}{5}$

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

Hence, 402 is not a term of the sequence.

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

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

$$\begin{aligned}\phantom{Common difference}= \frac{38 - 31}{4} \\[1em] \phantom{Common difference}= \frac{7}{4} \\[1em] \phantom{Common difference}= 1\frac{3}{4}\end{aligned}$$

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,

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

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

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.

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.

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

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

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

a_n = a + (n - 1)d

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

Hence, no. terms in the series = 10.

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.

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.

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.

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.

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, ...........

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.

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.

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.

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 = $\frac{5}{2}$.

Substituting d in (iv) we get,

⇒ a + $7 \times \frac{5}{2}$ = 17

⇒ a + $\frac{35}{2}$ = 17

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

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

$$\begin{aligned}= -\frac{1}{2}, -\frac{1}{2} + \frac{5}{2}, -\frac{1}{2} + 2 \times \frac{5}{2} ......... \\[1em] = -\frac{1}{2}, \frac{-1 + 5}{2}, -\frac{1}{2} + 5.......... \\[1em] = -\frac{1}{2}, 2, \frac{9}{2} ........\end{aligned}$$

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

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.

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.

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.

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.

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 = $\frac{72}{24}$ = 3.

Hence, Option 4 is the correct option.

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 = $\frac{105}{5}$ = 21.

⇒ n - 4 = 21 - 4 = 17.

Hence, Option 1 is the correct option.

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.

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.

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 .

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.

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.

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.

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.

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, .........

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.

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.

(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.

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

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 = $\frac{3d}{2}$ ........(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,

$$\begin{aligned}\Rightarrow -\frac{3d}{2} + 2d = 1 \\[1em] \Rightarrow \frac{-3d + 4d}{2} = 1 \\[1em] \Rightarrow \frac{d}{2} = 1 \\[1em] \Rightarrow d = 2.\end{aligned}$$

Substituting value of d in (i) we get,

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

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

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,........

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 = $\frac{m - n}{n - m} = \frac{m - n}{-(m - n)}$ = -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).

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.

No. of terms = 41

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

Given,

Middle term = 40

⇒ a₂₁ = 40

⇒ a + (21 - 1)d = 40

⇒ a + 20d = 40 ............(1)

By formula,

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

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

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

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

= 41 × 40 ..........[From (1)]

= 1640.

Hence, Option 2 is the correct option.

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 = $\frac{n}{2}[2a + (n - 1)d]$

Sum of the above A.P.

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

Hence, Option 4 is the correct option.

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 = $\frac{n}{2}[2a + (n - 1)d]$

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

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

Hence, Option 1 is the correct option.

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

First term (a) = 7

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

By formula,

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

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

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

Hence, Option 2 is the correct option.

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.

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

We know that,

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

Let sum of n terms be 78.

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

Hence, no. of terms = 4 or 13.

Given,

a_n = 8n - 5

So,

a₁ = 8(1) - 5 = 3,

a₂₈ = 8(28) - 5 = 219.

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

Hence, sum = 3108.

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,

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

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

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.

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

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

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.

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

Hence, sum of first 51 terms = 5610.

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,

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

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

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

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.

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

Hence, sum of n terms = n².

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

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

Given, l = a₄₀ = 45

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

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

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

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.

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

Hence, sum = 52542.

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.

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

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

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,

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

Substituting value of n in (i),

⇒ (33 - 1)d = -42

⇒ 32d = -42

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

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

Given,

18, a, (b - 3) are in A.P.

In an A.P., consecutive differences are equal.

⇒ a - 18 = (b - 3) - a

⇒ a - 18 = b - 3 - a

⇒ a + a - 18 + 3 = b

⇒ 2a - 15 = b

⇒ 2a - b = 15

Hence, the value of 2a - b = 15.

Given,

a₄ = 9

⇒ a + (4 - 1)d = 9

⇒ a + 3d = 9

⇒ a = 9 - 3d ........(1)

Given,

Sum of a₆ + a₁₃ = 40

⇒ a + (6 - 1)d + a + (13 - 1)d = 40

⇒ a + 5d + a + 12d = 40

⇒ 2a + 17d = 40 ......(2)

Substituting value of a from equation (1) in equation (2), we get :

⇒ 2(9 − 3d) + 17d = 40

⇒ 18 − 6d + 17d = 40

⇒ 18 + 11d = 40

⇒ 11d = 22

⇒ d = 2.

Substituting the value of d in equation (1), we get :

⇒ a = 9 - 3(2)

⇒ a = 9 - 6

⇒ a = 3.

So the A.P. is,

3, 5, 7, 9, 11, ....

Hence, A.P is 3, 5, 7, 9, 11, .....

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.

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, ...........

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

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

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.

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.

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.

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) = $\frac{440}{8}$

⇒ 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.

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°.

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.

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

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.

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

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

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

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

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

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.

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 or 71, 69, 67.

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.

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.

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

Hence, arithmetic mean between 3 and 13 = 8.

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 = $\frac{n}{2}[2 \times 120° + (n - 1) \times 5°]$

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

Hence, number of sides = 9 or 16.

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

Then,

T_p = a + (p − 1)d = q …(1)

T_q = a + (q − 1)d = p …(2)

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

⇒ a + (p − 1)d − [a + (q − 1)d] = q − p

⇒ (p − 1)d − (q − 1)d = q − p

⇒ pd - d - qd + d = q - p

⇒ pd - qd = q - p

⇒ (p − q)d = q − p

⇒ (p − q)d = −(p − q)

⇒ d = $\frac{−(p − q)}{(p − q)}$

⇒ d = -1.

Substituting d = −1 in equation (1), we get :

⇒ a + (p − 1)(−1) = q

⇒ a − p + 1 = q

⇒ a = p + q − 1.

Now, the (p + q)^th term :

T_{p + q} = a + (p + q − 1)d

= (p + q − 1) + (p + q − 1)(−1)

= (p + q − 1) − (p + q − 1)

= p - p + q - q - 1 + 1

= 0.

Hence, the (p + q)^th term of the A.P. is zero.

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

The nth term of an A.P is given by, a_n = t + (n - 1) d

pth term = t + (p - 1)d = a ....(1)

qth term = t + (q - 1)d = b ....(2)

rth term = t + (r - 1)d = c ....(3)

Subtracting (2) from (1), we obtain :

⇒ t + (p - 1)d - [t + (q - 1)d] = a - b

⇒ t + (p - 1)d - t - (q - 1)d = a - b

⇒ (p - 1 - q + 1)d = a - b

⇒ (p - q)d = a - b

⇒ d = $\frac{(a - b)}{(p - q)}$ ....(4)

Subtracting (3) from (2), we obtain :

⇒ t + (q - 1)d - [t + (r - 1)d] = b - c

⇒ t + (q - 1)d - t - (r - 1) d = b - c

⇒ (q - 1 - r + 1)d = b - c

⇒ (q - r)d = b - c

⇒ d = $\frac{b - c}{q - r}$ ....(5)

From (4) and (5), we get:

$\frac{(a - b)}{(p - q)} = \frac{b - c}{q - r}$

⇒ (a - b)(q - r ) = (b - c)( p - q)

⇒ aq - ar - bq + br = bp - bq - cp + cq

⇒ bp - cp + cq - aq + ar - br - bq + bq = 0

⇒ (-aq + ar) + (bp - br) + (-cp + cq) = 0

⇒ -a(q - r) - b(r - p) - c(p - q) = 0

⇒ a(q - r) + b(r - p) + c(p - q) = 0

Hence, proved that a(q - r) + b(r − p) + c(p − q) = 0.

Since $\frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b}$ are in A.P.,

$$\begin{aligned}\Rightarrow \frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a} \\[1em] \Rightarrow \frac{(b + c) - (c + a)}{(c + a)(b + c)} = \frac{(c + a) - (a + b)}{(a + b)(c + a)} \\[1em] \Rightarrow \frac{b - a}{(c + a)(b + c)} = \frac{c - b}{(a + b)(c + a)} \\[1em] \Rightarrow \frac{b - a}{b + c} = \frac{c - b}{a + b} \\[1em]\end{aligned}$$

⇒ (b − a)(a + b) = (c − b)(b + c)

⇒ ab + b² − a² − ab = bc + c² − b² − bc

⇒ b² − a² = c² − b²

⇒ b² − a² = c² − b²

We have,

b² − a² = c² − b²

Since consecutive differences are equal,

Therefore,

a², b², c² are in A.P.

Hence, a², b², c² are in A.P.

Total saving in 10 years = ₹ 7,65,000

Each year he saved ₹ 6,000 more than previous year

d = 6000

Let his first year saving be ₹ a.

Number of years (n) = 10

By formula,

$S_n = \frac{n}{2}[2a + (n - 1)d]$

Substituting values, we get :

$$\begin{aligned}\Rightarrow 765000 = \frac{10}{2}[2a + 9(6000)] \\[1em] \Rightarrow 765000 = 5[2a + 54000] \\[1em] \Rightarrow 765000 = 10a + 270000 \\[1em] \Rightarrow 10a = 765000 - 270000 \\[1em] \Rightarrow 10a = 495000 \\[1em] \Rightarrow a = \frac{495000}{10} \\[1em] \Rightarrow a = 49500\end{aligned}$$

Savings in the first seven years :

$$\begin{aligned}\Rightarrow S_7 = \frac{7}{2}[2(49500) + (7 - 1)(6000)] \\[1em] = 3.5[99000 + 6(6000)] \\[1em] = 3.5(99000 + 36000) \\[1em] = 3.5(135000) \\[1em] = ₹ 4,72,500.\end{aligned}$$

Hence, amount saved in first 7 years = ₹ 4,72,500

By formula,

$S_n = \frac{n}{2}[2a + (n - 1)d]$

where, a = first term, d = common difference

$$\begin{aligned}\Rightarrow S_{12} = \frac{12}{2}[2a + 11d] \\[1em] \Rightarrow S_{12} = 6(2a + 11d) \\[1em] \Rightarrow S_{12} = 12a + 66d \text{ ....(1)} \\[1em] \Rightarrow S_{8} = \frac{8}{2}[2a + 7d] \\[1em] \Rightarrow S_{8} = 4(2a + 7d) \\[1em] \Rightarrow S_{8} = 8a + 28d \text{ ....(2)} \\[1em] \Rightarrow S_{4} = \frac{4}{2}[2a + 3d] \\[1em] \Rightarrow S_{4} = 2(2a + 3d) \\[1em] \Rightarrow S_{4} = 4a + 6d \text{ ....(3)}\end{aligned}$$

Subtracting equation (3) from equation (2), we get :

⇒ S₈ - S₄ = (8a + 28d) - (4a + 6d)

⇒ S₈ - S₄ = 8a + 28d - 4a - 6d

⇒ S₈ - S₄ = 4a + 22d

Multiplying the above equation by 3, we get :

⇒ 3(S₈ - S₄) = 3(4a + 22d)

⇒ 3(S₈ - S₄) = 12a + 66d

⇒ 3(S₈ - S₄) = S₁₂

Hence, proved that S₁₂ = 3(S₈ − S₄).

We know that,

a_n = a + (n - 1)d

Given,

18^th term of an A.P. is equal to 4 times its 4^th term

a₁₈ = a + 17d

a₄ = a + 3d

⇒ a + 17d = 4(a + 3d)

⇒ a + 17d = 4a + 12d

⇒ a - 4a + 17d - 12d = 0

⇒ -3a + 5d = 0

⇒ 5d = 3a

⇒ a = $\frac{5d}{3}$ .....(1)

Given,

6th term exceeds twice 2nd term by 4

a₆ = a + 5d

a₂ = a + d

⇒ a + 5d = 2(a + d) + 4

⇒ a + 5d = 2a + 2d + 4

⇒ a + 5d - 2a - 2d = 4

⇒ 3d - a = 4

⇒ 3d - a = 4 ...(2)

Substituting value of a from equation (1) in equation (2), we get :

⇒ 3d - $\frac{5d}{3}$ = 4

⇒ $\frac{9d - 5d}{3}$ = 4

⇒ 4d = 12

⇒ d = $\frac{12}{4}$

⇒ d = 3

Substituting value of d in equation (1), we get :

⇒ a = $\frac{5d}{3}$

⇒ a = $\frac{5(3)}{3}$

⇒ a = 5

By formula,

$S_n = \frac{n}{2}[2a + (n - 1)d]$

where, a = first term, d = common difference

$$\begin{aligned}\Rightarrow S_{9} = \frac{9}{2}[2(5) + 8(3)] \\[1em] = \frac{9}{2}[10 + 24] \\[1em] = \frac{9}{2}[34] \\[1em] = 9 \times 17 \\[1em] = 153.\end{aligned}$$

Hence, the sum of the first 9 terms of this A.P. = 153.

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.

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 = $\frac{90}{18}$ = 5.

Hence, Option 2 is the correct option.

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.

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 = $\frac{n}{2}$[2a + (n - 1)d]

Substituting values we get :

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

Hence, Option 2 is the correct option.

Given, sequence

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

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.

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

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

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

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 = $4 - \frac{5}{2} = \frac{8 - 5}{2} = \frac{3}{2}$

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

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

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

So, Assertion is true.

The sequence $\frac{5}{2}, 4, \frac{11}{2}, 7,.................$ 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.

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 = $\frac{12}{6}$

⇒ 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.

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.

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 = $\frac{n}{2}[2a + (n - 1)d]$

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

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

Given, the sum of its first 15 term = 16

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

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

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

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.

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.

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.

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.

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

Hence, sum of first 20 terms = 600.

Let n terms be added to get sum of 45.

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

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

Hence, n = 3 or 10.

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.

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 = $\frac{164}{3} = 54\frac{2}{3}$.

Since, no. of terms cannot be in fraction.

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

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.

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.

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.

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 > $\frac{109}{4} = 27\frac{1}{4}.$

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

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.

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.

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.

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

We know that,

a_n = a + (n - 1)d

⇒ 31^st term = a₃₁

= a + (31 - 1)d

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

= 20 + 2800

= 2820.

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

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,

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

Hence, sum = 408.

Number of terms in A.P. (n) = 57

Given,

We know that,

a_n = a + (n - 1)d

⇒ a₇ = a + 6d

⇒ a + 6d = 13 ....(1)

⇒ a₅₇ = a + 56d

⇒ a + 56d = 138 ....(2)

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

⇒ (a + 56d) - (a + 6d) = 138 - 13

⇒ 50d = 125

⇒ d = $\frac{125}{50}$

⇒ d = 2.5

Substituting value of d in equation (1), we get :

⇒ a + 6(2.5) = 13

⇒ a + 15 = 13

⇒ a = 13 - 15

⇒ a = -2

45th term:

⇒ a₄₅ = a + (45 - 1)d

⇒ a₄₅ = -2 + (45 - 1)(2.5)

⇒ a₄₅ = -2 + 110

⇒ a₄₅ = 108.

Hence, the 45^th term of this A.P. = 108.

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

We know that,

a_n = a + (n - 1)d

Given,

Ten times the tenth term of an A.P. is equal to fifteen times its fifteenth term.

⇒ 10a₁₀ = 15a₁₅

⇒ 10(a + 9d) = 15(a + 14d)

⇒ 10a + 90d = 15a + 210d

⇒ 10a - 15a + 90d - 210d = 0

⇒ -5a - 120d = 0

⇒ 5a + 120d = 0

⇒ 5(a + 24d) = 0

⇒ a + 24d = 0

⇒ a = -24d .....(1)

25th term :

⇒ a₂₅ = a + (25 - 1)d

⇒ a₂₅ = -24d + (24)d [From equation 1]

⇒ a₂₅ = 0

Hence, the twenty-fifth term of this A.P. = 0

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,........

(a) Given,

23, $21\frac{1}{2}$, 20, .... is A.P.

a = 23

d = $21\frac{1}{2} - 23 = -1\frac{1}{2} = -\frac{3}{2}$.

We know that,

a_n = a + (n - 1)d

$$\begin{aligned}\Rightarrow a_n = 23 - \Big(\frac{3}{2}\Big)(n - 1) \\[1em] \Rightarrow a_n = 23 - \Big(\frac{3n - 3}{2}\Big) \\[1em] \Rightarrow a_n = \Big(\frac{46 - 3n + 3}{2}\Big) \\[1em] \Rightarrow a_n = \Big(\frac{49 - 3n}{2}\Big)\end{aligned}$$

Hence, a_n = $\Big(\frac{49 - 3n}{2}\Big)$ .

(b) The last positive term occurs when :

⇒ a_n > 0

⇒ $\frac{49 - 3n}{2} \gt 0$

⇒ 49 - 3n > 0

⇒ 3n < 49

⇒ n < $\frac{49}{3}$

⇒ n < $16\frac{1}{3}$

Thus, 16th term will be the last positive term of the sequence.

⇒ a₁₆ = $\Big(\frac{49 - 3 \times 16}{2}\Big) = \frac{49 - 48}{2} = \frac{1}{2}$ = 0.5

Hence, the 16th term is the last positive term.

(i) Given,

Total cable length = 5000 m

Distance of first pole from base (a) = 200 m

Distance between poles (d) = 150 m

So poles form an A.P. : 200, 350, 500, 650,…

Distance of 10th pole from base,

We know that,

a_n = a + (n - 1)d

⇒ a₁₀ = 200 + (9) × 150

= 200 + 1350

= 1550 m.

Hence, distance of 10th pole from base = 1550 m.

(ii) Distance between 15th and 25th poles,

We know that,

a_n = a + (n - 1)d

⇒ a₁₅ = 200 + (14)150

= 200 + 2100

= 2300 m.

⇒ a₂₅ = 200 + (24)150

= 200 + 3600

= 3800 m.

Distance between 15th and 25th poles = 3800 - 2300 = 1500 m.

Hence, distance between 15th and 25th poles is 1500 m.

(iii) Let the total number of poles be n.

The last pole is 300 m from the top (5000 m point), so its distance from the base is : 5000 - 300 = 4700 m.

∴ a_n = 4700

⇒ a + (n - 1)d = 4700

⇒ 200 + (n - 1)150 = 4700

⇒ 150(n - 1) = 4700 - 200

⇒ 150(n - 1) = 4500

⇒ n - 1 = $\frac{4500}{150}$

⇒ n - 1 = 30

⇒ n = 30 + 1 = 31.

Thus, there are 31 poles in total.

Thus, the 15th pole from the top = (31 - 15 + 1) = 17th pole from the base.

Distance between 31st and 17th pole = (31 - 17)d = 14d = 14 × 150 = 2100 m.

Total distance from top to the 15th pole from the top = 2100 + 300 = 2400 m

Speed = 5 m/s

Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{2400}{5}$ = 480 s or 8 mins.

Hence, time to reach 15th pole from top = 8 minutes.

(i) Given,

The first circular row has 30 seats.

a = 30

Each next row has 10 more seats (d) = 10

So the rows form an A.P.: 30, 40, 50, 60,…

Seats in 10th row :

We know that,

a_n = a + (n - 1)d

⇒ a₁₀ = 30 + (9) × 10

= 30 + 90

= 120.

Hence, there are 120 seats in the 10th row.

(ii) Let n rows be required for 1500 seats.

By formula,

$S_n = \frac{n}{2}[2a + (n - 1)d]$

Substituting values, we get :

$$\begin{aligned}\Rightarrow 1500 = \frac{n}{2}[2(30) + 10n - 10] \\[1em] \Rightarrow 1500 = \frac{n}{2}[60 + 10n - 10] \\[1em] \Rightarrow 1500 = \frac{n}{2}[50 + 10n] \\[1em] \Rightarrow 3000 = 10n[5 + n] \\[1em] \Rightarrow \frac{3000}{10} = n(n + 5) \\[1em] \Rightarrow 300 = n(n + 5) \\[1em] \Rightarrow 300 = n^2 + 5n \\[1em] \Rightarrow n^2 + 5n - 300 = 0 \\[1em] \Rightarrow n^2 + 20n - 15n - 300 = 0 \\[1em] \Rightarrow n(n + 20) - 15(n + 20) = 0 \\[1em] \Rightarrow (n - 15)(n + 20) = 0 \\[1em] \Rightarrow n - 15 = 0 \text{ or } n + 20 = 0 \\[1em] \Rightarrow n = -20 \text{ or } n = 15.\end{aligned}$$

Number rows cannot be negative, thus : n = 15

Hence, rows required for 1500 seats = 15.

(iii) If there are 17 rows, seats in the middle row :

Since 17 is odd,

Middle row number = $\frac{17 + 1}{2} = \frac{18}{2} = 9$.

9th term :

We know that,

a_n = a + (n - 1)d

⇒ a₉ = 30 + (8)10

= 30 + 80

= 110.

Hence, 110 seats will be there in the middle row.

(i) Number of triangles in rows are in A.P.:

1, 3, 5, 7, 9,....

We know that,

a_n = a + (n - 1)d

⇒ a₁₅ = 1 + (15 - 1).2

= 1 + 14.2

= 1 + 28

= 29.

Hence, 29 triangles will be there in the 15^th row.

(ii) We know that,

a_n = a + (n - 1)d

Let 47 triangles be there in n rows.

⇒ 47 = 1 + (n - 1).2

⇒ 47 - 1 = 2(n - 1)

⇒ $\frac{46}{2}$ = (n - 1)

⇒ 23 = (n - 1)

⇒ n = 23 + 1

⇒ n = 24.

Hence, 47 triangles occur in 24th row.

(iii) In first row, there is 1 dark shaded triangle, in 2nd row there are 2, in 3rd row there are 3.

Thus, the dark shaded triangle are in A.P. with first term (a) = 1 and common difference (d) = 1.

By formula,

$S_n = \frac{n}{2}[2a + (n - 1)d]$

Substituting values, we get :

$$\begin{aligned}\Rightarrow S_{15} = \frac{15}{2}[2(1) + (15 - 1)1] \\[1em] = \frac{15}{2} \times [2 + 14] \\[1em] = \frac{15}{2} \times 16 \\[1em] = 15 \times 8 \\[1em] = 120.\end{aligned}$$

Hence, total number of dark shaded triangles in the first 15 rows = 120.