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AP is 5, 15, 25,…

First term = a = 5

Common difference = d = 15 – 5 = 10

Find 31^st term:

a₃₁ = a + (n – 1)d

= 5 + (31 – 1) 10

= 5 + 30 x 10

= 305

Required term = 305 + 130 = 435

Now, say 435 be the nth term, then

a_n = a + (n – 1)d

435 = 5 + (n – 1)10

435 – 5 = (n – 1)10

n – 1 = 43

n = 44

The required term will be 44th term.

To Find : 28^th term from the end of the AP.

Given: The AP is 7, 2, –3, –8, –13, …., –113

a₁ = 7, a₂ = 2, d = 2–7 = –5 and l = –113

Formula Used: n^th term from the end = l– (n–1)d

(Where l is last term and d is common difference of given AP)

By using n^th term from the end = l– (n–1)d formula

16^th term from the end = (–113) – 15d → (–113)–15 × (–5) = –38

So 16^th term from the end is equal to –38.

Given AP is 3, 8, 13, 18,…

First term = a = 3

Common difference = d = 8 – 3 = 5

And n = 20 and a₂₀ be the 20^th term, then

a₂₀ = a + (n – 1)d

= 3 + (20 – 1) 5

= 3 + 95

= 98

The required term = 98 + 55 = 153

Now, 153 be the nth term, then

a_n = a + (n – 1)d

153 = 3 + (n – 1) x 5

153 – 3 = 5(n – 1)

150 = 5(n – 1)

n – 1 = 30

n = 31

Required term will be 31st term.

Given: p^th term is q and (p + q)^th term is 0.

To prove: q^th term is p.

p^th term is given by

q = a + (p - 1) × d……equation1

(p + q)^th term is given by

0 = a + (p + q - 1) × d

0 = a + (p - 1) × d + q × d

Using equation1

0 = q + q × d

d = - 1

Put in equation1 we get

a = q + p - 1

q^th term is

⟹ q + p - 1 + (q - 1) × ( - 1)

⟹ p

Hence proved.

The given AP is 5, \(4\frac{1}{2},4,3\frac{1}{2},3\),........

First term = 5

Common difference = \(4\frac{1}{2}-5\Rightarrow \frac{9}{2}-5\) \(\Rightarrow \frac{9\,-\,10}{2}=-\frac{1}{2}\)

∴ a = 5 and d = \(-\frac{1}{2}\)

Now, T₂₅ = a + (25 - 1) d = a + 24d

\(=5+24\times(-\frac{1}{2})=5-12=-7\)

∴ 25^th term = -7

Given: The sum of first n terms.

To Find: The r^th term.

Let the first term be a and common difference be d

Put n = 1 to get the first term

a = S₁ = 3 + 2 = 5

Put n = 2 to get a + (a + d)2a + d = 12 + 4 = 1610 + d = 16d = 6t_r = a + (r - 1)d t_r = 5 + (r - 1)6 = 5 + 6r - 6 = 6r - 1

The r^th term is given by 6r - 1.

Show that: 18^th term of the AP is zero.

Given: 7a₇ = 11a₁₁

(Where a₇ is Seventh term, a₁₁ is Eleventh term, a_n is n^th term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

7(a + 6d) = 11(a + 10d)

7a + 42d = 11a + 110d → 68d = (–4a)

a + 17d = 0 ….equation (i)

Now a₁₈ = a + (18 - 1)d

So a + 17d = 0 [by using equation (i)]

HENCE PROVED

[NOTE: If n times the n^th term of AP is equal to m times the m^th term of same AP then its (m + n)^th term is equal to zero]

The 35th term of the AP 20, 17, 14, 11, ……

Given: AP is 20, 17, 14, 11, ……

Here, first term = a = 20

Common difference = d = 17 – 20 = -3

n = 35

a_n = a + (n-1)d

a₃₅ = 20 + (35-1)(-3)

= -82

Given: AP is 9, 13, 17, 21, ……

Here, first term = a = 9

Common difference = d = 13 – 9 = 4

a_n = a + (n-1)d

a₂₀= 9 + (20-1)4

= 85

To Find: The three numbers which are in AP.

Given: Sum and product of three numbers are 27 and 648 respectively.

Let required number be (a - d), (a), (a + d).

Then,

(a - d) + a + (a + d) = 27

⇒ 3a = 27

⇒ a = 9

Thus, the numbers are (9 - d), 9 and (9 + d).

But their product is 648.

∴ (9 - d) × 9 × (9 + d)= 648

⇒ (9 - d)(9 + d)= 72

⇒ 81 – d² = 72

⇒ d² = 9

⇒ d = ± 3

When d = 3 numbers are 6, 9, 12

When d = (– 3) numbers are 12, 9, 6

So, Numbers are 6, 9, 12 or 12, 9, 6.

To Find: The angles of a quadrilateral.

Given: Angles of a quadrilateral are in AP with common difference = 10°.

Let the required angles be a, (a + 10°), (a + 20°) and (a + 30°).

Then, a + (a + 10°) + (a + 20°) + (a + 30°) = 360°

⇒ 4a + 60° = 360°

⇒ a = 75°

NOTE: Sum of angles of quadrilateral is equal to 360°

So Angles of a quadrilateral are 75°, 85°, 95° and 105°.

To Find: The three numbers which are in AP.

Given: Sum and sum of the squares of three numbers are 21 and 165 respectively.

Let required number be (a - d), (a), (a + d).

Then,

(a – d) + a + (a + d) = 21

⇒ 3a = 21 

⇒ a = 7

Thus, the numbers are (7 - d), 7 and (7 + d).

But their sum of the squares of three numbers is 165.

∴ (7 - d)² + 7² + (7 + d)² = 165

⇒ 49 + d² –14d + 49 + d² + 14d = 116

⇒ 2d² = 18

⇒ d² = 9

⇒ d = ± 3

When d = 3 numbers are 4, 7, 10

When d = (– 3) numbers are 10, 7, 4

So, Numbers are 4, 7, 10 or 10, 7, 4.

Let d be the common difference.

Given:

first term = a = -4

last term = l = 29

Sum of all the terms = S_n = 150

S_n = n/2[a + l]

150 = n/2[-4 + 29]

n = 12

There are 12 terms in total.

Therefore, 29 is the 12th term of the AP.

Now, 29 = -4 + (12 – 1)d

29 = -4 + 11d

d = 3

The common difference is 3.

(i) The cost of digging a well for the first metre = Rs. 150

The cost of digging a well for the second metre = 150 + 20 = Rs. 170

The cost of digging a well for the third metre = 170 + 20 = Rs. 190

The cost of digging a well for the fourth metre = 190 + 20 = Rs. 210

Difference between first and second metre = 170 – 150 = 20

Difference between second and third metre = 190 – 170 = 20

Difference between third and fourth metre = 210 – 190 = 20

Since, all the differences are equal.

Therefore, it’s an A.P.

(ii) Let the amount of air present in the cylinder first time = x

Amount of air present in the cylinder second time = x - \(\frac{x}{4}\) = \(\frac{3x}{4}\)

Amount of air present in the cylinder third time = \(\frac{3x}{4}\) - \(\frac{3x}{16}\) = \(\frac{9x}{16}\)

Difference in the amount of air present in the cylinder between first and the second time = \(\frac{3x}{4}\) – x = \(\frac{-x}{4}\)

Difference in the amount of air present in the cylinder between second and the third time = \(\frac{9x}{16}\) - \(\frac{3x}{4}\) = \(\frac{-3x}{16}\)

Since, the differences are unequal.

Therefore, it’s not an A.P.

 The given A.P is √2 + √8 + √18 + √32 +....

The simplified A.P.  is 

√2,2√2,3√2,4√2, .......

Here a = √2

d = 2√2 - √2 = √2

S_n = \(\frac{n}{2}\)(2a + (n–1) d) 

= \(\frac{n}{2}\)(2√2 + (n–1)√2) 

= √2n/2 (1 + n) 

= n (n + 1)/√2

S_n = 4n² + 2n 

First term, a = 4(1)² + 2(1) = 4 + 2 = 6 

Sum of first two terms, S₂ = 4(2)² + 2(2) 

a + a + d = 16 + 4 

6 + 6 + d = 20 

d = 8 

Now the n^th term, a_n = a + (n – 1) d 

= 6 + (n – 1) 8 

= 6 + 8n – 8 

= 8n - 2

Here, S_n = 3n² + 5n 

S₁ = a₁ = 3 + 5 = 8 

S₂ = a₁ + a₂ = 12 + 10 = 22 

⇒ a₂ = S₂ – S₁ = 22 – 8 = 14 

S₃ = a₁ + a₂ + a₃ = 27 + 15 = 42

⇒a₃ = S₃ – S₂ = 42 – 22 = 20 

∴ Given AP is 8, 14, 20, ..... Thus 

a = 8, d = 6 

Given t_m = 164. 

164 = [a + (n –1)d]

164 = [(8) + (m –1)6] 

164 = [8 + 6m – 6] 

164 = [2 + 6m] 

162 = 6m 

m = 162 / 6. 

∴ m = 27.

a_n= a + (n–1) d = 2n + 1 (given)…………(1) 

S_n= n/2 (2a + (n–1) d) 

By putting n=1 in (1) 

a=3 

similarly a₂= 5 

a₃= 7 

d= common difference = a₂ – a₁

= 2S_n = n/2 (6 + (n–1) 2) 

= n/2 (2n + 4) 

= n (n + 2)

Given that: 

S_q= 63q – 3q² 

Put q = 1 

S₁ = T₁ = 63 – 3 = 60 

Put q = 2 

S₂ = 63 x 2 – 3 x (2)² = 126 – 12 = 114 

T₂ = S₂ – S₁ = 114 – 60 = 54

Put q = 3 

S₃ = 63 x 3 – 3(3)² 

= 189 – 27 = 162 

T₃ = S₃ – S₂ 

= 162 – 114 = 48

Therefore, first term of this A.P. is 60 and Common difference is 54 - 60 = -6 

T_p = a + (p – 1) d 

-60 = 60 + (p – 1) (-6) 

-120 = (p – 1) (-6) 

(p – 1) = 20

p = 21 

Now 11^th term of this A.P 

T₁₁ = 60 + (11 – 1) (-6) 

= 60 – 60 = 0

S = 3n² + 5nS₁ 

= a₁ = 3 + 5 = 8S₂ 

= a₁ + a₂ = 12 + 10 = 22 

⇒ a₂ = S₂ – S₁ = 22 – 8 

= 14S₃ = a₁ + a₂ + a₃ 

= 27 + 15 = 42

⇒ a₃ = S₃ – S₂ 

= 42 – 22 = 20

∴ Given AP is 8,14,20,.....Thus a = 8, d = 6

Given t_m = 164.

164 = [a + (n –1)d]

164 = [(8) + (m –1)6]

164 = [8 + 6m – 6]

164 = [2 + 6m]

162= 6mm = 162/6.

∴ m = 27.

Given that 

S_n = 5n² + 3n 

Put n = 1 

S₁= T₁= 5 + 3 = 8 

Put n = 2 

S₂= 5(2)² + 3 + 2 = 26 

T₂= S₂ – S₁ = 26 – 8 = 18 

S₃= 5(3)² + 3 + 3 = 54 

T₃= S₃ – S₂ = 54 – 26

= 28 

Therefore, first term, a = 8 and common difference = 18 – 8 = 10 

T_m = a + (m – 1) d 

168 = 8 + (m – 1) 10 

168 = 8 + 10m – 10 

170 = 10m 

m = 17 

T₂₀ = 8 + (20 – 1) 10 

= 8 + 19 x 10 = 198

Let the three parts of the number 207 are

a₁ = a - d

a₂ = a

a₃ = a + d

Clearly a₁, a₂ and a₃ are in AP with common difference as d.

Now, by given condition,

Sum = 207

a₁ + a₂ + a₃ = 207

(a - d) + a + (a + d) = 207

3a = 207

a = 69

Also,

a₁a₂ = 4623

(a - d)a = 4623

(69 - d)69 = 4623

69 - d = 67

d = 69 - 67

d = 2

Hence, required three parts are 67, 69, 71.

Given,

Sum of 7 terms of an A.P. is 49

⟹ S₇ = 49

And, sum of 17 terms of an A.P. is 289

⟹ S₁₇ = 289

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

And, we know that the sum of n terms of an A.P is

S_n =\(\frac{ n}{2}\)[2a + (n − 1)d]

So,

S₇ = 49 = \(\frac{7}{2}\)[2a + (7 – 1)d]

= \(\frac{7}{2}\) [2a + 6d]

= 7[a + 3d]

⟹ 7a + 21d = 49

a + 3d = 7 ….. (i)

Similarly,

S₁₇ = \(\frac{17}{2}\)[2a + (17 – 1)d]

= \(\frac{17}{2}\) [2a + 16d]

= 17[a + 8d]

⟹ 17[a + 8d] = 289

a + 8d = 17 ….. (ii)

Now, subtracting (i) from (ii), we have

a + 8d – (a + 3d) = 17 – 7

5d = 10

d = 2

Putting d in (i), we find a

a + 3(2) = 7

a = 7 – 6 = 1

So, for the A.P: a = 1 and d = 2

For the sum of n terms is given by,

S_n  = \(\frac{n}{2}\)[2(1) + (n − 1)(2)]

= \(\frac{n}{2}\)[2 + 2n – 2]

= \(\frac{n}{2}\)[2n]

= n²

Therefore, the sum of n terms of the A.P is given by n².

a₂= 14 

a + d = 14 ..... (i) 

a₃= 18 

a + 2d = 18 ...... (ii)

Subtracting (i) from (ii), we get d = 4

Putting the value of d in (i), we get a = 14 – 4 = 10

\(S_{25} = \frac{25}{2}[2(a) + 24(d)]\)

= 25 [58] 

= 1450

The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :

23, 21, 19, . . ., 5

It forms an AP (Why?). Let the number of rows in the flower bed be n.

Then a = 23, d = 21 – 23 = – 2, a_n = 5

As, a_n = a + (n – 1) d

We have, 5 = 23 + (n – 1)(– 2)

i.e., – 18 = (n – 1)(– 2)

i.e., n = 10

So, there are 10 rows in the flower bed.

S₇= 49

\(\frac{7}{2}\)[2a + 6d] = 49

a + 3d = 7 (i) 

S₁₇= 289

\(\frac{17}{2}[2a + 16d] =289\)

a + 8d = 17 (ii) Subtract 

(i) from (ii), 

we get 

5d = 10 

d = 2 

Put d = 2 in (i), 

we get a = 7 – 6 = 1

Sn= \(\frac{n}{2}\)[2(1) + (n – 1)2]

= n [1 + n – 1] 

= n²

In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers: 147, 148, 149, 150, ...

a₂ – a₁ = 148 – 147 = 1

a₃ – a₂ = 149 – 148 = 1

a₄ – a₃ = 150 – 149 = 1

Here, the difference of any two consecutive terms in each case is -1. So, the given list is an AP whose first term a is 147 and common difference d is 1.

To Find: 23^rd term of the AP

Given: The series is 7, 5, 3, 1, –1, –3, …

a₁ = 7, a₂ = 5 and d = 3–5 = –2

(Where a = a₁ is first term, a₂ is second term, a_n is n^th term and d is common difference of given AP)

Formula Used: a_n = a + (n - 1)d

So put n = 23 in above formula, we have

a₂₃= a₁ + (23 - 1)(–2) = 7– 44 = –37

So 23^rd term of AP is equal to –37.

To Find: what amount will he pay in the 30^th instalment.

Given: first instalment = 10000 and it increases the instalment by 500 every month.

∴ So it form an AP with first term is 10000, common difference 500 and number of instalment is 30

Formula Used: T_n = a + (n - 1)d

(Where a is first term, T_n is n^th term and d is common difference of given AP)

∴ T_n = a + (n - 1)d

⇒ T_n = 10000 + (30 - 1)500

⇒ T_n = 10000 + 29 × 500

∴ T_n = 10000 + 14500

⇒ T_n = 24,500

So, he will pay 24,500 in the 30^th instalment.

Given: 1, 6, 11, 16, …

Here, a = 1

d = a₂ – a₁ = 6 – 1 = 5

and n = 16

We have,

t_n = a + (n – 1)d

So, t₁₆ = 1 + (16 – 1)5

= 1 + 15×5

t₁₆ = 1 +75

t₁₆ = 76

i. Given, t_n = 4n – 3

For n = 1, t₁ = 4(1) – 3 = 4 – 3 = 1

For n = 2, t₂ = 4(2) – 3 = 8 – 3 = 5

For n = 3, t₃ = 4(3) – 3 = 12 – 3 = 9

For n = 4, t₄ = 4(4) – 3 = 16 – 3 = 13

For n = 5, t₅ = 4(5) – 3 = 20 – 3 = 17 

\(\therefore\) The first five terms are 1, 5, 9, 13 and 17.

ii. Given, t_n = 2n – 5 For n = 1,

t₁ = 2(1) – 5 = –3

For n = 2, t₂ = 2(2) – 5 = –1

For n = 3, t₃ = 2(3) – 5 = 1

For n = 4, t₄ = 2(4) – 5 = 3

For n = 5, t₅ = 2(5) – 5 = 5

\(\therefore\) The first five terms are –3, –1, 1, 3 and 5.

iii. Given, t_n = n + 2

For n = 1, t₁ = 1 + 2 = 3

For n = 2, t₂ = 2 + 2 = 4

For n = 3, t₃ = 3 + 2 = 5

For n = 4, t₄ = 4 + 2 = 6

For n = 5, t₅ = 5 + 2 = 7

\(\therefore\) The first five terms are 3, 4, 5, 6 and 7.

iv. Given, t_n = n² – 2n

For n = 1, t₁ = (1)² – 2(1) = 1 – 2 = -1

For n = 2, t₂ = (2)² – 2(2) = 4 – 4 = 0

For n = 3, t₃ = (3)² – 2(3) = 9 – 6 = 3

For n = 4, t₄ = (4)² – 2(4) = 16 – 8 = 8

For n = 5, t₅ = (5)² – 2(5) = 25 – 10 = 15

\(\therefore\) The first five terms are –1, 0, 3, 8 and 15.

v. Given, t_n = n³

For n = 1, t₁ = (1)³ = 1

For n = 2, t₂ = (2)³ = 8

For n = 3, t₃ = (3)³ = 27

For n = 4, t₄ = (4)³ = 64

For n = 5, t₅ = (5)³ = 125

\(\therefore\) The first five terms are 1, 8, 27, 64 and 125.

The correct option is: (B) -490

Explanation:

We have, T_n = (7 -3n)

First term, T₁ = (7 -3 x 1) = 4

Second term, T₂ = 7 - 3 x 2= 1

Third term, T₃ = 7 -3 x 3 = -2

.'. Series is 4, 1, - 2, ..........

and common difference = -3

Sum of first 20 terms (S₂₀)

= 20/2[2 x 4 + (20 - 1)(-3)] = 10[8 - 57] = -490

Given, t_n = n³

For n = 1, t₁ = (1)³ = 1

For n = 2, t₂ = (2)³ = 8

For n = 3, t₃ = (3)³ = 27

For n = 4, t₄ = (4)³ = 64

For n = 5, t₅ = (5)³ = 125

\(\therefore\) The first five terms are 1, 8, 27, 64 and 125.

Given: a = 21 , d = –5

To find: t_n and t₂₅

We have,

t_n = a + (n – 1)d

t_n = 21 + (n – 1)(–5)

= 21 – 5n + 5

t_n = 26 – 5n

Now, n = 25

So, t₂₅ = 21 + (25 – 1)(–5)

= 21 + 24 × (–5)

t₂₅ = 21 – 120

t₂₅ = –99

Given, t_n = 3n – 4

For n = 1, t₁ = 3(1) – 4 = –1

For n = 2, t₂ = 3(2) – 4 = 6 – 4 = 2

Correct answer is

(B) 3.5

i. a = -1.25, d = 3 …[Given] 

t₁ = a = -1.25 

t₂ = t₁ + d = – 1.25 + 3 = 1.75 

t₃ = t₂ + d = 1.75 + 3 = 4.75

t₄ = t₃ + d = 4.75 + 3 = 7.75 

∴ The required A.P. is -1.25, 1.75, 4.75, 7.75,…

ii.  a = 6, d = -3 …[Given] 

∴ t₁ = a = 6 

t₂ = t₁ + d = 6 – 3 = 3 

t₃ = t₂ + d = 3 – 3 = 0 

t₄ = t₃ + d = 0- 3 = -3 

∴ The required A.P. is 6, 3, 0, -3,…

iii. a = -19, d = -4 …[Given] 

t₁ = a = -19 

t₂ = t₁ + d = -19 – 4 = -23 

t₃ = t₂ + d = -23 – 4 = -27 

t₄ = t₃ + d = -27 – 4 = -31 

∴ The required A.P. is -19, -23, -27, -31, …

Here, a = 3, d = 7 – 3 = 4 and l = 399

To find : n and 20^th term from the end

We have,

l = a + (n – 1)d

⇒ 399 = 3 + (n – 1) × 4

⇒ 399 – 3 = 4n – 4

⇒ 396 + 4 = 4n

⇒ 400 = 4n

⇒ n = 100

So, there are 100 terms in the given AP

Last term = 100^th

Second Last term = 100 – 1 = 99^th

Third last term = 100 – 2 = 98^th

And so, on

20^th term from the end = 100 – 19 = 81^st term

The 20^th term from the end will be the 81^st term.

So, t₈₁ = 3 + (81 – 1)(4)

t₈₁ = 3 + 80 × 4

t₈₁ = 3 + 320

t₈₁ = 323

Hence, the number of terms in the given AP is 100, and the 20^th term from the last is 323.

The even natural numbers are 2, 4, 6, 8, … 27 

The above sequence is an A.P.

∴ a = 2, d = 4 - 2 = 2, n = 123

         Now, S_n = n/2 [ 2a + (n - 1)d]

∴  S_n = 123/2 [2(2) + (123 - 1)(2)]

= 123/2 [2 (2) + 122 (2)]

= 123/2 x 2[2 + 122]

= 123 x 124

= 15252

∴ The sum of first 123 even natural numbers is 15252.

The three digit natural numbers divisible by 

5 are 100, 105, 110, …,995 

The above sequence is an A.P. 

∴ a = 100, d = 105 – 100 = 5 

Let the number of terms in the A.P. be n. Then, t_n = 995 

Since, t_n = a + (n – 1)d 

∴ 995 = 100 +(n – 1)5 

∴ 995 – 100 = (n – 1)5 

∴ 895 = (n – 1)5 

∴ n – 1 = 895/5

∴ n – 1 = 179 

∴ n = 179 + 1 = 180 

∴ There are 180 three digit natural numbers which are divisible by 5.

Correct answer is

(C) 137

a=(-3)

d=4-(-3)=7
t

Correct answer is

(A) 45

i. The given A.P. is 5, 1,-3,-7,… 

Here, t₁= 5, t₂= 1 

∴ a = t₁ = 5 and 

d = t_1 ^- t₂ = 1 – 5 = -4 

∴ first term (a) = 5, 

common difference (d) = -4

ii. The given A.P. is 0.6, 0.9, 1.2, 1.5,… 

Here, t₁ = 0.6, t₂ = 0.9 

∴ a = t₁ = 0.6 and 

d = t₂ – t₁ = 0.9 – 0.6 = 0.3 

∴ first term (a) = 0.6, 

common difference (d) = 0.3

iii. The given A.P. is 127, 135, 143, 151,… 

Here, t₁ = 127, t₂ = 135 

∴ a = t₁ = 127 and 

d = t₂ – t₁ = 135 – 127 = 8 

∴ first term (a) = 127, 

common difference (d) = 8

iv The given A,P is 1/4, 3/4, 5/4, 7/4, ...

Here, t₁ = 1/4, t₂ = 3/4

∴ a = t₁ = 1/4 and 

d = t₂ - t₁ = 3/4 - 1/4 = 2/4 = 1/2

∴ first terms (a) = 1/4,

common diference (d) = 1/2

The correct answer is : (C) 25 

Hints: 

t₁₈ - t₁₃= a + (18 - 1)d - [a + (13 - 1)d] 

= a + 17d - a - 12d 

= 5d = 5 x 5 = 25 

Correct answer is

(C) 25

The natural numbers from 1 to n are 1,2, 3, ……, n. 

The above sequence is an A.P. 

∴ a = 1, d = 2 – 1 = 1 

S_n = 36 …[Given] 

Now, S_n = [2a + (n – 1)d] 

∴ 36 = n/2 [2(1) + (n – 1)(1)] 

∴ 36 = (n/2) (2 + n – 1) 

∴ 36 × 2 = n (n + 1) 

∴ 72 = n (n + 1) 

∴ 72 = n² + n 

∴ n² + n – 72 = 0 

∴ n² + 9_n – 8n – 72 = 0 

∴ n(n + 9) – 8 (n + 9) = 0 

∴ (n + 9) (n – 8) = 0 

∴ n + 9 = 0 or n – 8 = 0 

∴ n = -9 or n = 8 But, n cannot be negative. 

∴ n = 8 

∴ The value of n is 8. 

According to the given condition,

mt_m = nt_n 

∴ m[a + (m – 1)d] = n[a + (n – 1)d] 

∴ ma + md(m – 1) = na + nd(n- 1) 

∴ ma + m²d – md = na + n²d – nd 

∴ ma + m²d – md – na – n²d + nd = 0 

∴ (ma – na) + (m²d – n²d) – (md – nd) = 0 

∴ a(m – n) + d(m² – n²) – d(m – n) = 0 

∴ a(m – n) + d(m + n) (m – n) – d(m – n) = 0 

∴ (m – n)[a + (m + n – 1) d] = 0 

∴ [a+ (m + n – 1)d] = 0 …[Dividing both sides by (m – n)] 

∴ t_{(m + n)} = 0 

∴ The (m + n)^th term of the A.P. is zero.

The correct answer is : (B) 38

Hints : 

S_n = n/2 (first terms + last terms)

∴ 399 = n/2 (1+20)

∴ 399 x 2 = 21n

∴ n = 798/21 = 38

The correct answer is : (C) 137

The correct answer is : (C) 25

Hints :  

t₁₈ - t₁₃ = a + (18 - 1)d - [a + (13 - 1)d]

= a + 17d - a - 12d

= 5d = 5 x 5 = 25

The correct answer is : (A) -75