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The sum of n terms of an A.P. is ${3 n}^{2} + 5 n$ then its 16th term is,

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100

answer is C.

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Detailed Solution

Given that,
The sum of n terms of an A.P. is

3 n 2 + 5 n .

The nth term of an arithmetic progression is given by

a n = a + n − 1 d

.
The first term is a.
The common difference is d.
The number of terms are n.
For

n = 1

S 1 = 312 + 51 ⇒ S 1 = 3 + 5 ⇒ S 1 = 8

Therefore the first term is,

a = 8

.
For

n = 2

S 2 = 322 + 52 ⇒ S 2 = 12 + 10 ⇒ S 2 = 22

Therefore the second term is,

a 2 = S 2 − S 1 ⇒ a 2 = 22 − 8 ⇒ a 2 = 14

For

n = 3

S 3 = 332 + 53 = 27 + 15 = 42

Therefore the third term is,

a 3 = S 3 − S 2 ⇒ a 3 = 42 − 22 ⇒ a 3 = 20

Thus, the arithmetic progression is, 8,14,20,... .
Here,

a = 8 d = 6

Therefore, the 16th term is,

a 16 = 8 + 16 − 1 6 ⇒ a 16 = 8 + 90 ⇒ a 16 = 98

The

16 t h

term is 98.
Hence, option 3 is correct.

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