# If the sum of n terms of an A.P of 3n^{2} + 5n and its m^{th} term is 164, find the value of m

**Solution:**

Let a and d be the first term and the common difference of the given arithmetic progression respectively.

a_{m} = a + (m - 1) d = 164 ....(1)

Sum of n terms, S_{n} = n/2 [2a + (n - 1) d]

Here,

⇒ n/2 [2a + nd - d] = 3n^{2} + 5n

⇒ an + d/2 n^{2} - d/2 n = 3n^{2} + 5n

d/2 n^{2} + (a - d/2) n = 3n^{2} + 5n ....(2)

Comparing the coefficient of n^{2} on both sides in (2) , we obtain

⇒ d/2 = 3

⇒ d = 6

Comparing the coefficient of n on both sides in (2) , we obtain

⇒ a - d/2 = 5

⇒ a - 3 = 5

⇒ a = 8

Therefore, from (1)

⇒ 8 + (m - 1)6 = 164

⇒ (m - 1)6 = 164 - 8

⇒ (m - 1) = 156/6

⇒ m = 26 + 1

⇒ m = 27

Thus, the value of m = 27

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.2 Question 13

## If the sum of n terms of an A.P of 3n^{2} + 5n and its m^{th} term is 164 , find the value of m

**Summary:**

We knew that the m^{th} term of this series is 164, therefore the value of m is 27

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