Q. 134.2( 113 Votes )

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

Answer :

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

Given, Sum of n terms of A.P. = S_{n} = 3n^{2} + 5n ………….(I)

m^{th} term of A.P. = t_{m} = 164 ……………………….(III)

⇒ t_{m} = a + (m – 1)d …………………..(IV)

Equating (I) and (II)

⇒ 2an + n^{2}d – dn = 6n^{2} + 5n

⇒ (2a – d)n + n^{2}d = 6n^{2} + 10n

Comparing the coefficients of n^{2}, we get

d = 6

Comparing the coefficients of n, we get

2a – d = 10

Putting the value d = 6

⇒ 2a – 6 = 10 ⇒ 2a = 10 + 6

⇒ 2a = 16

⇒ a = 8

m^{th} term = t_{m} = 164 = 8 + (m – 1)6

⇒ 164 – 8 = (m – 1)6

⇒ m – 1 = 156/6

⇒ m – 1 = 26

⇒ m = 26 + 1

∴ m = 27

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