The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its

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Here, the sum of first n terms is given by the expression, S_n = 3n^2 + 5n We ned to find which term of the A.P. is 164. Let us take 164 as the nth term. So we know that the nthterm of an A.P. is given by, a_n = S_n - S_( n-1) So, 164 = S_n - S_( n-1) `164 = 3n^2 + 5n - [ 3(n-1)^2 + 5(n - 1 ) ] Using the property, ( a - b)^2 = a^2 + n^2 - 2ab We get, 164 = 3n^2 + 5n - [3 (n^2 + 1 - 2n ) + 5 (n -1)] 164 = 3n^2 + 5n - [ 3n^2 + 3 - 6n + 5n - 5 ] 164 = 3n^2 + 5n - (3n^2 - n - 2) 164 = 3n^2 + 5n - 3n^2 + n + 2 ` 164 = 6n + 2 Further solving for n, we get 6n = 164 - 2 n = $\frac{162}{6}$ n = 27 Therefore, 164 is the 27th term of the given A.P.

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