The sum of the first $n$ terms of an A.P is $(3 n^{2} + 6 n)$ . Find the $n t h$ term and $15 t h$ term of this A.P.

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Solution

We know that ,

Sum of the AP $= \frac{n}{2} [2 a + (n - 1) d] = 3 n^{2} + 6 n$ ....Where, ( $a = f i r s t t e r m$ and $d = C o m m o n d i f f .$ )

$⟹ \frac{n^{2} d}{2} + n (a - \frac{d}{2}) = 3 n^{2} + 6 n$

On comparing coefficients , we get,

$a = 9$ and $d = 6$

So, $n t h$ term $= a + (n - 1) d = 9 + (n - 1) (6)$

$⟹ n t h t e r m = 6 n + 3$

And $15 t h$ term $= a + 14 d = 9 + 14 (6) = 93$

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