The sum of n terms of an A.P. is $4 n^{2}$ + 2n. Find the nth term of this A.P.

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Solution

$S n = 4 n^{2} + 2 n$
$S_{1} = a_{1} = 4 + 2 = 6$
$S 2 = 4 x 2^{2} + 2 * 2$
$= 16 + 4 = 20$
$a 2 = S 2 - a 1$
$a 2 = 20 - 6$
$= 14$
$d = a 2 - a 1$
$= 14 - 6 = 8$
$a_{n} = a_{1} + (n - 1) d = 6 + (n - 1) 8 = 6 + 8 n - 8 = 8 n - 2 = 2 (4 n - 1)$

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