The sum of first n terms of an AP is $(4 n^{2} + 2 n)$ The nth term of this AP is
(a) (6n-2) (b) (7n-3) (c) (8 n - 2) (d) (8n + 2)

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Solution

$S_{n} = 4 n^{2} + 2 n S_{1} = 4 (1)^{2} + 2 (1) = 4 + 2 = 6 = a_{1} S_{2} = 4 (2)^{2} + 2 (2) = 16 + 4 = 20 S_{2} = a_{2} + a_{1} a_{2} = S_{2} - a_{1} = 20 - 6 = 14 d = a_{2} - a_{1} = 14 - 6 = 8 n^{t h} t e r m, a_{n} = a + (n - 1) d = 6 + (n - 1) 8 = 6 + 8 n - 8 = 8 n - 2$

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