The sum of first n terms of an A.P is $3 n^{2} + 4 n$ . Find its $n^{t h}$ term and A.P

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Solution

Given: $S_{n} = 3 n^{2} + 4 n$

$S_{n - 1} = 3 {(n - 1)}^{2} + 4 (n - 1)$

$n^{t h}$ term $= t_{n} = S_{n} - S_{n - 1}$

$= 3 n^{2} + 4 n - 3 {(n - 1)}^{2} - 4 (n - 1)$

$= 3 (n^{2} - {(n - 1)}^{2}) + 4 (n - n + 1)$

$= 3 (n - n + 1) (n + n - 1) + 4$

$= 3 (2 n - 1) + 4$

$= 6 n - 3 + 4 = 6 n + 1$

$∴ t_{n} = 6 n + 1$

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