The sum of n terms of an A.P. is (2n+3n sq). Determine the AP and find its nth term.

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Solution

$S_{n} = 3 n^{2} + 2 n$
$n = 1$ , $s_{1} = 3 \times 2 = 5$
$n = 2$ $s_{2} = 3 \times 4 + 2 \times 2 = 16$
sum of $1^{s t}$ term and $2^{n d} t e r m = 16$
$s + 2^{n d} t e r m = 16 2^{n d} t e r m = 16 - 5 = 11$
Now, d = 11 – 5 = 6
Hence, AP = 5, 11, 17, 23....

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