the sum of the first n terms of an AP is $(3 n^{2} + 6 n) .$ Find the nth term and the 15th terms of this AP.

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Solution

Given the sum of n terms, $S_{n} = 3 n^{2} + 6 n$

$S_{1} = 3 (1)^{2} + 6 (1) = 3 + 6 = 9 = a$ ,the first term

$S_{2} = 3 (2)^{2} + 6 (2) = 12 + 12 = 24$

$S_{2} = a_{1} + a_{2}$

$a_{2} = S_{2} - a_{1} = 24 - 9 = 15$

common difference, $d = a_{2} - a_{1} = 15 - 9 = 6$

nth term of an AP, $a_{n} = a + (n - 1) d$

ie, $a_{n} = 9 + (n - 1) 6 = 9 + 6 n - 6 = 6 n + 3$

15th term of the AP, $a_{15} = 6 n + 3 = 6 (15) + 3 = 90 + 3 = 93$

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