If the nth term of an AP is $(4 n + 1)$ , find the sum of the first $15$ terms of this AP. Also find the sum of is n terms.

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Solution

$n^{t h}$ term of $A P$ is $(4 n + 1)$
$1^{t h}$ term of $A P$ is $(4 (1) + 1) = 5$
$2^{t h}$ term of $A P$ is $(4 (2) + 1) = 9$
$3^{r d}$ term of $A P$ is $(4 (3) + 1) = 13$
$S_{15} = \frac{15}{2} [2 \times 5 + (15 - 1) 4]$
$=> \frac{15}{2} [10 + 56] = \frac{15}{2} \times 6 = 225$

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