Let a_1,a_2,a_3,,a_n be in A.P.

Question

Let $a_1,a_2,a_3,\ldots,a_n$ be in A.P. If $a_3+a_7+a_{11}+a_{15}=72$, then the sum of its first 17 terms is equal to:

Accepted Answer

$a_3+a_{15}=2a_9$ and $a_7+a_{11}=2a_9$ (each pair is symmetric about the 9th term), so $4a_9=72\Rightarrow a_9=18$.

$$S_{17}=\frac{17}{2}(a_1+a_{17})=17a_9=17\times18=306.$$

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