If a_1,a_2,,a_n are in A.P.

Question

If $a_1,a_2,\ldots,a_n$ are in A.P. and $a_1+a_4+a_7+\cdots+a_{16}=114$, then $a_1+a_6+a_{11}+a_{16}$ is equal to:

Accepted Answer

$a_1+a_4+a_7+\cdots+a_{16}$ (6 terms, common gap $3d$) sums via pairing: $3(a_1+a_{16})=114 \Rightarrow a_1+a_{16}=38$.

$$a_1+a_6+a_{11}+a_{16} = 2(a_1+a_{16}) = 76.$$

Solutions · 1