Question

Let $a_1,a_2,a_3,\dots \dots a_n$ be an increasing arithmetic progression of positive integers. If $a_3=13$, then the the maximum value of $\sum _{n=1}^5{a}_{a_n}$ is $M$. Find the value of $\frac{M}{73}$.

Answer 1

Answer: 5

To find maximum value of $S=a_{a_1}+a_{a_2}+\dots \dots \dots +a_{a_5}$
Let $a_1=13-2d;a_2=13-d;a_3=13;a_4=13+d;a_5=13+2d$ $a_n=(13-2d)+(n-1)d=(13-3d)+nd$
$\Rightarrow \sum _{n=1}^5{a}_{a_a}=a_{13-2d}+a_{13-d}+a_{13}+a_{13+d}+a_{13+2d}=5(13+10d)$
Now $d\le 6$ as terms are positive.
$\therefore S_{\max }$ occurs at $d=6\Rightarrow S=5\times 73=365$. ?