Question

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

Answer 1

To find maximum value of $S = a _{ a _1}+ a _{ a _2}+\ldots \ldots \ldots+ a _{ a _5}$
Let $a _1=13-2 d ; a _2=13- d ; a _3=13 ; a _4=13+ d ; a _5=13+2 d$ $a_n=(13-2 d)+(n-1) d=(13-3 d)+n d$
$\Rightarrow \displaystyle\sum_{ n =1}^5 a _{ a _{ a }}= a _{13-2 d }+ a _{13- d }+ a _{13}+ a _{13+ d }+ a _{13+2 d }=5(13+10 d )$
Now $d \leq 6$ as terms are positive.
$\therefore S _{\max }$ occurs at $d =6 \Rightarrow S =5 \times 73=365$. ?