Question

Let $a_1, a_2, a_3 \ldots \ldots a_n$ be an increasing A.P. of positive integers such that $a_3=10$, then if the maximum value of $S=\displaystyle\sum_{n=1}^4 a_{a_n}$ is $M$. Find the sum of digits of $\left(\frac{M}{10}\right)$.

Answer 1

$a_1=10-2 d \geq 1 \Rightarrow d \leq \frac{9}{2} d>0 $
$S=a_{a_1}+a_{a_2}+a_{a_3}+a_{a_4}$
$=a_1+\left(a_1-1\right) d+a_1+\left(a_2-1\right) d+a_1+\left(a_3-1\right) d+a_1+\left(a_4-1\right) d $
$=4 a_1+d\left(a_1+a_2+a_3+a_4-4\right) $
$=4 a_1+d\left(4 a_1+6 d-4\right)$
$=4 a_1(1+d)+6 d^2-4 d=4(10-2 d)(1+d)+6 d^2-4 d$
$=40-28 d-2 d^2$
$=40-2\left(d^2-14 d+48\right)+98$
$=138-2(d-7)^2 \text { for maximum sum } d=4$
for maximum sum $d=4$
$M = 120$
$\therefore\left[\frac{ M }{10}\right]=12 \Rightarrow$ sum of digits $=3$