Question

Four number are in arithmetic progression. If their sum is $20$ and sum of their square is $120$, then find the number.

Answer 1

Let the terms of AP be $a-3d,a-d,a+d,a+3d$

$a-3d+a-d+a+d+a+3d=20$

$4a=20$

$a=5$

${(a-3d)}^2+{(a-d)}^2+{(a+d)}^2+{(a+3d)}^2=1204a^2+20d^2=120100+20d^2=120d^2=1d=\pm 1$

So the numbers are $5,4,3,2$ or $5,6,7,8$