Question

The sum of four consecutive terms of an A.P. is $20$ and the sum of their squares is $120$. Find those numbers

Answer 1

Let the terms be $a-3d,a-d,a+d,a+3d$.
Given, Sum $=20$
$\Rightarrow a-3d+a-d+a+d+a+3d=20$
$\Rightarrow 4a=20$
$\Rightarrow a=5$
Also given, sum of squares $=120$
$\Rightarrow (a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120$
$\Rightarrow (5-3d{)}^2+(5-d{)}^2+(5+d{)}^2+(5+3d{)}^2=120$
$\Rightarrow 25-30d+9d2+25-10d+d^2+25+10d+d^2$
$\Rightarrow 25+30d+9d^2=120$
$\Rightarrow 100+20d^2=120$
$\Rightarrow 20d^2=20$
$\Rightarrow d^2=1$
$\Rightarrow d=\pm 1$
The terms are $a-3d,a-d,a+d,a+3d$
Thus, $5-3,5-1,5+1,5+3$
(i.e.) $2,4,6,8$ or $8,6,4,2$