Question

Sum of squares of four numbers in $\text{A.P.}$ is $120.$ and there sum is $20.$Then what are the numbers if common difference is non-negative?

Answer 1

Let the four numbers in $\text{A.P.}$ be $(a-3d),(a-d),(a+d),(a-3d).$

Now, given that there sum is $20$.
$\Rightarrow a-3d+a-d+a+d+a+3d=20$
$\Rightarrow 4a=20$
$\Rightarrow a=5$

Also, sum of there squares is $120$.
$\Rightarrow (a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120$
$\Rightarrow (a-3d{)}^2+(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2=120$

$\Rightarrow 2(a^2+9d^2)+2(a^2+d^2)=120$
$\Rightarrow 4a^2+20d^2=120$
$\Rightarrow a^2+5d^2=30$
$\Rightarrow 25+5d^2=30$
$\Rightarrow d=\pm 1$

Since, common difference $d$ is non negative, $\Rightarrow d=1$

Hence, the numbers are $2,4,6,8.$