Question

The sum of four consecutive numbers in an A.P. with common difference $d>0$ is 20. If the sum of their squares is 120, then find the numbers.

Answer 1

Let the numbers be $a-3d,a-d,a+d$ and $a+3d.$ These numbers form an AP with first term $a-3d$ and common difference $2d>0.$
Also, given that $(a-3d)+(a-d)+(a+d)+(a+3d)=20.$
$\Rightarrow 4a=20$
$\Rightarrow a=5$
Also, ${(a-3d)}^2+{(a-d)}^2+{(a+d)}^2+{(a+3d)}^2=120$
$\Rightarrow 4a^2+20d^2=120$
$\Rightarrow 4(5{)}^2+20d^2=120$
$\Rightarrow d^2=1$
$\Rightarrow d=\pm 1$
SIince $d>0,$ we must have $d=1.$
Hence, the numbers are 2, 4, 6, and 8.