Question

The sum of four consecutive numbers of an $A.P.$ is $20$ and sum of their squares is $120,$ then the absolute value of the common difference is

Answer 1

Let the required numbers be
$(a-3d),(a-d),(a+d),(a+3d)$

Now,
$(a-3d)+(a-d)+(a+d)+(a+3d)=20\Rightarrow 4a=20\Rightarrow a=5$
Also,
$(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120\Rightarrow 4a^2+20d^2=120\Rightarrow 100+20d^2=120\Rightarrow d=\pm 1$

Hence, the required numbers are $2,4,6,8$ or $8,6,4,2$, so the absolute value of common difference is $2$.