Question

Find four numbers in A.P. whose sum is $20$ and the sum of whose squares is $120$.

Answer 1

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

Then, Sum of numbers $=20$
$⟹(a-3d)+(a-d)+(a+d)+(a+3d)=20⟹4a=20⟹a=5$

It is given that, sum of the squares $=120$
$⟹(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120$
$⟹4a^2+20d^2=120$
$⟹a^2+5d^2=30$
$⟹25+5d^2=30$
$⟹5d^2=5⟹d=\pm 1$
If $d=1$, the, the numbers are $2,4,6,8$.

If $d=-1$, then the numbers are $8,6,4,2$.

Thus, the numbers are $2,4,6,8$ or $8,6,4,2$.