Question

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

Answer 1

Let the four numbers in AP be,

$a-3d,a-d,a+d,a+3d$

Their sum => $4a=20$

=> $a=5$

Sum of their squares = $(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=120$

=> $4a^2+20d^2=120$

On putting value of $a$,

We get,

=> $d^2=1$

=> $d=1$ or $d=-1$

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