Question

Find four numbers in A.P. whose sum is $32$ and sum of squares is $276$.

Answer 1

Let the numbers be: $a-3d,a-d,a+d,a+3d$

$a-3d+a-d+a+d+a+3d=32$

$4a=32$

$\therefore a=8$

$(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=276$

$4a^2+20d^2=276$

$20d^2=276-4(8{)}^2=20$

$\therefore d=1$

Therefore the 4 numbers are: $a-3d,a-d,a+d,a+3d=8-3,8-1,8+1,8+3=\{5,7,9,11\}$