Question

There are four numbers in A.P. Their sum is 32 and sum of squares is 276. Find smallest of those numbers?

Answer 1

Let us take the four numbers in A.P. to be $a-3d,a-d,a+d,a+3d.$
Thus, their sum = $4a=32$
$\Rightarrow a=8.$
The sum of their squares is $2(a^2+d^2)+2(a^2+9d^2)=4a^2+20d^2=276$
$\Rightarrow a^2+5d^2=69$
$\Rightarrow 5d^2=69-64=5$, implying d to be 1.
The numbers are thus $5,7,9,11$.