Question

Find four numbers in AP whose sum is $28$ and the sum of whose squares is $216$.

Answer 1

Let the four numbers be $(a-3d),(a-d),(a+d)$ and $(a+3d)$.
Given:
1. Their sum is $28$
$(a-3d)+(a-d)+(a+d)+(a+3d)=28$
$\Rightarrow 4a=28$
$\therefore a=7$.

2.Their sum of square $216$
$(a-3d{)}^2+(a-d{)}^2+(a+d{)}^2+(a+3d{)}^2=216$
$\Rightarrow a^2+9d^2-6ad+a^2+d^2-2ad+a^2+d^2+2ad+a^2+9d^2+6ad=216$
$\Rightarrow 4a^2+20d^2=216$
$\Rightarrow a^2+5d^2=54$
Putting the value of $a$
$49+5d^2=54$
$5d^2=5$
$d^2=1$
$\therefore d=\pm 1$.
For $d=1$:
The numbers are $7+3,7+1,7-1,7-3\Rightarrow 10,8,6,4$.
For $d=-1$:
The numbers are $7+3,7+1,7-1,7+3\Rightarrow 4,6,8,10$.