Question

Find four numbers in $A.P.$ whose sum is $28$ and sum of whose square is $216.$

Answer 1

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

Given,

$a-3d+a-d+a+3d+a+d=28$

$\Rightarrow 4a=28,$

$\therefore a=7$

$(a-3d{)}^2+(a-d{)}^2+(a+3d{)}^2+(a+d{)}^2=216$

$2(a^2+9d^2)+2(a^2+d^2)=216$

$4a^2+20d^2=216$

$4\left(7^2\right)+20d^2=216$

$\Rightarrow d=\pm 1$

for $d=-1$

the series is $7-3(-1),7-(-1),7-1,7-3\Rightarrow 10,8,6,4$

For $d=1$

the series is $7-3,7-1,7+1,7+3\Rightarrow 4,6,8,10$

Therefore the numbers are, $4,6,8,10$