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

Solution:

Let the four numbers be

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

According to question:

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

$\Rightarrow 4 a-4 d+4 d=28$

$\Rightarrow a=7$

And

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

$a^{2}+9 d^{2}-6 a d+a^{2}+d^{2}-2 a d+a^{2}+d^{2}+2 a d+a^{2}+9 d^{2}+6 a d=216$

$\Rightarrow 4 a^{2}+20 d^{2}=216$

$\Rightarrow 4(7)^{2}+20 d^{2}=216$

$\Rightarrow d=\pm 1$

Four numbers are :

$d=1, a=7$

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

$7-3(1), 7-1,7+1,7+3(1)$

$4,6,8,10$

And

$d=-1, a=7$

$10,8,6,4$