The sum of three consecutive terms of an AP

To Find: The three numbers which are in AP.

Given: Sum and sum of the squares of three numbers are 21 and 165 respectively.

Let required number be $(a-d),(a),(a+d) .$ Then,

$(a-d)+a+(a+d)=21 \Rightarrow 3 a=21 \Rightarrow a=7$

Thus, the numbers are $(7-d), 7$ and $(7+d)$.

But their sum of the squares of three numbers is 165.

$\therefore(7-\mathrm{d})^{2}+7^{2}+(7+\mathrm{d})^{2}=165$

$\Rightarrow 49+d^{2}-14 d+49+d^{2}+14 d=116$

$\Rightarrow 2 d^{2}=18 \Rightarrow d^{2}=9 \Rightarrow d=\pm 3$

When $\mathrm{d}=3$ numbers are $4,7,10$

When $d=(-3)$ numbers are $10,7,4$

So, Numbers are $4,7,10$ or $10,7,4$.