Question

Four different integers form an increasing AP. One of these numbers is equal to the sum of the squares of the first three numbers, then the sum of all the four numbers is,

• A. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Let the number be $a-d,a,a+d,a+2d$

where $a,dϵZ$ and $d>0$

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

$\Rightarrow {2d}^2-2d+{3a}^2-a=0$

$\therefore d=\frac{1}{2}[1\pm \sqrt{1+2a-{6a}^2}]$

Since $d$ is positive integer $\therefore 1+2a-{6a}^2>0$

$\Rightarrow \left(\frac{1-\sqrt{7}}{6}\right)<a<\left(\frac{1+\sqrt{7}}{6}\right)$

Since $a$ is an integer

$\therefore a=0$

then $d=\frac{1}{2}[1\pm 1]=1$ or $0$

Hence numbers are $-1,0,1,2$

Therefore the sum of all four numbers is $2$.