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 product of all numbers is

• A. -2
• B. 1
• C. 0
• D. 2

Answer 1

The correct option is C 0

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 product of numbers is $0$