Question

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

• A. 1
• B. 2
• C. 3
• D. -3

Answer 1

Answer: (A)

1

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 $0$