Question

Let four different integers form an increasing A.P. If one of these numbers is equal to the sum of squares of the other three numbers, then which of the following is (are) CORRECT?

• A. The product of all four numbers is $0$.
• B. The sum of all four numbers is $0$.
• C. The common difference of the A.P. is $1$.
• D. The smallest number among the four numbers is $-1$.

Answer 1

Answer: (A), (C), (D)

The correct options are
A The product of all four numbers is $0$.
C The common difference of the A.P. is $1$.
D The smallest number among the four numbers is $-1$.

Let the terms of the A.P. be
$a-d,a,a+d,a+2d,$
where $a$ and $d$ are integers and $d>0$

Now, $a+2d=(a-d{)}^2+a^2+(a+d{)}^2$
$\Rightarrow a+2d=3a^2+2d^2\Rightarrow 2d^2-2d+3a^2-a=0\cdots \left(1\right)$
As $d\in R$
$D\ge 0\Rightarrow 4-24a^2+8a\ge 0$
$\Rightarrow 1-6a^2+2a\ge 0\Rightarrow 6a^2-2a-1\le 0\therefore \frac{1-\sqrt{7}}{6}\le a\le \frac{1+\sqrt{7}}{6}$
$\Rightarrow a=0(\because a\text{is an integer})$
Using equation $\left(1\right)$,
$2d^2-2d=0\Rightarrow 2d(d-1)=0\Rightarrow d=1(\because d>0)$

Therefore, the four numbers are $-1,0,1,2$