Four different integers form an increasing A.P .One of these numbers is equal to the sum of the squares of the other three numbers. Then
The product of all numbers is

Options

A. `-2`
B. 1
C. 0
D. 2

Answer

C

Explanation

Let the four integers be a-d,a,a+d, and a+2d, where a and d are integers and `dgt0`. Now,
`a+2d=(a-d)^(2)+a^(2)+(a+d)^(2)`
or `2d^(2)-2d+3a^(2)-a=0` (1)
`therefored=1/2[1pmsqrt(1+2a-6a^(2))]` (2)
Since d is a positive integer, so
`1+2a-6a^(2)gt0`
or `6a^(2)-2a-1lt0`
or `(1-sqrt7)/6ltalt(1+sqrt7)/6`
or a=0 (`because` a is an integer)
Hence, from (2),
d=1 or 0
But since `dgt0`,
`therefore` d=1
Hence, the four numbers are -1,0,1,2.