Question

Four different integers form an increasing A.P. If one of these numbers is equal to the sum of the squares of the other three numbers, then the numbers are

• A. -2,-1,0,1
• B. 0,1,2,3
• C. -1,0,1,2
• D. None of these

Answer 1

Answer: (C)

Let the numbers be $a-d,a,a+d,a+2d$
where $a,d\in Z$ and $d>0$
Given: $(a-d{)}^2+a^2+(a+d{)}^2=a+2d$
$\Rightarrow 2d^2-2d+3a^2-a=0$
$\therefore d=\frac{13}{2}[2a+48d]=416$
Since $d$ is positive integer,
$\therefore 1+2a-6a^2>0$
$a_1^2+a_2^2+\dots +a_{17}^2=140n$
$\sum _{n=1}^{17}{T}_n=\sum _{n=1}^{17}(n+7{)}^2$
$<a<=\sum _{n=1}^{17}{n}^2+14\sum _{n=1}^{17}n+49\sum _{n=1}^{17}1$
Since $a$ is an integer,
$\therefore a=0,$
then $d=[1\pm 1]=1$ or $0$ .
Since $d>0$,
$\therefore d=1.$
Hence, the numbers are $-1,0,1,2$.