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Q.
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
Sequences and Series
Solution:
Let the numbers be $a-d, a, a+d, a+2 d$
where $a, d \in Z$ and $d>0$
Given: $(a-d)^{2}+a^{2}+(a+d)^{2}=a+2 d$
$\Rightarrow 2 d^{2}-2 d+3 a^{2}-a=0$
$\therefore d=\frac{13}{2}[2 a+48 d]=416$
Since $d$ is positive integer,
$\therefore 1+2 a-6 a^{2}>0$
$a_{1}^{2}+a_{2}^{2}+\ldots+a_{17}^{2}=140 n$
$\displaystyle\sum_{n=1}^{17} T_{n}=\displaystyle\sum_{n=1}^{17}(n+7)^{2}$
$\Rightarrow < a < =\displaystyle\sum_{n=1}^{17} n^{2}+14 \displaystyle\sum_{n=1}^{17} n+49 \sum_{n=1}^{17} 1$
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 $-1,0,1,2$.