Question

Four distinct integers from an increasing $A.P$. Such that one of them is sum of square of remaining numbers, then the largest no. is?

Answer 1

Let, the four distinct integers are,

$a-3d,a-d,a+d,a+3d$

According to the question,

$a+3d={(a-3d)}^2+{(a-d)}^2+{(a+d)}^2\Rightarrow {3a}^2-(6d+1)a+(11d^2-3d)=0⟶\left(1\right)$

for $a$ to be real,

${[-(6d+1\left)\right]}^2-4\times 3({11d}^2-3d)\ge 0\Rightarrow {-96d}^2+48d+1\ge 0\Rightarrow {96d}^2-48d-1\le 0$

If $d\ge 1$ the expression is greater than zero.

So, $0<d<1$

Now, by hit and trial method,

$d=\frac{1}{2}$

Putting $d=\frac{1}{2}$ in equation $\left(1\right)$,

${3a}^2-\left[\right(6\times \frac{1}{2})+1]a+11{\left(\frac{1}{2}\right)}^2-3\left(\frac{1}{2}\right)=0\Rightarrow {12a}^2-16a+5=0\Rightarrow a=\frac{1}{2}ora=\frac{5}{6}$

but for $a=\frac{5}{6}$

$a-3d,a-d,a+d,a+3d$ are not integers.

$\therefore a=\frac{1}{2}$

and the integers are $-1,0,1,2$

Hence, the largest number is $2.$