How to Solve this Arithmetic Progression Question?

Please help- Four different integers form an increasing AP.One of these numbers is equal to the sum of the squares of the other three numbers.Then- find all the four numbers.

I assumed the numbers to be $a,a+d,a+2d,a+3d$ and wlog let $a+3d$ this number and according to question-

$a^2+(a+d)^2+(a+2d)^2=a+3d$

but I could not solve the above equation.

Solution 1:

$$(a-d)^2+a^2+(a+d)^2=a+2d\iff3a^2+2d^2=a+2d$$

$$\iff3\left(a-\frac16\right)^2+2\left(d-\frac12\right)^2=\frac1{12}+\frac12=\frac7{12}$$

$$\iff(6a-1)^2+6(2d-1)^2=7$$

As $a,d$ are integers, $$6(2d-1)^2\le7\implies(2d-1)^2\le\dfrac76\implies d=0,1$$

Both cases $\implies (6a-1)^2=1\iff6a-1=\pm1\implies a=0$ as $a$ is an integer

Solution 2:

It might help to remember that the $n$th term of an AP is $a_1 + (n-1)d$ where $a_1$ is the first term of the AP and $d$ is the common difference. Setting $a_1^2 + (a_1 + d)^2+(a_1 +2d)^2=a_1 + 3d$ should help you get the answer.