An arithmetic progression where square of the fourth term is sum of the first three terms Here is the  given question:

Show that there can be infinite number of sets of four distinct real numbers in AP which are such that the square of the last term is equal to the sum of the first three terms.

What I did:  I chose 4 terms in AP like 
$a-d, a, a+d, a+2d$
and then I made the equation according to the question and got the result
$d = (-1) + \sqrt{2} \cdot a$.
So my question is this relation enough to show that there can be infinite sets of 4 real numbers. 

It depends. You did not specify exactly how you "got the result". If you have merely proven that the result holds, then you have not shown that there are infinitely many solutions. However, if you have proven that that equation is equivalent to all the original conditions, then yes because it gives you one solution for each choice of $a$.
