Summing to a perfect square

There are $81$ terms in the series, so, using the formula $\frac { 1 }{ 2 } n(a+l)$ for an arithmetic series: $S=\frac { 81 }{ 2 } (x+20+x+100)$

= $81(x+60)$ .

Now 81 is a perfect square, so $S$ is a perfect square if and only if $x+60$ is a perfect square. As x is a positive integer, the smallest possible value of x is $\boxed { 4 }$ .

Another way of solving this question is to find the sum of AP 20, 21, 22, ..., 100 = 4860

4860 + 81x = Perfect square. Any perfect square never ends with 2, 3, 7, or 8. Hence, x is ending with either 1, 4, 5, 6, 9, 0. Substituting x as 1 doesn't give you the answer, whereas substituting 4 gives it. Julian Uy has given a better solution.