Sum to n terms in GP is $$a\cdot\frac{\left(r^n-1\right)}{r-1}$$

The sum of the first 18 terms is equal to 22 terms.

So $$a\cdot\frac{\left(r^{18}-1\right)}{r-1} = a\cdot\frac{\left(r^{24}-1\right)}{r-1}$$

This will be reduced to $$\left(r^{18}-1\right)=\ r^{22}-1$$

Reducing this we get $$r^{4\ }=1$$

Now r has 4 posibilities 1, -1, i,-i.

1 is impossible as that would lead to a zero denominator in the sum of terms formulae.

In all the other cases of r, the sum of the first four values would be zero.