Some Specific Sum

Relevant wiki: Geometric Progression Sum

Let the first term be $a$ and the common ratio be $r$ . We know that:

$a+ar=7 \implies$ Eq.(1)

$a+ar+ar^2+ar^3+ar^4+ar^5=91 \implies$ Eq.(2)

Eq.(2) $\div$ Eq.(1)

$\dfrac{a+ar+ar^2+ar^3+ar^4+ar^5}{a+ar}=\dfrac{91}{7}\\ \dfrac{a(1+r+r^2+r^3+r^4+r^5)}{a(1+r)}=13\\ \dfrac{1(1+r) + r^2(1+r) + r^4(1+r)}{1+r} = 13\\ \dfrac{(1+r)(1+r^2+r^4)}{1+r}=13\\ 1+r^2+r^4=13\\ r^4+r^2-12=0\\ (r^2+4)(r^2-3)=0\\ r^2=-4 \quad r^2=3$

The first factor does not give a real value for $r$ , so we reject it.

Therefore, $r^2=3 \implies r=\pm\sqrt{3}$

$S_4 = a+ar+ar^2+ar^3\\ =a+ar+r^2(a+ar)\\ =7+3(7)\\ =\boxed{28}$

Note: It doesn't matter which value of $r$ we use. The answer will still be the same