In a geometric sequence of real numbers, the sum of the first $2$ terms is $7$, and the sum of the first $6$ terms is $91$. The sum of the first $4$ terms is
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84$
Denote the sum of the first $2$ terms as $x$. Since we know that the sum of the first $6$ terms is $91$ which is $7 \cdot 13$, we have $x$ + $xy$ + $xy^2$ = $13x$ because it is a geometric series. We can quickly see that $y$ = $3$, and therefore, the sum of the first $4$ terms is $4x = 4 \cdot 7 = \boxed {(A) 28}$
~Arcticturn