Sequences

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.

What is the sum of the first 4 terms?

$\begin{array}{|rcll|} \hline s_2 &=& a_1 + a_2 \\ \mathbf{s_2} &=& \mathbf{a_1 + a_1r} \\ \hline \end{array} \begin{array}{|rcll|} \hline s_4 &=& a_1 + a_2 + a_3 + a_4 \\ s_4 &=& s_2 + a_3 + a_4 \\ s_4 &=& s_2 + a_1r^2 + a_1r^3 \\ s_4 &=& s_2 + (a_1+a_1r)r^2 \\ s_4 &=& s_2 + s_2r^2 \\ s_4 &=& s_2(1+r^2) \quad | \quad s_2 = 7 \\ \mathbf{s_4} &=& \mathbf{ 7(1+r^2) } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline s_6 &=& a_1 + a_2 + a_3 + a_4 + a_5 + a_6 \\ s_6 &=& s_4 + a_5 + a_6 \\ s_6 &=& s_4 + a_1r^4 + a_1r^5 \\ s_6 &=& s_4 + (a_1+a_1r)r^4 \\ s_6 &=& s_4 + s_2r^4 \quad | \quad s_6 = 91,~s_2=7\\ 91 &=& s_4 +7r_4 \\ \mathbf{s_4} &=& \mathbf{ 91-7r_4 } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline s_4 = 7(1+r^2) &=& 91-7r_4 \\ 7(1+r^2) &=& 91-7r_4 \quad | \quad : 7 \\ 1+r^2 &=& 13-r^4 \\ \mathbf{r^4+r^2 -12} &=& \mathbf{ 0 } \\ \hline r^2 &=& \dfrac{ -1\pm\sqrt{1+4*12} }{2} \\\\ r^2 &=& \dfrac{ -1\pm 7 }{2} \\\\ r^2 &=& \dfrac{ -1 {\color{red}+} 7 }{2} \\\\ \mathbf{r^2} &=& \mathbf{3} \\ \hline s_4 &=& 7(1+r^2) \quad | \quad r^2 = 3 \\ s_4 &=& 7(1+3) \\ \mathbf{s_4} &=& \mathbf{28} \\ \hline \end{array}$