geometric progression

$a \dfrac{1-r^6}{1-r} = 91 = 13 \cdot 7 = 13 a\dfrac{1-r^2}{1-r}\\ \dfrac{1-r^6}{1-r^2}=13\\ r^6-13r^2+12=0\\ r=\pm 1,~\pm \sqrt{3}\\ \text{We only need consider $r=\sqrt{3}$}\\ a \dfrac{1-3}{1-\sqrt{3}}=7\\ a = \dfrac 7 2 (\sqrt{3}-1)\\ a \dfrac{1-r^4}{1-r} = 28$

    (1 - r^2)                                          (1 - r^6) 

a*_______  =  7          and      a  *  _________  =  91

        1 - r                                            (1   - r)

a * (1  -r^2) =  7 (1 - r)                a ( 1 -r^6)  =  91 ( 1 - r)

13 a ( 1 - r^2) = 91 (1 - r)          a ( 1 - r^6)  = 91 ( 1 - r)

So

13a ( 1 -r^2)  =  a ( 1 - r^6)

13 ( 1 - r^2)  = ( 1 -r^6)

13 - 13r^2 = 1 - r^6

r^6  - 13r^2  + 12  =   0

Note that   r = 1   and  r  = -1  are roots   

So  (r + 1) ( r -1)  =  r^2  - 1

Dividing we have

                 r^4  + r^2  - 12

r^2  -1   [  r^6  - 13r^2 +  12]

                r^6                          - 1r^4

              _____________________________________

                                      r^4  -13r^2 + 12

                                      r^4 -  r^2

                                   _______________

                                              -12r^2  + 12

                                              -12r^2  +  12

                                             ____________

                                                                  0

The remaining polynomial  =   r^4  + r^2  - 12

Set to 0  and factor

(r^2 + 4) ( r^2 - 3)   = 0

Since the  sum of the  first n terms increases, then  √3  is the only possible value  for  r

So

7 =  a ( 1 - 3) / ( 1  -√3)

7 = -2a / (1 - √3)

7=  2a / (√3 - 1)

a = (7/2)(√3 - 1)

And the sum of the 1st  4 terms   =

(7/2) (√3 -1)  ( 1 - 9) / ( 1 - √3)  =

(7/2) (√3  -1) (-8) / ( 1 - √3)   =

(7/2) (√3 -1) (8) / ( √3 - 1)  =

(7/2) (8)  =

28

In a geometric progression of real numbers, the sum of the first two terms is 7,

and the sum of the first six term is 91. 

Find the sum of the first four terms.

$\begin{array}{|rclrcl|} \hline \mathbf{s_4 = a \dfrac{1-r^4}{1-r}} && \mathbf{s_2 = a \dfrac{1-r^2}{1-r}} \\ \hline \\ \dfrac{s_4}{s_2} &=& a \dfrac{1-r^4}{1-r} \above 1pt a \dfrac{1-r^2}{1-r} \\\\ \dfrac{s_4}{s_2} &=& \dfrac{(1-r^4)}{1-r^2} \\\\ \dfrac{s_4}{s_2} &=& \dfrac{(1-r^2)(1+r^2)}{1-r^2} \\\\ \dfrac{s_4}{s_2} &=& 1+r^2 \\\\ s_4 &=& s_2(1+r^2) \quad | \quad s_2 = 7 \\\\ \mathbf{s_4} &=& \mathbf{7(1+r^2)} \qquad (1) \\ \hline \end{array} \begin{array}{|rclrcl|} \hline \mathbf{s_2} &=& \mathbf{a+ ar} \\\\ s_2 &=& a(1+r) \quad | \quad s_2 = 7 \\\\ 7 &=& a(1+r) \\\\ \mathbf{a} &=& \mathbf{\dfrac{7}{1+r} } \qquad (2) \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline s_6 &=& S_2 + ar^2+ar^3+ar^4+ar^5 \quad | \quad s_2 = 7, s_6 = 91 \\ 91 &=& 7 + ar^2+ar^3+ar^4+ar^5 \\ 84 &=& ar^2+ar^3+ar^4+ar^5 \\ 84 &=& ar^2 (1 +r+r^2+r^3) \quad | \quad \mathbf{a=\dfrac{7}{1+r} } \\\\ 84 &=& \dfrac{7r^2(1 +r+r^2+r^3)}{1+r} \quad | \quad 1 +r+r^2+r^3 = \dfrac{1-r^4}{1-r} \\\\ 84 &=& \dfrac{7r^2\dfrac{1-r^4}{1-r}}{1+r} \\\\ 84 &=& \dfrac{7r^2(1-r^4)}{(1-r)(1+r)} \\\\ 84 &=& \dfrac{7r^2(1-r^4)}{(1-r^2)} \\\\ 84 &=& \dfrac{7r^2(1-r^2)(1+r^2)}{(1-r^2)} \\\\ 84 &=& 7r^2(1+r^2) \quad | \quad : 7 \\\\ 12 &=& r^2(1+r^2) \\\\ 12 &=& r^2 + r^4 \\\\ r^4+r^2 -12 &=& 0 \\\\ r^2 &=& \dfrac{-1\pm \sqrt{1-4*(-12)} }{2} \\ r^2 &=& \dfrac{-1\pm \sqrt{49} }{2} \\ r^2 &=& \dfrac{-1\pm 7 }{2} \\ \\ r^2 &=& \dfrac{-1+7 }{2} \\ r^2 &=& \dfrac{6 }{2} \\ r^2 &=& \mathbf{3} \\\\ r^2 &=& \dfrac{-1- 7 }{2} \\ r^2 &=& \dfrac{-8 }{2} \\ r^2 &=& -4 \qquad \text{r is a complex number! No solution} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{s_4} &=& \mathbf{7(1+r^2)} \quad | \quad r^2 = 3 \\ s_4 &=& 7(1+3) \\ s_4 &=& 7*4 \\ \mathbf{s_4} &=& \mathbf{28} \\ \hline \end{array}$

The sum of the first four terms is 28