SOLUTION: In a geometric sequence of real numbers, the sum of the first two terms is 7 and the sum of the first six terms is 91. What is the sum of the first four terms?

We are given

     +  = 7    (1)

     +  +  +  +  +  = 91      (2)   


In terms of "a" (the first term) and "r" (the common ratio) these equalities take the form

    a + ar = 7,       (3)

    a + ar +  +  +  +   = 91    (4)


In (4), group the terms

    (a + ar) + ( + ) + ( + )  = 91.    (5)


Re-write (5) in an equivalent form

    (a + ar) +  + ) = 91.    (6)


In (6), replace (a+ar) by the value of 7, based on (3).  You will get

    7 +  +  = 91.   (7)


In (7), divide both sides by 7

    1 +  +  = 13,

or

     +  - 12 = 0.    (8)


This biquadratic equation (8) is the quadratic equation relative .  Solve it by factoring

     = 0


Since we are given that the progression is in real numbers,   can not be zero 
(it is positive at any real r), we conclude that only possible value of    is

     = 3.


Then for the sum of the first four terms of this geometric progression    we have
     = a + ar +  +  = (a + ar) + ( + ) = (a+ar) + .


We substitute here  a+ar = 7 and  = 3, and we get
     = 7 + 3*7 = 7 + 21 = 28.


ANSWER.  The sum of the first 4 terms of this GP is 28.