what is answer??????????????????

We have

7 = a [1 - r^2] / [1 - r]   and

91 = a[1 - r^6]/ [1 - r]         where a is the first term and r is the common ratio

Using the first equation, we can simplify and rearrange it thusly:

7 = a[(1 - r) (1 + r)] / (1 -r)   →   7 = a (1 + r)  →  a = 7/(1 + r)

Substituting this into the second equation, we have

91 = 7 (1 - r^6)/ [(1 - r)(1 + r)] →   [ divide both sides by 7]

13 = (1 - r^6)/(1 - r^2)    factoring, we have

13 = [(1 - r^3)(1 + r^3)]/[(1 + r)  ( 1 -r)]

13 = [ (1 -r)(r^2 + r + 1)(1 + r)(r^2 - r  + 1)] / [(1 +r) (1 - r)]

13 = (r^2 + r + 1)(r^2 - r + 1)

13 = r^4 + r^2 + 1

r^4 + r^2 - 12  = 0   factor

(r^2 + 4) (r^2 - 3)  = 0      only the second eqaution yields "real" solutons of  r = ±√3

The a = 7 / (1 - √3)   or    a = 7/ (1 + √3)

Proof  using  r =√3 and    a = 7/ (1 + √3) 

7 = [7/ (1 + √3)]* [1 - 3] / (1 - √3) = 7(1 -3)/ (1 -3)  = 7    

and

91 =  [7/ (1 + √3)]* [1 - 27]] / (1 - √3) = 7(-26)/ (-2) = 91

So the sum of the first  4 terms when a = [7/ (1 + √3)] and r = √3  is given by :

 [7/ (1 + √3)][1 - (√3)^4]/ (1 - √3) = 28

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What if r = -√3   ???

The sum of the first two terms is : 7/ (1 - √3)*(1 - 3)/(1 + √3) = 7

and

The sum of the first six terms is:

[7/ (1 - √3)]*[1 -27]/(1 + √3) = 91

So...the sum of the first four terms is : [7/ (1 - √3)]*[1 -9]/(1 + √3) is also 28!!!

So we actually have two series:

One has the first term of  [7/ (1 + √3)]  with a common ratio of  √3......and the other has a first term of [7/ (1 - √3)] and a common difference of -√3

This looks like it was a doosey of a question Chris ;)

Thanks, Melody..it was a "bear" to calculate all of that !!!...I'm not totally sure about my answer, but it seemed to "work".....maybe someone else will have a different "take"......????