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A geometric series has first term 1,the sum of the first 5 terms is twice that of the sum of the 6th to 15th term inclusive. Prove that [tex] r^5= \frac{1}{2} \sqrt {3-1}[/tex]

What i did was.....

[tex] 2s_5=s_{15}-s_5 [/tex]

using the formula for the sum of a GS, i got...

[tex] 2r^4 -1 =r^{14} -r^4 [/tex]

and things when downhill from there.

I've tried expressing it in terms of x^5 because the answer seems to suggest the quadratic formula. But i don't seem to be getting anywhere. Can anyone help?