Need algebra help with the Sum of the first n terms of an geometric sequence in fractions

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Dec 18th, 2015

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Sum of n terms of G.P. is given by

Sn = a(r^n - 1)/(r-1)

First question:

a=1/3 , r=1/3 , n=5

Putting these values in the equation, we get

S5 = (1/3)*((1/3)^5 - 1)/(1/3 - 1)

S5 = (1/3)*(-242/243)*(-3/2)

S5 = 121/243

Second question:

a= -6, r = -2, n=9

So S9 = (-6)*((-2)^9 - 1)/(-2-1)

S9 = (-6)*(-513)/(-3)

S9 = -1026

Please let me know if you need any clarification. I'm always happy to answer your questions.
Dec 18th, 2015

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