AP ICET 2017 Shift 2 Paper

Question : 129 of 200

Marks: +1, -0

If the sum of the first six terms of a geometric progression is 9 times the sum of the first three terms then the common ratio

r > 1

Solution:

{S}_{n}=a{r^{n}-1}/{r-1}

for

r > 1

S_{6}=a({r^{6}-1}/{r-1})

and

{S}_{3}=a({r^{3}-1}/{r-1})

a({r^{6}-1}/{r-1})=9({r^{3}-1}/{r-1})

⇒{r^{6}-1}/{r^{3}-1}=9⇒{(r^{3}+1)(r^{3}-1)}/{r^{3}-1}=9

⇒r^{3}+1=9⇒r^{3}=8=2^{3}

∴r=2

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