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Mathematics 20 Online
OpenStudy (anonymous):

Help

OpenStudy (anonymous):

The sum of an infinitely many terms of a GP is 3 times the sum of even terms.. The common ratio of the GP is

OpenStudy (anonymous):

I am getting 2 as the answer

OpenStudy (anonymous):

@gorv

OpenStudy (gorv):

lets the gp a ar ar^2 ar^3 ar^4............ sum of infinite series=a/(1-r) even term ar ar^3 ar^5................ it will also be infinite so sum = ar/(1-r^2)

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

i did the same thing

OpenStudy (gorv):

\[\frac{ a }{ 1-r }=3*\frac{ ar }{ 1-r^{2} }\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

then r^2 -3r +2 = 0

OpenStudy (gorv):

\[\frac{ a }{ 1-r }=3*\frac{ ar }{ (1+r)(1-r) }\]

OpenStudy (anonymous):

r=1 or r=2

OpenStudy (gorv):

1-r will get cancel from both side

OpenStudy (gorv):

and a also get cancelled

OpenStudy (gorv):

so 1=3r/(1+r) 1+r=3r 1=2r r=1/2

OpenStudy (anonymous):

why my method won't work

OpenStudy (anonymous):

@gorv

OpenStudy (gorv):

can u sjow me what u did???

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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