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Mathematics 17 Online
OpenStudy (anonymous):

The first term of an infinite G.P is 1 and any term is equal to the sum of all succeeding terms. Find the series

ganeshie8 (ganeshie8):

\[\large S_{\infty} = \dfrac{1}{1-r}\]

ganeshie8 (ganeshie8):

you're given : \[\large 1 = S_{\infty } - 1\]

ganeshie8 (ganeshie8):

you can solve \(r\), right ?

OpenStudy (anonymous):

hey sorry i was afk

OpenStudy (anonymous):

My textbook has a sollution i don't get it

ganeshie8 (ganeshie8):

\[\large a_n = S_{\infty} - S_{n}\]

OpenStudy (anonymous):

Let the GP be 1,r , r^2 , r^3 ........ give condition:- \[r=\frac{ r ^{2} }{ 1-r }\] Thus r = 1/2

OpenStudy (anonymous):

i don't get how they got this \[r=\frac{ r ^{2} }{ 1-r }\]

OpenStudy (anonymous):

\[\frac{ a }{ 1-r }\] i know this

ganeshie8 (ganeshie8):

they have just taken part of the given constraint : Notice that \(r\) is the second term of given GP

ganeshie8 (ganeshie8):

1, `r` , r^2 , r^3 ...

ganeshie8 (ganeshie8):

what does the constraint tell us ?

ganeshie8 (ganeshie8):

this second term equals sum of all the next terms, right ?

OpenStudy (anonymous):

yes

ganeshie8 (ganeshie8):

r = r^2 + r^3 + ...

OpenStudy (anonymous):

yes got it .

ganeshie8 (ganeshie8):

good :)

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