Mathematics
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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
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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
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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
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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 + ...
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OpenStudy (anonymous):
yes got it .
ganeshie8 (ganeshie8):
good :)