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What is sum of an infinite geometric progression(GP)

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S =a/1-r' where a is first term and r is common difference and less than 1
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it is a/(1-r) provided |r|<1 otherwise series can't converge.
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a/1-r
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a/(1-r).......where a is the first term, r is the common ratio(ratio between any two consecutive numbers)
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S=a/1-r, if r(common ratio)1
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Sn =a/1-r ,where a- first term and r- is common ratio
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Sn =a/(1-r),where a is first term and r is common ratio.
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Sum of an infinite GP equals a/1 - r where a is the first term and r is the common ration S=a/1-r, if r(common ratio)1
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The sum of infinite G.P is a/(1-r) where a is the first term and r is the common ratio
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Sum of an infinite GP equals a/1 - r where a is the first term and r is the common ration.
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