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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Dedicated Teacher With Lots of Teaching Experience In Maths For All Boards

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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S?=a/(1-r ) only if -1
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10 Years+ Experienced Maths & Science Teacher

a=1st term ; r=common ratio Sum=a/(r-1), if r>1 Sum=a/(1-r), if r<1
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Maths teacher

Sorry earlier i told what is the general term of a GP. You are asking sum of nth term of a GP. Well is it equal to a(1-(r)n)/1-r for r1 then it will be a((r) n-1)/r-1 If r=1 then it will be na
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S= a /1 - r where a is first term and r is common ration
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Mathematics teacher for the learners

sum of finite terms is: s= a1(1-r^n)/1-r. for infinte terms, r^n tends to zero. so s = a1/1-r
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Sn =a/1-r ,where a- first term and r- is common ratio
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Mathematics And Physics Tutor

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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a/(1-r) ; a= first term r=common ratio
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