wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

If S1,S2 and S3 and the sum of first n,2n and 3n terms of a geometric series respectively, then prove that S1(S3S2)=(S2S1)2

Open in App
Solution

Let 'a' be the first term.
'r' be the common ratio.
Given, S1,S2,S3 are sum of first n,2n and 3n
S1=a(rn1)r1
S2=a(r2n1)r1
S3=a(r3n1)r1
S3S2=a(r3n1)r1a(r2n1)r1
=ar1(r3n1(r2n1))
=ar1(r3n1r2n+1)=ar1(r3nr2n)
=ar1r2n(rn1)
S1(S2S3)=a(rn1)r1ar1r2n(rn1)
=a2r2n(rn1)2(r1)2 ..........(1)
S2S1=a(r2n1)r1a(rn1)r1
=ar1(r2n1rn+1)=ar1(r2nrn)
=ar1rn(rn1)
=(S2S1)2=a2(r1)2r2n(rn1)2 ..........(2)
From (1) and (2), we have
S1(S3S2)=(S2S1)2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Summation by Sigma Method
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon