Question:
If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to
Answer:
Given first n terms are in AP
So Sn = n*(n+1)/2
and S2n = 2n*(2n+1)/2
and S3n = 3n*(3n+1)/2
Now S2n = 3*Sn
=> 2n*(2n+1)/2 = 3n*(n+1)/2
=> 2(2n+1) = 3(n+1)
=> 4n+2 = 3n + 3
=> 4n - 3n = 3-2
=> n = 1
Now
S3n /Sn = {3n*(3n+1)/2}/{n*(n+1)/2}
= {3*1*(3*1+1)/2}/{1*(1+1)/2}
= (3*4/2)/(1*2/2)
= 12/2
= 6
So S3n /Sn = 6
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