213=S2nSn = 2 (2a + (2n-1)d)(2a + (n-1)d) = 2(1+ nd2a+ (n-1)d) so, 2nd2a+ (n-1)d = 1
Now, S3nSn= 3(2a+ (3n-1)d2a+ (n-1)d)= 3(1+ 2nd2a+ (n-1)d) = 3(1+1) = 6
Let Sn denote sum of first n terms of an AP.If S2n = 3 Sn then ratio S3n:Sn=?
had been trying it for whole day today but unable to solve it so finally sending it here!
-
UP 0 DOWN 0 0 3
3 Answers
21
1S2n = 3Sn
=> n[2a+(2n-1)d] = 3n2[2a+(n-1)d]
=> 2a = (n+1)d
=> 2a + (n-1)d = 2nd
and, 2a + (3n-1)d = 4nd
now, S3nSn = 3*4nd2nd = 6
341I have found that the amount of manipulation needed reduces when you use the fact that the sum to n terms of an AP is of the form an^2+bn
The given condition then translates to 4an^2+2bn = 3an^2+3bn \Rightarrow an^2=bn
Hence
\frac{S_{3n}}{S_n} = \frac{9an^2+3bn}{an^2+bn} = \frac{12bn}{2bn}=6