Sequence And Series Question 82
Question: If $ S_{n} $ denotes the sum of first n terms of an A.P. and $ \frac{S_{3n}-{S_{n-1}}}{S_{2n}-{S_{2n-1}}}=31 $ , then the value of n is
Options:
A) 21
B) 15
C) 16
D) 19
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ S_{3n}=\frac{3n}{2}[2a+(3n-1)d] $ $ {S_{n-1}}=\frac{n-1}{2}[2a+(n-2)d] $
$ \Rightarrow S_{3n}-{S_{n-1}}=\frac{1}{2}[2a(3n-n+1)] $ $ +\frac{d}{2}[3n(3n-1)-(n-1)(n-2)] $ $ =\frac{1}{2}[2a(2n+1)+d(8n^{2}-2)] $ $ =a(2n+1)+d(4n^{2}-1) $ $ =(2n+1)[a+(2n-1)d] $ $ S_{2n}-{S_{2n-1}}=T_{2n}=a+(2n-1)d $
$ \Rightarrow \frac{S_{3n}-{S_{n-1}}}{S_{2n}-{S_{2n-1}}}=(2n+1) $ Given,
$ \Rightarrow \frac{S_{3n}-{S_{n-1}}}{S_{2n}-{S_{2n-1}}}=31\Rightarrow n=15 $
BETA
