Sequence And Series Question 376
Question: If $ S_{n} $ denotes the sum of first n terms of an A.P. whose first term is a and $ \frac{S_{nx}}{S_{x}} $ is independent of x, then $ S_{p}= $
Options:
A) $ P^{3} $
B) $ P^{2}a $
C) $ Pa^{2} $
D) $ a^{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \frac{S_{nx}}{S_{x}}=\frac{\frac{nx}{2}[2a+( nx-1 )d]}{\frac{x}{2}[2a+( x-1 )d]} $ $ =\frac{n[(2a-d)+nxd]}{( 2a-d )+xd} $ For $ \frac{S_{nx}}{S_{x}} $ to be independent of x $ 2a-d=0 $
$ \therefore ,2a=d $ now, $ S_{p}=\frac{P}{2}[2a+(P-1)d]=P^{2}a $
BETA
