Sequence And Series Question 634
Question: If the sum of first $ n $ terms of an A.P. be equal to the sum of its first $ m $ terms, $ (m\ne n) $ , then the sum of its first $ (m+n) $ terms will be
[MP PET 1984]
Options:
A) 0
B) $ n $
C) $ m $
D) $ m+n $
Show Answer
Answer:
Correct Answer: A
Solution:
As given $ \frac{n}{2}{ 2a+(n-1)d }=\frac{m}{2}{ a+(m-1)d } $
$ \Rightarrow $ $ 2a(m-n)+d(m^{2}-m-n^{2}+n)=0 $
$ \Rightarrow $ $ (m-n){ 2a+d(m+n-1) }=0 $
$ \Rightarrow $ $ 2a+(m+n-1)d=0 $ , $ (\because \ m\ne n) $
$ \therefore $ $ {S_{m+n}}=\frac{m+n}{2}{ 2a+(m+n-1)d }=\frac{m+n}{2}{ 0 }=0 $ .
BETA
