Sequence And Series Question 484
Question: The ratio of sum of $ m $ and $ n $ terms of an A.P. is $ m^{2}:n^{2} $ , then the ratio of $ m^{th} $ and $ n^{th} $ term will be
[Roorkee 1963; MP PET 1995; Pb. CET 2001]
Options:
A) $ \frac{m-1}{n-1} $
B) $ \frac{n-1}{m-1} $
C) $ \frac{2m-1}{2n-1} $
D) $ \frac{2n-1}{2m-1} $
Show Answer
Answer:
Correct Answer: C
Solution:
Given that $ \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^{2}}{n^{2}} $
$ \Rightarrow $ $ \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n} $
$ \Rightarrow $ $ \frac{a+\frac{1}{2}(m-1)d}{a+\frac{1}{2}(n-1)d}=\frac{m}{n} $
$ \Rightarrow $ $ an+\frac{1}{2}(m-1)nd=am+\frac{1}{2}(n-1)md $
$ \Rightarrow $ $ i.e. $
$ \Rightarrow $ $ a(n-m)+\frac{d}{2}(m-n)=0 $
$ \Rightarrow $ $ a=\frac{d}{2} $ or $ d=2a $ So, required ratio, $ \frac{T_{m}}{T_{n}}=\frac{a+(m-1)d}{a+(n-1)d}=\frac{a+(m-1)2a}{a+(n-1)2a} $ $ =\frac{1+2m-2}{1+2n-2}=\frac{2m-1}{2n-1} $ . Trick: Replace $ m $ by $ 2m-1 $ and $ n $ by $ 2n-1 $ . Obviously if $ S_{m} $ is of degree 2, then $ T_{m} $ is of $ a $ $ i.e. $ linear.
BETA
