Sequence And Series Question 582
Question: If $ {{(p+q)}^{th}} $ term of a G.P. be $ m $ and $ {{(p-q)}^{th}} $ term be $ n $ , then the $ p^{th} $ term will be
[RPET 1997; MP PET 1985, 99]
Options:
A) $ m/n $
B) $ \sqrt{mn} $
C) $ mn $
D) 0
Show Answer
Answer:
Correct Answer: B
Solution:
Given that $ m=a{r^{p+q-1}} $ and $ n=a{r^{p-q-1}} $ $ {r^{p+q-1-p+q+1}}=\frac{m}{n}\Rightarrow r={{( \frac{m}{n} )}^{1/(2q)}} $ and $ a=\frac{m}{{{( \frac{m}{n} )}^{(p+q-1)/2q}}} $ Now $ p^{th} $ term $ =a{r^{p-1}}=\frac{m}{{{( \frac{m}{n} )}^{(p+q-1)/2q}}}{{( \frac{m}{n} )}^{(p-1)/2q}} $ $ =m{{( \frac{m}{n} )}^{(p-1)/2q-(p+q-1)/(2q)}}=m{{( \frac{m}{n} )}^{-1/2}}={m^{1-1/2}}{n^{1/2}} $ $ ={m^{1/2}}{n^{1/2}}=\sqrt{mn} $ . Aliter : As we know each term in a G.P. is geometric mean of the terms equidistant from it. Here $ {{(p+q)}^{th}} $ and $ {{(p-q)}^{th}} $ terms are equidistant from $ p^{th} $ term $ i.e. $ at a distance of $ q $ . Therefore, $ p^{th} $ term will be G.M. of $ {{(p+q)}^{th}} $ and $ {{(p-q)}^{th}} $ $ i.e. $ $ \sqrt{mn} $ .
BETA
