Sequence And Series Question 236
Question: If the $ p^{th},\ q^{th} $ and $ r^{th} $ term of a G.P. and H.P. are $ a,\ b,\ c $ , then $ a(b-c)\log a+b(c-a) $ $ \log b+c(a-b)\log c= $
[Dhanbad Engg. 1976]
Options:
A) $ -1 $
B) 0
C) 1
D) Does not exist
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ A $ and $ R $ be the first term and common ratio of the G.P. Then $ a=A{R^{p-1}},\ b=A{R^{q-1}} $ and $ c=A{R^{r-1}} $ ?..(i) Again if $ x $ and $ d $ be the first and common difference of the A.P. corresponding to the given H.P. Then $ \frac{1}{a}=x+(p-1)d,\ \frac{1}{b}=x+(q-1)d $ , $ \frac{1}{c}=x+(r-1)d $ ?..(ii) From (i), $ \frac{a}{b}={R^{p-q}} $ or $ {{( \frac{a}{b} )}^{1/c}}={{({R^{p-q}})}^{1/c}}=R^{k} $ , where $ k=\frac{p-q}{c} $ From (ii), $ k=(p-q){ x+(r-1)d } $ $ =(p-q)x-(p-q)(r-1)d $ $ =(p-q)x-(p-q)d+(rp-rq)d $ ?..(iii) Similarly, $ {{( \frac{b}{c} )}^{1/a}}={{({R^{q-r}})}^{1/a}}=R^{n} $ , where $ n=\frac{q-r}{a} $
$ \Rightarrow $ $ n=(q-r)\times { x+(p-1)d } $
$ \Rightarrow $ $ n=(q-r)x-(q-r)d+(pq-pr)d $ ?..(iv) and $ {{( \frac{c}{a} )}^{1/b}}={{({R^{r-p}})}^{1/b}}=R^{m} $ Where $ m=\frac{r-p}{b}=(r-p){ x+(q-1)d } $ $ =(r-p)x-(r-p)d+(rq-qp)d $ ?..(v) Hence $ {{( \frac{a}{b} )}^{1/c}}{{( \frac{b}{c} )}^{1/a}}{{( \frac{c}{a} )}^{1/b}}=R^{k}R^{m}R^{n}={R^{m+n+k}} $ $ =R^{0}=1 $ {since $ k+m+n=0 $ }, Adding (iii), (iv), (v) Taking logarithm of both sides, we get $ \frac{1}{c}({\log_{e}}a-{\log_{e}}b)+\frac{1}{a}({\log_{e}}b-{\log_{e}}c) $ $ +\frac{1}{b}({\log_{e}}c-{\log_{e}}a)={\log_{e}}(1) $
$ \Rightarrow $ $ ( \frac{1}{c}-\frac{1}{b} ){\log_{e}}a+( \frac{1}{a}-\frac{1}{c} ){\log_{e}}b+( \frac{1}{b}-\frac{1}{a} ){\log_{e}}c=0 $
$ \Rightarrow $ $ ( \frac{b-c}{bc} ){\log_{e}}a+( \frac{c-a}{ac} ){\log_{e}}b+( \frac{a-b}{ab} ){\log_{e}}c=0 $
$ \Rightarrow $ $ a(b-c){\log_{e}}a+b(c-a){\log_{e}}b+c(a-b){\log_{e}}c=0 $ . Note: Such type of questions $ i.e. $ containing term with cyclic coefficient associated with negative sign reduce to 0 mostly.
BETA
