A) \[-1\]
B) 0
C) 1
D) Does not exist
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 \[{{\left( \frac{a}{b} \right)}^{1/c}}={{({{R}^{p-q}})}^{1/c}}={{R}^{k}}\], where \[k=\frac{p-q}{c}\] From (ii), \[k=(p-q)\left\{ x+(r-1)d \right\}\] \[=(p-q)x-(p-q)(r-1)d\] \[=(p-q)x-(p-q)d+(rp-rq)d\] ?..(iii) Similarly, \[{{\left( \frac{b}{c} \right)}^{1/a}}={{({{R}^{q-r}})}^{1/a}}={{R}^{n}}\], where \[n=\frac{q-r}{a}\] \[\Rightarrow \]\[n=(q-r)\times \left\{ x+(p-1)d \right\}\] \[\Rightarrow \]\[n=(q-r)x-(q-r)d+(pq-pr)d\] ?..(iv) and \[{{\left( \frac{c}{a} \right)}^{1/b}}={{({{R}^{r-p}})}^{1/b}}={{R}^{m}}\] Where \[m=\frac{r-p}{b}=(r-p)\left\{ x+(q-1)d \right\}\] \[=(r-p)x-(r-p)d+(rq-qp)d\] ?..(v) Hence \[{{\left( \frac{a}{b} \right)}^{1/c}}{{\left( \frac{b}{c} \right)}^{1/a}}{{\left( \frac{c}{a} \right)}^{1/b}}={{R}^{k}}{{R}^{m}}{{R}^{n}}={{R}^{m+n+k}}\] \[={{R}^{0}}=1\]{since \[k+m+n=0\]}, Adding (iii), (iv), (v) Taking logarithm of both sides, we get \[\frac{1}{c}({{\log }_{e}}a-{{\log }_{e}}b)+\frac{1}{a}({{\log }_{e}}b-{{\log }_{e}}c)\] \[+\frac{1}{b}({{\log }_{e}}c-{{\log }_{e}}a)={{\log }_{e}}(1)\] \[\Rightarrow \]\[\left( \frac{1}{c}-\frac{1}{b} \right){{\log }_{e}}a+\left( \frac{1}{a}-\frac{1}{c} \right){{\log }_{e}}b+\left( \frac{1}{b}-\frac{1}{a} \right){{\log }_{e}}c=0\] \[\Rightarrow \]\[\left( \frac{b-c}{bc} \right){{\log }_{e}}a+\left( \frac{c-a}{ac} \right){{\log }_{e}}b+\left( \frac{a-b}{ab} \right){{\log }_{e}}c=0\] \[\Rightarrow \]\[a(b-c){{\log }_{e}}a+b(c-a){{\log }_{e}}b+c(a-b){{\log }_{e}}c=0\]. Note: Such type of questions \[i.e.\] containing term with cyclic coefficient associated with negative sign reduce to 0 mostly.
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