A) \[\frac{a}{c}-\frac{c}{a}\]
B) \[\frac{a}{c}+\frac{c}{a}\]
C) \[\frac{b}{q}+\frac{q}{b}\]
D) \[\frac{b}{q}-\frac{q}{b}\]
Correct Answer: B
Solution :
[b] \[\frac{p}{r}+\frac{r}{p}=\frac{{{p}^{2}}+{{r}^{2}}}{pr}=\frac{{{(p+r)}^{2}}-2pr}{pr}\] \[=\frac{\frac{4{{p}^{2}}{{r}^{2}}}{{{q}^{2}}}-2pr}{pr}\] \[\left[ \begin{align} & \because p,q, r are\,in\,H.P. \\ & \therefore q=\frac{2pr}{p+r} \\ \end{align} \right]\] \[=\frac{4pr}{{{q}^{2}}}-2=\frac{4{{b}^{2}}}{ac}-2\] \[[\because ap,bq,cr\,are\,in\,A.P\Rightarrow {{b}^{2}}{{q}^{2}}=acpr]\] \[=\frac{{{(a+c)}^{2}}}{ac}-2\]\[[a,b,c\,are\,in\,A.P\Rightarrow 2b=a+c]\] \[=\frac{a}{c}+\frac{c}{a}\]
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