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# If $j,k,l$ are in AP $p,q,r$ in HP and $jp,kq,lr$ in $GP$ , then $\;\frac{p}{r}+\frac{r}{p}\;is\;equal\;to:$

$(a)\;\frac{j}{l}+\frac{l}{j}\qquad(b)\;\frac{j}{l}-\frac{l}{j}\qquad(c)\;\frac{k}{q}+\frac{q}{k}\qquad(d)\;\frac{k}{q}-\frac{j}{p}$

Answer : $(a) \;\frac{j}{l}+\frac{l}{j}$
Explanation : $2k=j+l$
$q=\frac{2pr}{p+r}$
$k^2q^2=jplr$
$(\frac{j+l}{2})^2\;.\frac{4p^2r^2}{(p+r)^2}=(jl)(pr)$
$\frac{(j+l)^2}{jl}=\frac{(p+r)^2}{pr}$
$\frac{j}{l}+\frac{l}{j}=\frac{p}{r}+\frac{r}{p}.$