Question

In an HP pth term is qr and the qth term is pr, show that the rth term is pq

Answer 1

$$\begin{gathered} \mathrm{Given}\mathrm{pth}\mathrm{term}\mathrm{of}\mathrm{HP}=\mathrm{qr} \\ \mathrm{So}\mathrm{pth}\mathrm{term}\mathrm{of}\mathrm{AP}=\frac{1}{\mathrm{qr}} \\ \mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}=\frac{1}{\mathrm{qr}}....\left(1\right) \\ \mathrm{and}\mathrm{qth}\mathrm{term}\mathrm{of}\mathrm{HP}=\mathrm{pr} \\ \mathrm{so}\mathrm{qth}\mathrm{term}\mathrm{of}\mathrm{AP}=\frac{1}{\mathrm{pr}} \\ \mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}=\frac{1}{\mathrm{pr}}.....\left(2\right) \\ \mathrm{subtracting}\mathrm{equation}1\mathrm{and}2\mathrm{we}\mathrm{get}, \\ \left(\mathrm{p}-\mathrm{q}\right)\mathrm{d}=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{pqr}} \\ \mathrm{d}=\frac{1}{\mathrm{pqr}} \\ \mathrm{Now}\mathrm{from}\mathrm{equation}1, \\ \mathrm{a}=\frac{1}{\mathrm{qr}}-\frac{\mathrm{p}-1}{\mathrm{pqr}}=\frac{\mathrm{p}-\mathrm{p}+1}{\mathrm{pqr}}=\frac{1}{\mathrm{pqr}} \\ \mathrm{So} \\ \mathrm{rth}\mathrm{term}\mathrm{of}\mathrm{AP}=\mathrm{a}+\left(\mathrm{r}-1\right)\mathrm{d} \\ =\frac{1}{\mathrm{pqr}}+\frac{\mathrm{r}-1}{\mathrm{pqr}}=\frac{1}{\mathrm{pq}} \\ \mathrm{So},\mathrm{rth}\mathrm{term}\mathrm{of}\mathrm{HP}\hspace{0.17em}=\mathrm{pq} \end{gathered}$$