Mathematics - Sequence & Series Question with Solution | TestHub

Let H.P. be $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{a}+\mathrm{d}}+\frac{1}{\mathrm{a}+2\mathrm{d}}+\dots \dots$ $\begin{array}{rl}& \therefore \mathrm{u}=\frac{1}{\mathrm{a}+(\mathrm{p}-1)\mathrm{d}},\mathrm{v}=\frac{1}{\mathrm{a}+(\mathrm{q}-1)\mathrm{d}},\\ & \mathrm{w}=\frac{1}{\mathrm{a}+(\mathrm{r}-1)\mathrm{d}}\Rightarrow \mathrm{a}+(\mathrm{p}-1)\mathrm{d}=\frac{1}{\mathrm{u}}\\ & \mathrm{a}+(\mathrm{q}-1)\mathrm{d}=\frac{1}{\mathrm{v}},\mathrm{a}+(\mathrm{r}-1)\mathrm{d}=\frac{1}{\mathrm{w}}\\ & \Rightarrow (\mathrm{q}-\mathrm{r})\{\mathrm{a}+(\mathrm{p}-1)\mathrm{d}\}+(\mathrm{r}-\mathrm{p})\{\mathrm{a}+(\mathrm{q}-1)\mathrm{d}\}+\dots \dots \\ & =\frac{1}{\mathrm{u}}(\mathrm{q}-\mathrm{r})+\frac{1}{\mathrm{v}}(\mathrm{r}-\mathrm{p})+\dots .\Rightarrow (q-r)\mathrm{vw}+\dots ..=0\end{array}$