Question

If the $(m+1)\mathrm{th},(n+1)\mathrm{th}\mathrm{and}(r+1)\mathrm{th}$ terms of an A.P. are G.P. and $\mathrm{m},\mathrm{n},\mathrm{r}$ are in H.P., then the value of the ratio of the common difference to the first term of the A.P. will be:

Answer 1

Compute the required ratio:

Let the first term of the AP is $\mathrm{a}$ and the common difference is $\mathrm{d}$.

AP is the arithmetic progression having a common difference d in every two terms.

GP is the geometric progression having a common ratio r in every two terms.

$AM\ge GM$ where AM is the arithmetic mean and GM is the geometric mean.

Since $(m+1)\mathrm{th},(n+1)\mathrm{th}\mathrm{and}(r+1)\mathrm{th}$ terms of an AP are in GP.

$n\mathrm{th}\mathrm{term}\mathrm{of}\mathrm{an}\mathrm{AP}=\mathrm{a}+(\mathrm{n}-1)\mathrm{d}$

$\Rightarrow \mathrm{a}+\mathrm{md},\mathrm{a}+\mathrm{nd}\mathrm{and}\mathrm{a}+\mathrm{rd}$ are terms of AP.

$$\begin{gathered} \Rightarrow {\left(\mathrm{a}+\mathrm{nd}\right)}^2=\left(\mathrm{a}+\mathrm{md}\right)(\mathrm{a}+\mathrm{rd}) \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{n}}^2{\mathrm{d}}^2+2\mathrm{and}={\mathrm{a}}^2+\mathrm{ad}\left(\mathrm{m}+\mathrm{r}\right)+{\mathrm{mrd}}^2 \\ \Rightarrow {\mathrm{n}}^2{\mathrm{d}}^2+2\mathrm{and}=\mathrm{ad}\left(\mathrm{m}+\mathrm{r}\right)+{\mathrm{mrd}}^2[\mathrm{Cancel}{\mathrm{a}}^2\mathrm{from}\mathrm{both}\mathrm{sides}] \\ \Rightarrow {\mathrm{d}}^2\left({\mathrm{n}}^2-\mathrm{mr}\right)=\mathrm{ad}\left(\mathrm{m}+\mathrm{r}-2\mathrm{n}\right) \\ \Rightarrow \mathrm{d}\left({\mathrm{n}}^2-\mathrm{mr}\right)=\mathrm{a}\left(\mathrm{m}+\mathrm{r}-2\mathrm{n}\right)[\mathrm{Cancel}\mathrm{d}\mathrm{from}\mathrm{both}\mathrm{sides}] \\ \Rightarrow \frac{\mathrm{d}}{\mathrm{a}}=\frac{\left(\mathrm{m}+\mathrm{r}-2\mathrm{n}\right)}{\left({\mathrm{n}}^2-\mathrm{mr}\right)}-\left[1\right] \end{gathered}$$

Since $\mathrm{m},\mathrm{n},\mathrm{r}$ are in HP

$\Rightarrow \mathrm{n}=\frac{2\mathrm{mr}}{\mathrm{m}+\mathrm{r}}-\left[2\right]$

After solving the equations $\left[1\right]$ and $\left[2\right]$ we get

$\frac{\mathrm{d}}{\mathrm{a}}=\frac{-2}{\mathrm{n}}$

Hence, the required ratio is $-\frac{2}{n}$.