Mathematics - Sequence & Series Question with Solution | TestHub

Let the coefficients of rth, $(r+1)$ th, and (r $+$ 2) th terms be in HP. Then, $\frac{2}{{}^n{C}_r}=\frac{1}{{}^n{C}_{r-1}}+\frac{1}{{}^n{C}_r+1}$ $$\begin{array}{l}\Rightarrow 2=\frac{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}-1}}=\frac{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}+1}}\\ \Rightarrow 2=\frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}+\frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}}\\ \Rightarrow {\mathrm{n}}^2-4\mathrm{nr}+4{\mathrm{r}}^2+\mathrm{n}=0\\ \Rightarrow (\mathrm{n}-2\mathrm{r}{)}^2+\mathrm{n}=0\end{array}$$ which is not possible for any value for $n$.