$\text{ If }{x}_1,x_2,\dots ,x_n\text{ are in H.P, then }\sum _{r=1}^{n-1} x_r{x}_{r+1}\text{ is equal to }$

$\begin{array}{l}\text{ Clearly }\frac{1}{x_1},\frac{1}{x_2},---\frac{1}{x_n}\text{ will be in A.P }\\ \Rightarrow \frac{1}{x_2}-\frac{1}{x_1}=\frac{1}{x_3}-\frac{1}{x_2}=---=\frac{1}{x_{r+1}}-\frac{1}{x_r}=--\\ \Rightarrow \frac{1}{x_n}-\frac{1}{x_{n-1}}=\lambda \left(say\right)\\ \Rightarrow \sum _{r=1}^{n-1} x_r{x}_{r+1}=\frac{1}{\lambda }\sum _{r=1}^{n-1} \left(x_r-x_{r+1}\right)=\frac{1}{\lambda }\left(x_1-x_n\right)\\ \underset{\_}{\underset{\_}{\text{ Now }}},\frac{1}{x_n}=\frac{1}{x_1}+(n-1)\lambda \Rightarrow \frac{x_1-x_n}{x_1{x}_n}=(n-1)\lambda \\ =\sum _{r=1}^{n-1} x_r{x}_{r+1}=(n-1)x_1{x}_n\end{array}$