If $a_{1}, a_{2}, \ldots, a_{n}$ are in H.P., then the expression $a_{1} a_{2}+a_{2} a_{3}+\ldots+a_

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If $a_{1}, a_{2}, \ldots, a_{n}$ are in H.P., then the expression $a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{n-1} a_{n}$ is equal to\

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(d) : Given $a_{1}, a_{2}, \ldots, a_{n}$ are in H.P.\ $\Rightarrow \quad \frac{1}{a_{1}}, \frac{1}{a_{2}}, \ldots ., \frac{1}{a_{n}} \in$ A.P.\ $\Rightarrow \quad \frac{1}{a_{2}}-\frac{1}{a_{1}}=d$ $$\begin{equation*}\Rightarrow \quad a_{1} a_{2}=\frac{a_{1}-a_{2}}{d}=\frac{a_{1}}{d}-\frac{a_{2}}{d} 	ag{i} \end{equation*}$$ $$\begin{equation*}a_{2} a_{3}=\frac{a_{2}}{d}-\frac{a_{3}}{d} 	ag{ii} \end{equation*}$$ $$\begin{equation*}a_{n-1} a_{n}=\frac{a_{n-1}}{d}-\frac{a_{n}}{d} 	ag{n} \end{equation*}$$ Adding (i), (ii) ( $n$ ) equations we get $a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+\ldots . a_{n-1} a_{n}$\ $=\frac{a_{1}}{d}-\frac{a_{n}}{d}$\ Also $\frac{1}{a_{n}}=\frac{1}{a_{1}}+(n-1) d$\ $\Rightarrow \frac{a_{1}-a_{n}}{d}=(n-1) a_{1} a_{n}$\ $	herefore \quad a_{1} a_{2}+a_{2} a_{3}+\ldots a_{n-1} a_{n}=(n-1) a_{1} a_{n}$.\