If a1,a2,…an are in H.P. then the expression a1a2+a2a3+…+an−1an is equal to

A
$(n - 1) a_{1} a_{n}$
B
$n (a_{1} - a_{n})$
C
$(n - 1) (a_{1} - a_{n})$
D
$n a_{1} a_{n}$

Solution:

$a_{1}, a_{2}, \dots a_{n}$ are in H.P.
$\Rightarrow \frac{1}{a_{1}}, \frac{1}{a_{2}}, \dots \frac{1}{a_{n}}$ are in A.P.
$\Rightarrow \frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - \frac{1}{a_{2}} = \dots \frac{1}{a_{n}} - \frac{1}{a_{n - 1}} = d$ (say)
$\begin{aligned} \Rightarrow a_{1} a_{2} = \frac{1}{d} (a_{1} - a_{2}), a_{2} a_{3} = \frac{1}{d} (a_{2} - a_{3}), \dots \\ a_{n - 1} a_{n} = \frac{1}{d} (a_{n - 1} - a_{n}) \end{aligned}$
Thus, $a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n}$
$\begin{array}{l} = \frac{1}{d} [a_{1} - a_{2} + a_{2} - a_{3} + \dots + a_{n - 1} - a_{n}] \\ = \frac{1}{d} (a_{1} - a_{n}) \end{array}$
But $\frac{1}{a_{n}} = \frac{1}{a_{1}} + (n - 1) d \Rightarrow \frac{a_{1} - a_{n}}{a_{n} a_{1}} = (n - 1) d$
$∴ a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n} = (n - 1) a_{1} a_{n}$

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

Solution:

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