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

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$n a_{1} a_{n}$

$n (a_{1} - a_{n})$

$(n - 1) a_{1} a_{n}$

$(n - 1) (a_{1} - a_{n})$

answer is A.

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Detailed Solution

$\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} ..... \frac{1}{a_{n - 1}}, \frac{1}{a_{n}}$ AP

$\frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - \frac{1}{a_{2}} = .... = \frac{1}{a_{n}} - \frac{1}{a_{n - 1}} = d$

$\frac{a_{1} - a_{2}}{a_{1} a_{2}} = d, \frac{a_{2} - a_{1}}{a_{1} a_{2}} = d ...... \frac{a_{n - 1} - a_{n}}{a_{n - 1} a_{n}} = d$

$\frac{a_{1} - a_{2}}{d} + \frac{a_{2} - a_{3}}{d} + .... + \frac{a_{n - 1} - a_{n}}{d}$

$\frac{1}{d} [a_{1} - a_{n}] = \frac{a_{1} a_{n}}{d} [\frac{1}{a_{n}} - \frac{1}{a_{1}}]$

$= \frac{a_{1} a_{n}}{d} (n - 1) d = (n - 1) a_{1} a_{n}$

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