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

n a_{1} a_{n}

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

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

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

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Solution

The correct option is D $(n - 1) a_{1} a_{n}$
$a_{1}, a_{2}, a_{3}, . . . . . . . a_{n} \Rightarrow H P$
$\frac{1}{_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, . . . . . . . . . \frac{1}{a_{n}} \Rightarrow A P$
$\frac{1}{a_{1}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - . . . . . . . . . . . \frac{1}{a n} \Rightarrow A P - \frac{1}{a_{n} - 1} =$ common difference $= K$
$\frac{a_{1}}{a_{1} a_{2}} = \frac{a_{2} - a_{3}}{a_{2} a_{3}} = . . . . . . . . . . = \frac{a_{n} - 1 - a n}{a_{n} a_{n} - 1} = K$
$∴ a_{1} a_{2} = \frac{a_{1} - a_{2}}{K}, a_{2} a_{3} = \frac{a_{2} - a_{3}}{K}, . . . . . . . a_{n} a_{n} - 1 = \frac{a_{n - 1} - a_{n}}{K}$
$∴ a_{1} a_{2} + . . . . . . . a_{n - 1} a_{n}$
$S = (\frac{a_{1} - a_{2}}{K}) + (\frac{a_{2} - a_{3}}{K}) + . . . . . . (\frac{a_{n - 1} - a_{n}}{K})$
$\Rightarrow S = \frac{a_{1} - a_{n}}{K}$
Also, $K = \frac{\frac{1}{a n} - \frac{1}{a_{1}}}{(n - 1)} = \frac{a_{1} a_{n}}{(n - 1) a_{1} a_{n}}$
$∴ S = (n - 1) a_{1} a_{n}$

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Q. If

a_{1}

a_{2} . . . . .

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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