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

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 $(n - 1) a_{1} a_{n}$
Since $a_{1}, a_{2}, . . . . . a_{n}$ are in HP
$\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} . . . . . . \frac{1}{a_{n}}$ are in AP.

Let $d$ is the common difference of the AP
$\frac{1}{a_{2}} - \frac{1}{a_{1}} = d$
$a_{1} - a_{2} = a_{1} a_{2} d$
Similarly $a_{2} - a_{3} = a_{2} a_{3} d$
........................................................
..................................................
$a_{n - 1} - a_{n} = a_{n - 1} a_{n} d$

On adding all these we will get
$a_{1} - a_{n} = d (a_{1} a_{2} + a_{2} a_{3} + . . . . . . . . . . . . + a_{n - 1} a_{n})$

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

On putting the value from the above equation we get
$a_{1} - a_{n} = \frac{a_{1} - a_{n}}{a_{1} a_{n} (n - 1)} (a_{1} a_{2} + a_{2} a_{3} + . . . . + a_{n - 1} a_{n})$
$(a_{1} a_{2} + a_{2} a_{3} + . . . . + a_{n - 1} a_{n}) = a_{1} a_{n} (n - 1)$
So, option D is the correct answer.

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Similar questions

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

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