If $a_{1}, a_{2}, a_{3}, . . . a_{n}$ are in H.P then $a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . . + a_{n - 1} a_{n} =$

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

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(n + 1) a_{1} a_{n}

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(n - 2) a_{1} a_{n}

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(n + 4) a_{1} a_{n}

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Solution

The correct option is A $(n - 1) a_{1} a_{n}$
Given:
$a_{1}, a_{2}, a_{3}, . . . ., a_{n}$ are in H.P.
$∴ a_{2} = \frac{2 a_{1} a_{3}}{a_{1} + a_{3}}$ .....( 1 )
$\Rightarrow a_{1} a_{2} + a_{2} a_{3} = 2 a_{1} a_{3}$
$\Rightarrow a_{1} a_{2} + a_{2} a_{3} = (3 - 1) a_{1} a_{3}$ .....( 2 )
Now,
$∴ a_{3} = \frac{2 a_{2} a_{4}}{a_{2} + a_{4}}$
$\Rightarrow a_{3} (a_{2} + a_{4}) = 2 a_{4} . \frac{2 a_{1} a_{3}}{a_{1} + a_{3}}$ [From eq ( 1 )]
$\Rightarrow (a_{1} + a_{3}) (a_{2} + a_{4}) = 4 a_{1} a_{4}$
$\Rightarrow a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} = (4 - 1) a_{1} a_{4}$ .....( 3 )

Comparing eq (2) and (3) for upto n terms:
$\Rightarrow a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . . + a_{n - 1} a_{n} = (n - 1) a_{1} a_{n}$

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