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Logarithmic Differentiation

If b1,b2,b3...

Question

If $b_{1}, b_{2}, b_{3}, \dots, b_{n} ϵ H . P .$ then $b_{1} b_{2} + b_{2} b_{3} + b_{3} b_{4} + \dots, b_{n - 1} b_{n} =$

A

n b_{1} b_{n}

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B

(n + 1) b_{1} b_{n}

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C

(n - 1) b_{1} b_{n}

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D

None of these

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Solution

The correct option is C $(n - 1) b_{1} b_{n}$
Given that, $b_{1}, b_{2}, b_{3}, \dots, b_{n} ϵ H . P .$
$\Rightarrow \frac{1}{b_{1}}, \frac{1}{b^{2}}, . . . \frac{1}{b_{n}} ϵ A . P .$
$∴ d = \frac{b_{1} - b_{2}}{b_{1} b_{2}} = \frac{b_{2} - b_{3}}{b_{2} b_{3}} = . . . = \frac{b_{n - 1} - b_{n}}{b_{n - 1} b_{n}} = \frac{b_{1} - b_{n}}{(n - 1) b_{1} b_{n}}$
$∴ d = \frac{b_{1} - b_{n}}{b_{1} b_{2} + b_{2} b_{3} + . . . + b_{n - 1} b_{n}}$
$\Rightarrow \frac{b_{1} - b_{n}}{(n - 1) b_{1} b_{n}} = \frac{b_{1} b_{n} (\frac{1}{b_{n}} - \frac{1}{b_{1}})}{b_{1} b_{2} + b_{2} b_{3} + . . . + b_{n - 1} b_{n}}$
Therefore, $b_{1} b_{2} + b_{2} b_{3} + . . . + b_{n - 1} b_{n} = (n - 1) b_{1} b_{n}$
Ans: C

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