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Harmonic Mean

If b1,b2,b3...

Question

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

A

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

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B

n b_{1} b_{n}

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C

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

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D

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

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Solution

The correct option is A $(n - 1) b_{1} b_{n}$
$\frac{1}{b_{1}} - \frac{1}{b_{2}} = d$
$\Rightarrow \frac{b_{1} - b_{2}}{b_{1} b_{2}} = d$
$\Rightarrow \frac{b_{1} - b_{2}}{d} = b_{1} b_{2}, . . . \frac{b_{n - 1} - b_{n}}{d} = b_{n - 1} b_{n}$
$\Rightarrow \frac{b_{1} - b_{2}}{d} + \frac{b_{2} - b_{3}}{d} + \frac{b_{3} - b_{4}}{d} + . . . + \frac{b_{n - 1} - b_{n}}{d} = b_{1} b_{2} + b_{2} b_{3} + b_{3} b_{4} + . . . + b_{n - 1} b_{n}$ ... $(1)$
Also, $(n - 1) = \frac{b_{1} - b_{n}}{b_{1} b_{n}}$ .... $(2)$
From eqns $(1)$ and $(2)$
$∴ b_{1} b_{2} + b_{2} b_{3} + b_{3} b_{4} + . . . + b_{n - 1} b_{n} = (n - 1) b_{1} b_{n}$

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Similar questions

b_{1}, b_{2}, b_{3}, \dots, b_{n} ϵ H . P .

b_{1} b_{2} + b_{2} b_{3} + b_{3} b_{4} + \dots, b_{n - 1} b_{n} =

b_{n + 1} = \frac{1}{1 - b_{n}}

n \geq 1

b_{1} = b_{3}

2001 \sum r = 1 b_{r}

Q. Show that in the continued fraction

\frac{b_{1}}{a_{1}} - \frac{b_{2}}{a_{2}} - \frac{b_{3}}{a_{3}} - \dots

p_{n} = a_{n} p_{n - 1} - b_{n} p_{n - 2}, q_{n} = a_{n} q_{n - 1} - b_{n} q_{n - 2}

Q. Amit wants to use Bayes’ formula

P (\frac{B i}{A}) = \frac{P (B i) P (\frac{A}{B i})}{\sum (B i) P (\frac{A}{B i})}

To use this formula,

Write a general polynomial $q (z)$ of degree n with coefficients that are
$b_{0}, b_{1}, b_{2}, . . . . . . . . . b_{n}$ What are the conditions on
$b_{0}, b_{1}, b_{2}, . . . . . . . . . b_{n}$ .

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