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Implicit Differentiation

If a1,a2,a3...

Question

If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are in HP, then $a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + a_{4} a_{5}$ is equal to

A

2 a_{1} a_{5}

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B

3 a_{1} a_{5}

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C

4 a_{1} a_{5}

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D

- 4

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Solution

The correct option is D $4 a_{1} a_{5}$
If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are in HP, then their reciprocal will be in AP.
$\frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - \frac{1}{a_{2}}$
$\frac{2}{a_{2}} = \frac{a_{1} + a_{3}}{a_{1} a_{3}}$
$2 (a_{1} a_{3}) = a_{1} a_{2} + a_{2} a_{3}$
Similarly, $2 (a_{3} a_{5}) = a_{3} a_{4} + a_{4} a_{5}$
$a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + a_{4} a_{5} = 2 (a_{1} a_{3} + a_{3} a_{5})$
Applying that condition that $\frac{1}{a_{1}}, \frac{1}{a_{3}}, \frac{1}{a_{5}}$ will be in AP.
$(a_{1} a_{3} + a_{3} a_{5}) = 2 a_{1} a_{5}$
$a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + a_{4} a_{5} = 2 (2 a_{1} a_{5})$
$= 4 a_{1} a_{5}$

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Similar questions

a_{1}, a_{2}, a_{3}, . . . a_{n}

a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . + a_{n - 1} a_{n} =

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} =

a_{1}, a_{2}, a_{3}, . . . .,

an are in AP , then

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . . . + \frac{1}{a_{n - 1} a_{n}}

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