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Quantum Numbers

If a1, a2, ...

Question

If $a_{1}, a_{2}, . . ., a_{n}$ are in H.P. and $d$ is the common difference of the corresponding A.P., then the expression $a_{1} a_{2} + a_{2} a_{3} + . . . . . + a_{n - 1} a_{n}$ is equal to

A

\frac{a_{1} - a_{n}}{d}

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B

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

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C

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

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D

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

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Solution

The correct options are
A $\frac{a_{1} - a_{n}}{d}$
D $(n - 1) a_{1} a_{n}$
$a_{1} a_{2} = \frac{a_{1} a_{2} d}{d} = \frac{a_{1} a_{2}}{d} (\frac{1}{a_{2}} - \frac{1}{a_{1}}) = \frac{a_{1} - a_{2}}{d}$
Similarly, $a_{2} a_{3} = \frac{a_{2} - a_{3}}{d}, a_{3} a_{4} = \frac{a_{3} - a_{4}}{d}, . . ., a_{n - 1} a_{n} = \frac{a_{n - 1} - a_{n}}{d}$
Now,
$a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . . . + a_{n - 1} a_{n}$
$= \frac{1}{d} [a_{1} - a_{2} + a_{2} - a_{3} + . . . . . + a_{n - 1} - a_{n}]$
$= \frac{a_{1} - a_{n}}{d} . . . . . (i)$
We have,
$\frac{1}{a_{n}} = \frac{1}{a_{1}} + (n - 1) d$
$\Rightarrow \frac{\frac{1}{a_{n}} - \frac{1}{a_{1}}}{n - 1} = d$
Substituting the value of $d$ in (i), we get,
$a_{1} a_{2} + a_{2} a_{3} + . . . . . + a_{n - 1} a_{n} = (n - 1) a_{1} a_{n}$ .
Hence, options A and D are correct.

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Similar questions

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}

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

are in HP then the expression

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

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

is an A.P. with common difference d then

t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}})] + t a m^{- 1} (\frac{d}{1 + a_{2} a_{3}})] + . . . + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

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

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

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