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Arithmetic Progression

Let a n , nϵ ...

Question

Let $a_{n}, n ϵ N$ is an A.P. with common difference $^{'} d^{'}$ and whose all terms are non-zero. If $n$ approaches infinity, then the sum $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . + \frac{1}{a_{n} a_{n + 1}}$ will approach

$\frac{1}{a_{1} d}$

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$\frac{2}{a_{1} d}$

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$\frac{1}{2 a_{1} d}$

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$a_{1} d$

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Solution

The correct option is A
$\frac{1}{a_{1} d}$

The A.P. is $a_{1}, a_{1} + d, a_{1} + 2 d, . . .$

The given expression can be written as

$\frac{1}{a_{1} (a_{1} + d)} + \frac{1}{(a_{1} + d) (a_{1} + 2 d)} + . . . + \frac{1}{(a_{1} + (n - 1) d) (a_{1} + n d)}$

$= \frac{1}{d} [\frac{a_{1} + d - a_{1}}{a_{1} (a_{1} + d)} + \frac{(a_{1} + 2 d) - (a_{1} + d)}{(a_{1} + d) (a_{1} + 2 d)} + . . . + \frac{(a_{1} + n d) - (a_{1} + (n - 1) d)}{(a_{1} + n d) (a_{1} + (n - 1) d)}]$

$= \frac{1}{d} [\frac{1}{a_{1}} - \frac{1}{a_{1} + d} + \frac{1}{a_{1} + d} - \frac{1}{a_{1} + 2 d} + . . . + \frac{1}{a_{1} + (n - 1) d} - \frac{1}{a_{1} + n d}]$

$= \frac{1}{d} [\frac{1}{a_{1}} - \frac{1}{a_{1} + n d}]$

$= \frac{1}{a_{1} d}$

Suggest Corrections

Similar questions

a_{n}, n ϵ N

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a_{n}, n \in N

^{'} d^{'}

n

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

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S = {tan}^{- 1} \frac{d}{1 + a_{1} a_{2}} + {tan}^{- 1} \frac{d}{1 + a_{2} a_{3}} +

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θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

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tan [{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \dots + {tan}^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

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