Let $a_{n}, n \in N$ is an AP with common difference $^{'} d^{'}$ and all whose 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 C $\frac{1}{a_{1} d}$
$\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . . . + \frac{1}{a_{n} a_{n + 1}} = \frac{1}{d} (\frac{a_{2} - a_{1}}{a_{1} a_{2}} + \frac{a_{3} - a_{2}}{a_{2} a_{3}} + . . . . .) = \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + . . . . .) ⟹ = \frac{1}{a_{1} d}$

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Similar questions

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

Q. Let

a_{n}, n ϵ N

is an A.P. with common difference 'd' and all whose 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}}

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Q. If

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Let $S_{n}$ denote the sum of $n$ terms of an A.P. whose first term is $a$ . If the common difference $d$ is given by $d = S_{n} - k S_{n} - 1 + S_{n - 2}$ , then $k = ?$

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Let an,n∈N is an AP with common difference ′d′ and all whose terms are non-zero.If n approaches infinity, then the sum 1a1a2+1a2a3+.....+1anan+1 will approach

The correct option is C 1a1d1a1a2+1a2a3+.....+1anan+1=1d(a2−a1a1a2+a3−a2a2a3+.....)=1d(1a1−1a2+1a2−1a3+.....)⟹=1a1d

Let $a_{n}, n \in N$ is an AP with common difference $^{'} d^{'}$ and all whose 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