Ltn→∞1aa+d1a+da+2d+.....+1a+n-1da+nd=

Ltn→∞1aa+d+1a+da+2d+.......+1a+n−1da+nd=Ltn→∞1ddaa+d+da+da+2d......+da+n−1da+nd=Ltn→∞1d1a−1a+d+1a+d−1a+2d+.......+1a+n−1d−1a+nd=Ltn→∞1d1a−1a+nd=1d1a−0=1ad

Ltn→∞1aa+d1a+da+2d+.....+1a+n-1da+nd=

Ltn→∞1aa+d+1a+da+2d+.......+1a+n−1da+nd=Ltn→∞1ddaa+d+da+da+2d......+da+n−1da+nd=Ltn→∞1d1a−1a+d+1a+d−1a+2d+.......+1a+n−1d−1a+nd=Ltn→∞1d1a−1a+nd=1d1a−0=1ad

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$\underset{n \to \infty}{Lt} (\frac{1}{a (a + d)} \frac{1}{(a + d) (a + 2 d)} + . . . . . + \frac{1}{[a + (n - 1) d] [a + nd]}) =$

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$\frac{1}{a}$

$\frac{1}{d}$

$\frac{1}{a d}$

$a d$

answer is C.

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Detailed Solution

$\begin{array}{l} \underset{n \to \infty}{L t} \frac{1}{a (a + d)} + \frac{1}{(a + d) (a + 2 d)} + .... . . . + \frac{1}{(a + (n - 1) d)} (a + n d) \\ = \underset{n \to \infty}{L t} \frac{1}{d} [\frac{d}{a (a + d)} + \frac{d}{(a + d) (a + 2 d)} ...... + \frac{d}{(a + (n - 1) d) (a + n d)}] \end{array}$
$\begin{array}{l} = \underset{n \to \infty}{L t} \frac{1}{d} [\frac{1}{a} - \frac{1}{a + d} + \frac{1}{a + d} - \frac{1}{a + 2 d} + ....... + \frac{1}{a + (n - 1) d} - \frac{1}{a + (n d)}] \\ = \underset{n \to \infty}{L t} \frac{1}{d} [\frac{1}{a} - \frac{1}{a + n d}] = \frac{1}{d} [\frac{1}{a} - 0] = \frac{1}{a d} \end{array}$