If a_1, a_2, a_3, ..., a_n is an arithmetic p | vedclass

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(D) Let the common difference of the given arithmetic progression be $d$.Then,$a_2 - a_1 = a_3 - a_2 = ... = a_n - a_{n-1} = d$.Now,consider the sum $

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$\frac{n - 1}{a_1 a_n}$

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(D) Let the common difference of the given arithmetic progression be $d$.Then,$a_2 - a_1 = a_3 - a_2 = ... = a_n - a_{n-1} = d$.Now,consider the sum $S = \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + ... + \frac{1}{a_{n-1} a_n}$.Multiplying and dividing by $d$,we get $S = \frac{1}{d} [\frac{d}{a_1 a_2} + \frac{d}{a_2 a_3} + ... + \frac{d}{a_{n-1} a_n}]$.Substituting $d = a_{k+1} - a_k$,we get $S = \frac{1}{d} [(\frac{1}{a_1} - \frac{1}{a_2}) + (\frac{1}{a_2} - \frac{1}{a_3}) + ... + (\frac{1}{a_{n-1}} - \frac{1}{a_n})]$.This is a telescoping sum,so $S = \frac{1}{d} [\frac{1}{a_1} - \frac{1}{a_n}] = \frac{1}{d} [\frac{a_n - a_1}{a_1 a_n}]$.Since $a_n = a_1 + (n-1)d$,we have $a_n - a_1 = (n-1)d$.Therefore,$S = \frac{1}{d} [\frac{(n-1)d}{a_1 a_n}] = \frac{n-1}{a_1 a_n}$.

If $a_1, a_2, a_3, ..., a_n$ is an arithmetic progression,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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