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Q.
If $a_1, a_2,...... a_{n+1}$ are in A.P. then $\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}+.......+\frac{1}{a_{n}a_{n+1}}$ is
Sequences and Series
Solution:
$a_1, a_2, a_3, ........., a_{n+1}$ are in A.P. and common difference $= d$
Let $S=\frac{1}{a_{1}a_{2}}+\frac{1}{a_{2}a_{3}}+.......+\frac{1}{a_{n}a_{n+1}}$
$\Rightarrow S=\frac{1}{d}\left\{\frac{d}{a_{1}a_{2}}+\frac{d}{a_{2}a_{3}}+.......+\frac{d}{a_{n}a_{n+1}}\right\}$
$\Rightarrow S=\frac{1}{d}\left\{\frac{a_{2}-a_{1}}{a_{1}a_{2}}+\frac{a_{3}-a_{2}}{a_{2}a_{3}}+......+\frac{a_{n+1}-a_{n}}{a_{n}a_{n+1}}\right\}$
$\Rightarrow S=\frac{1}{d}\left\{\frac{1}{a_{1}}-\frac{1}{a_{n+1}}\right\}=\frac{1}{d}\left\{\frac{a_{n+1}-a_{1}}{a_{1}a_{n+1}}\right\}$
$\Rightarrow S=\frac{1}{d}\left\{\frac{nd}{a_{1}a_{n+1}}\right\}=\frac{n}{a_{1}a_{n+1}}$