Question

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

• A. $\frac{n-1}{a_{1}a_{n+1}}$
• B. $\frac{1}{a_{1}a_{n+1}}$
• C. $\frac{n+1}{a_{1}a_{n+1}}$
• D. $\frac{n}{a_{1}a_{n+1}}$

Answer 1

Answer: (D)

$\frac{n}{a_{1}a_{n+1}}$

Answer 2

$a_1, a_2, a_3, ........., a_{n+1}$ 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}}