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)

$a_1,a_2,a_3,.........,a_{n+1}$ are in A.P. and common difference $=d$
Let $S=\frac{1}{a_1{a}_2}+\frac{1}{a_2{a}_3}+.......+\frac{1}{a_n{a}_{n+1}}$
$\Rightarrow S=\frac{1}{d}\left\{\frac{d}{a_1{a}_2}+\frac{d}{a_2{a}_3}+.......+\frac{d}{a_n{a}_{n+1}}\right\}$
$\Rightarrow S=\frac{1}{d}\left\{\frac{a_2-a_1}{a_1{a}_2}+\frac{a_3-a_2}{a_2{a}_3}+......+\frac{a_{n+1}-a_n}{a_n{a}_{n+1}}\right\}$
$\Rightarrow S=\frac{1}{d}\left\{\frac{1}{a_1}-\frac{1}{a_{n+1}}\right\}=\frac{1}{d}\left\{\frac{a_{n+1}-a_1}{a_1{a}_{n+1}}\right\}$
$\Rightarrow S=\frac{1}{d}\left\{\frac{nd}{a_1{a}_{n+1}}\right\}=\frac{n}{a_1{a}_{n+1}}$