Question

if {a \index{n}} be an Arithmetic sequence , calculate \frac{1}{a1a2} + \frac{1}{a2a3} +...+ \frac{1}{a39a40}​

Answer 1

Given:

$a_n$ is an arithmetic sequence.

To find:

The value of $\dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+...+\dfrac{1}{a_{39}a_{40}}$.

Solution:

We have,

$\dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+...+\dfrac{1}{a_{39}a_{40}}$

It can be written as:

$=\dfrac{d}{d}\left[\dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+...+\dfrac{1}{a_{39}a_{40}}\right]$

$=\dfrac{1}{d}\left[\dfrac{d}{a_1a_2}+\dfrac{d}{a_2a_3}+...+\dfrac{d}{a_{39}a_{40}}\right]$

$=\dfrac{1}{d}\left[\dfrac{a_2-a_1}{a_1a_2}+\dfrac{a_3-a_2}{a_2a_3}+...+\dfrac{a_{40}-a_{39}}{a_{39}a_{40}}\right]$

$=\dfrac{1}{d}\left[\dfrac{1}{a_1}-\dfrac{1}{a_2}+\dfrac{1}{a_2}-\dfrac{1}{a_3}+....+\dfrac{1}{a_{39}}-\dfrac{1}{a_{40}}\right]$

$=\dfrac{1}{d}\left[\dfrac{1}{a_1}-\dfrac{1}{a_{40}}\right]$

$=\dfrac{1}{d}\left[\dfrac{a_{40}-a_1}{a_1a_{40}}\right]$

$=\dfrac{1}{d}\left[\dfrac{a_1+(40-1)d-a_1}{a_1a_{40}}\right]$       $[\because a_n=a_1+(n-1)d]$

$=\dfrac{1}{d}\left[\dfrac{39d}{a_1a_{40}}\right]$

$=\dfrac{39}{a_1a_{40}}$

Therefore, the value of given expression is $\dfrac{39}{a_1a_{40}}$.