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
`(1)/(a_1a_2)+(1)/(a_2a_3)+...+(1)/(a_(39)a_(40))`.

Solution:

We have,

`(1)/(a_1a_2)+(1)/(a_2a_3)+...+(1)/(a_(39)a_(40))`

It can be written as:

`=(d)/(d)\left[(1)/(a_1a_2)+(1)/(a_2a_3)+...+(1)/(a_(39)a_(40))\right]`

`=(1)/(d)\left[(d)/(a_1a_2)+(d)/(a_2a_3)+...+(d)/(a_(39)a_(40))\right]`

`=(1)/(d)\left[(a_2-a_1)/(a_1a_2)+(a_3-a_2)/(a_2a_3)+...+(a_(40)-a_(39))/(a_(39)a_(40))\right]`

`=(1)/(d)\left[(1)/(a_1)-(1)/(a_2)+(1)/(a_2)-(1)/(a_3)+....+(1)/(a_(39))-(1)/(a_(40))\right]`

`=(1)/(d)\left[(1)/(a_1)-(1)/(a_(40))\right]`

`=(1)/(d)\left[(a_(40)-a_1)/(a_1a_(40))\right]`

`=(1)/(d)\left[(a_1+(40-1)d-a_1)/(a_1a_(40))\right]`
`[\because a_n=a_1+(n-1)d]`

`=(1)/(d)\left[(39d)/(a_1a_(40))\right]`

`=(39)/(a_1a_(40))`

Therefore, the value of given expression is
`(39)/(a_1a_(40))`.