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

1 Answer

3 votes

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)).

User Eduard Dumitru
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