Let `a_1, a_2, a_3, ,a_(101)` are in G.P. with `a_(101)=25a n dsum_(i=1)^(201)a_1=625.` Then the value of `sum_(i=1)^(201)1/(a_1)` equals____________.

Answer

1

Explanation

Let a be the first fterm and r be the common ratio of G.P. Then
`a(1-r^(201))/(1-r)=625` (1)
Now `sum_(r=1)^(201)1/(a_(i))=1/(a_(1))+1/(a_(2))+..+1/(a_(201))`
`=1/a+1/(ar)+..+1/(ar^(200))`
`=(1/a((1/r)^(201)-1))/((1/r-1))`
`=1/a((1-r^(201))/(1-r))1/r^(200)`
`=1/axx625/axx1/r^(200)` [from (1)]
`=625/((ar^(100))^(2))`
`=625/(a_(101))^(2)`
`=625/625=1`