Let `a_(1),a_(2),a_(3),"......"a_(10)` are in GP with `a_(51)=25` and `sum_(i=1)^(101)a_(i)=125 " than the value of " sum_(i=1)^(101)((1)/(a_(i)))` eq

Correct Answer - B
Let 1st term be a and common ratio be r, then
`sum_(i=1)^(101)(1)/(a_(i))=125`
` implies (a_(1)+a_(1)r+a_(1)r^(2)+"......"+a_(1)r^(100))=125`
` implies (a_(1)(1-r^(101)))/(1-r)=125 [" let " 0ltrlt1 ] ".....(i)"`
` :. sum_(i=1)^(101)(1)/(a_(i))= (1)/(a_(i))+(1)/(a_(i)r)+(1)/(a_(i)r^(2))+"....."+(1)/(a_(i)r^(101))=((1)/(a_(i))[((1)/(r))^(101)-1])/(((1)/(r)-1)) " " [" here " (1)/(r)gt1]`
` =((1-r^(101)))/(a_(1)r^(100)(1-r))=(1)/(a_(1)r^(100))xx(125)/(a_(1)) " " [ "from Eq. (i)"]`
`=(125)/((a_(1)r^(50))^(2))=(125)/((a^(51))^(2))=(125)/((25)^(2))=(1)/(5)`