$\text{ Let }{a}_1,a_2,a_3,\dots ,a_{10}\text{ be in G.P. with }{a}_{51}=25\text{ and }\sum _{i=1}^{101} a_i=125.\text{ Then the value of }\sum _{i=1}^{101} \left(\frac{1}{a_i}\right)\text{ is }$

$\begin{array}{l}\text{ Let the }{1}^{\text{st }}\text{ term be }a\text{ and common ratio be }r\text{ . }\\ \text{ Then }\frac{a\left(1-r^{101}\right)}{1-r}=125\end{array}$

$\begin{array}{l}\sum _{r=1}^{101} \frac{1}{a_i}=\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots +\frac{1}{a_{101}}\right)\\ =\frac{1}{a}\left(\frac{1-r^{101}}{1-r}\right)\cdot \frac{1}{r^{100}}=\frac{125}{{\left(a{r}^{50}\right)}^2}=\frac{125}{625}=\frac{1}{5}\end{array}$