$\text{ Let }{\mathrm{a}}_1,{\mathrm{a}}_2,{\mathrm{a}}_3,\dots {\mathrm{a}}_{101}\text{ are in G.P with }{\mathrm{a}}_{51}=25\text{, and }\sum _{\mathrm{i}=1}^{101} {\mathrm{a}}_{\mathrm{i}}=125. \text{ Then the value of }\sum _{i=1}^{101} \left(\frac{1}{\mathrm{ai}}\right)\text{ is }$

$\begin{array}{l} \sum _{i=1}^{101} \mathrm{ai}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{101}\right)}{1-\mathrm{r}}=125\\ Now \sum _{\mathrm{i}=1}^{101} \frac{1}{\mathrm{ai}}=\left(\frac{1}{{\mathrm{a}}_1}+\frac{1}{{\mathrm{a}}_2}+\dots +\frac{1}{{\mathrm{a}}_{101}}\right)\\ =\frac{1}{\mathrm{a}}\left(\frac{1-{\mathrm{r}}^{101}}{1-\mathrm{r}}\right)\frac{1}{{\mathrm{r}}^{100}}\end{array}$

                   $=\frac{125}{{\left({\text{ar}}^{\text{50}}\right)}^2}=\frac{125}{625}=1/5$