Question

Let $a_1,a_2,a_3,\dots ,a_{10}$ be in G.P. with $a_{51}=25$ and $\sum _{i=1}^{101}{a}_i=125.$ Then the value of $\sum _{i=1}^{101}\left(\frac{1}{a_i}\right)$ is ______.

Answer 1

Answer: 0.2

Let the $1^{\text{st }}$ term be $a$ and common ratio be $r$.
Then $\frac{a\left(1-r^{101}\right)}{1-r}=125$
$\sum _{r=1}^{101}\frac{1}{a_i}=\left(\frac{1}{a_1}+\frac{1}{a_2}+\dots +\frac{1}{a_{101}}\right)$
$=\frac{1}{a}\left(\frac{1-r^{101}}{1-r}\right)\cdot \frac{1}{r^{100}}=\frac{125}{{\left(a{r}^{50}\right)}^2}=\frac{125}{625}=\frac{1}{5}$