Question

Let $a_{1}, a_{2}, a_{3}, \ldots, a_{10}$ be in G.P. with $a_{51}=25$ and $\displaystyle\sum_{i=1}^{101} a_{i}=125 .$ Then the value of $ \displaystyle\sum_{i=1}^{101}\left(\frac{1}{a_{i}}\right)$ is ______.

Answer 1

Let the $1^{\text {st }}$ term be $a$ and common ratio be $r$.
Then $ \frac{a\left(1-r^{101}\right)}{1-r}=125 $
$ \displaystyle\sum_{r=1}^{101} \frac{1}{a_{i}}=\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{101}}\right) $
$=\frac{1}{a}\left(\frac{1-r^{101}}{1-r}\right) \cdot \frac{1}{r^{100}}=\frac{125}{\left(a r^{50}\right)^{2}}=\frac{125}{625}=\frac{1}{5}$