Question

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

• A. 5
• B. $\frac{1}{5}$
• C. $\frac{1}{25}$
• D. $\frac{1}{125}$

Answer 1

Answer: (B)

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