First slide

Geometric progression

Question

$Let a_{1}, a_{2}, a_{3}, \dots a_{101} are in G.P with a_{51} = 25, and \sum_{i = 1}^{101} a_{i} = 125 . Then the value of \sum_{i = 1}^{101} (\frac{1}{ai}) is$

Moderate

A

5
B

$\frac{1}{5}$
C

$\frac{1}{25}$
D

$\frac{1}{125}$

Solution

$\begin{array}{l} \sum_{i = 1}^{101} ai = \frac{a (1 - r^{101})}{1 - r} = 125 \\ N o w \sum_{i = 1}^{101} \frac{1}{ai} = (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \dots + \frac{1}{a_{101}}) \\ = \frac{1}{a} (\frac{1 - r^{101}}{1 - r}) \frac{1}{r^{100}} \end{array}$
$= \frac{125}{{({ar}^{50})}^{2}} = \frac{125}{625} = 1 / 5$

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