First slide

Geometric progression

Question

Let a_n be the nth term of the G.P. of positive numbers. $Let \sum_{n = 1}^{100} a_{2 n} = α and \sum_{n = 1}^{100} a_{2 n - 1} = β, such that α \neq β, then$ the common ratio is

Moderate

A

$\frac{α}{β}$
B

$\frac{β}{α}$
C

$\sqrt{\frac{α}{β}}$
D

$\sqrt{\frac{β}{α}}$

Solution

Let a be the first term and r, the common ratio of the given G.P. Then
$α - \sum_{n = 1}^{100} a_{2 n} \Rightarrow α - a_{2} + a_{4} + \dots + a_{200}$
$\begin{array}{ll} \Rightarrow α = ar + {ar}^{3} + \dots + {ar}^{199} \\ \Rightarrow α = ar (1 + r^{2} + r^{4} + \dots + r^{198}) (1) \\ and β = \sum_{n = 1}^{100} a_{2 n - 1} \Rightarrow β = a_{1} + a_{3} + \dots + a_{199} \\ \Rightarrow β = a + {ar}^{2} + \dots + {ar}^{198} (2) \\ \Rightarrow β = a (1 + r^{2} + \dots + r^{198}) \end{array}$
From Eqs (1) and (2), we get
$\frac{α}{β} = r$ .

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