First slide

Geometric progression

Question

If $A = 1 + r^{a} + r^{2 a} + r^{3 a} + \dots \infty$ and If $B = 1 + r^{b} + r^{2 b} + r^{3 b} + \dots \infty,$ then $\frac{a}{b}$ is equal to

Moderate

A

$\log_{B} A$
B

$\log_{1 - B} (1 - A)$
C

$\log_{\frac{B - 1}{B}} (\frac{A - 1}{A})$
D

None of these

Solution

$\begin{aligned} A = \frac{1}{1 - r^{a}} \Rightarrow 1 - r^{a} = \frac{1}{A} \Rightarrow r^{a} = 1 - \frac{1}{A} = \frac{A - 1}{A} \\ B = \frac{1}{1 - r^{b}} \Rightarrow 1 - r^{b} = \frac{1}{B} \Rightarrow r^{b} = 1 - \frac{1}{B} = \frac{B - 1}{B} \end{aligned}$
$\begin{array}{l} ∴ alog r = \log (\frac{A - 1}{A}) and blog r = \log (\frac{B - 1}{B}) \\ \frac{a}{b} = \frac{\log (\frac{A - 1}{A})}{\log (\frac{B - 1}{B})} \\ = \log_{\frac{B - 1}{B}} (\frac{A - 1}{A}) \end{array}$

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