The sum of infinite terms of the geometric pr | vedclass

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Question

(A) The given geometric progression is $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{\sqrt{2}(\sqrt{2} - 1)}, \frac{1}{2}, \dots$First term $a = \frac{

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$\sqrt{2}(\sqrt{2} + 1)^2$

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(A) The given geometric progression is $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{\sqrt{2}(\sqrt{2} - 1)}, \frac{1}{2}, \dots$First term $a = \frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{(\sqrt{2} + 1)^2}{2 - 1} = (\sqrt{2} + 1)^2 = 3 + 2\sqrt{2}$.Common ratio $r = \frac{1}{\sqrt{2}(\sqrt{2} - 1)} \div \frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \frac{1}{\sqrt{2}(\sqrt{2} + 1)} = \frac{1}{2 + \sqrt{2}}$.The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.$S = \frac{(\sqrt{2} + 1)^2}{1 - \frac{1}{\sqrt{2}(\sqrt{2} + 1)}} = \frac{(\sqrt{2} + 1)^2}{\frac{\sqrt{2}(\sqrt{2} + 1) - 1}{\sqrt{2}(\sqrt{2} + 1)}} = \frac{(\sqrt{2} + 1)^2 \cdot \sqrt{2}(\sqrt{2} + 1)}{2 + \sqrt{2} - 1} = \frac{\sqrt{2}(\sqrt{2} + 1)^3}{1 + \sqrt{2}} = \sqrt{2}(\sqrt{2} + 1)^2$.

The sum of infinite terms of the geometric progression $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{2 - \sqrt{2}}, \frac{1}{2}, \dots$ is

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