Question:medium

The sum of infinite terms of the GP $\dfrac{\sqrt{2}+1}{\sqrt{2}-1},\;\dfrac{1}{2-\sqrt{2}},\;\dfrac{1}{2},\;\ldots$ is

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Rationalise the first term: $\dfrac{\sqrt{2}+1}{\sqrt{2}-1} \times \dfrac{\sqrt{2}+1}{\sqrt{2}+1} = (\sqrt{2}+1)^2$. Then find $r$ by dividing consecutive terms.

Updated On: Apr 8, 2026

$\sqrt{2}(\sqrt{2}+1)^2$
$(\sqrt{2}+1)^2$
$5\sqrt{2}$
$3\sqrt{2}+\sqrt{5}$

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The Correct Option is A

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Questions Asked in MET exam

Let $ D = \begin{vmatrix} n & n^2 & n^3 \\ n^2 & n^3 & n^5 \\ 1 & 2 & 3 \end{vmatrix} $. Then $ \lim_{n \to \infty} \frac{M_{11} + C_{33}}{(M_{13})^2} $ is: