Which term of the following sequence: 2, 2sqr | vedclass

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(A) The given sequence is $2, 2\sqrt{2}, 4, \ldots$ Here,the first term $a = 2$ and the common ratio $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$. Let the $n^

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$13$

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(A) The given sequence is $2, 2\sqrt{2}, 4, \ldots$ Here,the first term $a = 2$ and the common ratio $r = \frac{2\sqrt{2}}{2} = \sqrt{2}$. Let the $n^{	ext{th}}$ term of the sequence be $128$. The formula for the $n^{	ext{th}}$ term of a geometric progression is $a_n = a r^{n-1}$. Substituting the values,we get: $2(\sqrt{2})^{n-1} = 128$ Dividing both sides by $2$: $(\sqrt{2})^{n-1} = 64$ Expressing in terms of base $2$: $(2^{1/2})^{n-1} = 2^6$ $2^{\frac{n-1}{2}} = 2^6$ Equating the exponents: $\frac{n-1}{2} = 6$ $n-1 = 12$ $n = 13$ Thus,the $13^{	ext{th}}$ term of the sequence is $128$.

Which term of the following sequence: $2, 2\sqrt{2}, 4, \ldots$ is $128$ (in $^{\text{th}}$)?

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