Which term of the geometric sequence frac{1}{ | vedclass

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(A) The given sequence is a geometric progression where the first term $a = \frac{1}{3}$ and the common ratio $r = \frac{1/9}{1/3} = \frac{1}{3}$.Let

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$9$

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(A) The given sequence is a geometric progression where the first term $a = \frac{1}{3}$ and the common ratio $r = \frac{1/9}{1/3} = \frac{1}{3}$.Let the $n^{th}$ term be $a_n = \frac{1}{19683}$.The formula for the $n^{th}$ term of a geometric sequence is $a_n = a \cdot r^{n-1}$.Substituting the values,we get $\frac{1}{19683} = \frac{1}{3} \cdot (\frac{1}{3})^{n-1}$.This simplifies to $\frac{1}{19683} = (\frac{1}{3})^n$.Since $3^9 = 19683$,we have $(\frac{1}{3})^9 = \frac{1}{19683}$.Comparing the powers,we get $n = 9$.

Which term of the geometric sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$?

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