Which term of the sequence frac{1}{3}, frac{1 | vedclass

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

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$9$

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

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

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