Which term of the sequence31,91,271...........is196831?

Here a=31,r=319‘1=31 Let Tn=196831⇒arn−1=196831⇒ 31(31)n−1=(31)9

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Which term of the sequence {1}{3} , {1}{9} , {1}{27} .........

Question

Which term of the sequence
$\frac{1}{3}, \frac{1}{9}, \frac{1}{27} ........... i s \frac{1}{19683} ?$

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Which term of the sequence

\frac{1}{3}, \frac{1}{9}, \frac{1}{27} ........... i s \frac{1}{19683} ?

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Here

a = \frac{1}{3}, r = \frac{\frac{‘1}{9}}{\frac{1}{3}} = \frac{1}{3}

Let

T_{n} = \frac{1}{19683}

\Rightarrow a r^{n - 1} = \frac{1}{19683} \Rightarrow \frac{1}{3} (\frac{1}{3})^{n - 1} = (\frac{1}{3})^{9}

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Question Text	Which term of the sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27} ........... i s \frac{1}{19683} ?$
Answer Type	Text solution:1
Upvotes	150