8. Sequences and Series
easy

अनुक्रम का कौन सा पद.

$\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots ; \frac{1}{19683}$ है ?

A

$9^{\text {th }}$

B

$9^{\text {th }}$

C

$9^{\text {th }}$

D

$9^{\text {th }}$

Solution

The given sequence is $\frac{1}{3}, \frac{1}{9}, \frac{1}{27} \dots$

Here, $a=\frac{1}{3}$ and $r=\frac{1}{9} \div \frac{1}{3}=\frac{1}{3}$

Let the $n^{t h}$ term of the given sequence be $\frac{1}{19683}$

$a_{n}=a r^{n-1}$

$\therefore a r^{n-1}=\frac{1}{19683}$

$\Rightarrow\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{n-1}=\frac{1}{19683}$

$\Rightarrow\left(\frac{1}{3}\right)^{n}=\left(\frac{1}{3}\right)^{9}$

$\Rightarrow n=9$

Thus, the $9^{\text {th }}$ term of the given sequence is $\frac{1}{19683}$

Standard 11
Mathematics

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