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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
