Question

15th term of the series $243,81,27,....$ is

• A. ${\left(\frac{1}{3}\right)}^9$
• B. ${\left(\frac{1}{3}\right)}^{10}$
• C. ${\left(\frac{1}{3}\right)}^{10}{\left(\frac{1}{3}\right)}^{-1}$
• D. ${\left(\frac{1}{3}\right)}^{10}\left(\frac{1}{3}\right)$

Answer 1

Answer: (A), (C)

${\left(\frac{1}{3}\right)}^9$
${\left(\frac{1}{3}\right)}^{10}{\left(\frac{1}{3}\right)}^{-1}$

Given series is $243,81,27,.....$

Common ratio $=\frac{81}{243}=\frac{27}{81}=\frac{1}{3}$
Hence, the above series is in G.P.
$t_{15}=243{\left(\frac{1}{3}\right)}^{15-1}=\frac{3^5}{3^{14}}=\frac{1}{3^9}=\frac{3}{3^9\left(3\right)}={\left(\frac{1}{3}\right)}^{10}{\left(\frac{1}{3}\right)}^{-1}$