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

  • A
    $8$
  • B
    $9$
  • C
    $10$
  • D
    $11$

Explore More

Similar Questions

The arithmetic mean of two numbers $b$ and $c$ is $a$,and $g_1$ and $g_2$ are two geometric means between them. If $g_1^3 + g_2^3 = kabc$,then $k = \dots$

Difficult
View Solution

If $G_1$ and $G_2$ are two geometric means between two numbers and $A$ is the arithmetic mean between them,then find the value of $\frac{G_1^2}{G_2} + \frac{G_2^2}{G_1}$.

The sum of three numbers in a Geometric Progression $(GP)$ is $38$ and their product is $1728$. The largest of these numbers is.......

Let $a_{n}$ be the $n^{\text{th}}$ term of a $G$.$P$. of positive terms. If $\sum_{n=1}^{100} a_{2n+1} = 200$ and $\sum_{n=1}^{100} a_{2n} = 100$,then $\sum_{n=1}^{200} a_{n}$ is equal to:

If the sum of three terms of a Geometric Progression $(GP)$ is $19$ and their product is $216$,then the common ratio of this $GP$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo