Find the sum of the first $n$ terms and the sum of the first $5$ terms of the geometric series $1 + \frac{2}{3} + \frac{4}{9} + \dots$

  • A
    $S_n = 3[1 - (\frac{2}{3})^n], S_5 = \frac{211}{81}$
  • B
    $S_n = 3[1 - (\frac{2}{3})^n], S_5 = \frac{205}{81}$
  • C
    $S_n = 2[1 - (\frac{2}{3})^n], S_5 = \frac{211}{81}$
  • D
    $S_n = 3[1 - (\frac{3}{2})^n], S_5 = \frac{211}{81}$

Explore More

Similar Questions

In an increasing geometric progression of positive terms,the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is $49$. Then the sum of the $4^{\text{th}}$,$6^{\text{th}}$,and $8^{\text{th}}$ terms is:

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of the squares of the first three terms is $33033$,then the sum of these three terms is equal to

In a $GP$ series consisting of positive terms,each term is equal to the sum of the next two terms. Then,the common ratio of this $GP$ series is

If the $4^{th}, 7^{th}$ and $10^{th}$ terms of a $G.P.$ are $a, b, c$ respectively,then the relation between $a, b, c$ is

If the sum of the second,fourth and sixth terms of a $G.P.$ of positive terms is $21$ and the sum of its eighth,tenth and twelfth terms is $15309$,then the sum of its first nine terms 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