Sequences And Series — Easy JEE math MCQ
The sum of the infinite geometric series \(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\) is:
- A. \(\frac{3}{2}\)
- B. \(\frac{2}{3}\)
- C. 3
- D. \(\frac{1}{2}\)
Solution
For an infinite GP with first term \(a\) and common ratio \(r\) where \(|r| < 1\):
\[
S_\infty = \frac{a}{1-r}
\]
Here \(a = 1\) and \(r = \frac{1}{3}\):
\[
S_\infty = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}
\]
So the correct option is A.