Question

The sum of the infinite geometric series \( 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots \) (rounded off to one decimal place) is \(\underline{\hspace{1cm}}\).

Hint:

For infinite geometric series, the sum exists only if \( |r| < 1 \).

Answer 1

Correct Answer: 1.5

The given series is geometric with first term \( a = 1 \) and common ratio \( r = \frac{1}{3} \).
The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2} = 1.5 \]
Thus, the required sum is \( 1.5 \).