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Find the sum of the infinite geometric series.\newline3+95+2725+81125+…3 + \frac{9}{5} + \frac{27}{25} + \frac{81}{125} + \dots\newlineWrite your answer as an integer or a fraction in simplest form.\newline______

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Q. Find the sum of the infinite geometric series.\newline3+95+2725+81125+…3 + \frac{9}{5} + \frac{27}{25} + \frac{81}{125} + \dots\newlineWrite your answer as an integer or a fraction in simplest form.\newline______
  1. Identify Terms and Ratio: To find the sum of an infinite geometric series, we need to identify the first term aa and the common ratio rr of the series. The formula for the sum of an infinite geometric series is S=a1−rS = \frac{a}{1 - r}, where ∣r∣<1|r| < 1.
  2. Calculate Common Ratio: The first term aa of the series is 33. To find the common ratio rr, we divide the second term by the first term: 95/3=95×13=35\frac{9}{5} / 3 = \frac{9}{5} \times \frac{1}{3} = \frac{3}{5}. So, the common ratio rr is 35\frac{3}{5}.
  3. Apply Sum Formula: Now we can apply the formula for the sum of an infinite geometric series: S=a1−rS = \frac{a}{1 - r}. Plugging in the values we have S=31−35S = \frac{3}{1 - \frac{3}{5}}.
  4. Simplify Denominator: Simplify the denominator: 1−35=55−35=251 - \frac{3}{5} = \frac{5}{5} - \frac{3}{5} = \frac{2}{5}.
  5. Calculate Final Sum: Now, calculate the sum: S=3(25)=3×(52)=152S = \frac{3}{\left(\frac{2}{5}\right)} = 3 \times \left(\frac{5}{2}\right) = \frac{15}{2}.

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