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Find the sum of the infinite geometric series.\newline−1−13−19−127+-1 - \frac{1}{3} - \frac{1}{9} - \frac{1}{27} + \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.\newline−1−13−19−127+-1 - \frac{1}{3} - \frac{1}{9} - \frac{1}{27} + \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. In the given series, the first term is −1-1 and the common ratio is the ratio between any two consecutive terms, which is 13\frac{1}{3} divided by 11 or −13\frac{-1}{3} divided by −1-1, which gives us 13\frac{1}{3}.
  2. Apply Sum Formula: Now we apply the formula for the sum of an infinite geometric series: S=a(1−r)S = \frac{a}{(1 - r)}. Here, a=−1a = -1 and r=13r = \frac{1}{3}. So, S=−1(1−13)S = \frac{-1}{(1 - \frac{1}{3})}.
  3. Calculate Denominator: We calculate the denominator of the fraction: 1−13=33−13=231 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}.
  4. Substitute Denominator: Now we substitute the denominator back into the formula: S=−1/(23)S = -1 / (\frac{2}{3}).
  5. Multiply by Reciprocal: To divide by a fraction, we multiply by its reciprocal. So, S=−1×(32)S = -1 \times \left(\frac{3}{2}\right).
  6. Perform Multiplication: We perform the multiplication: S=−32S = -\frac{3}{2}. This is the sum of the infinite geometric series in simplest fractional form.

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