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Find the sum of the infinite geometric series.\newline−8−325−12825−512125+-8 - \frac{32}{5} - \frac{128}{25} - \frac{512}{125} + \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−8−325−12825−512125+-8 - \frac{32}{5} - \frac{128}{25} - \frac{512}{125} + \newlineWrite your answer as an integer or a fraction in simplest form.\newline______
  1. Identify terms and ratio: First, we need to identify the first term aa and the common ratio rr of the geometric series.\newlineThe first term aa is −8-8.\newlineTo find the common ratio rr, we divide the second term by the first term: r=(−32/5)/(−8)=4/5r = (-32/5) / (-8) = 4/5.
  2. Calculate common ratio: Now that we have the first term and the common ratio, we can use the formula for the sum of an infinite geometric series, which is S=a1−rS = \frac{a}{1 - r}, provided that ∣r∣<1|r| < 1. In this case, ∣45∣<1|\frac{4}{5}| < 1, so we can use the formula.
  3. Use formula for sum: Let's plug the values into the formula: S=−81−(45)S = \frac{-8}{1 - \left(\frac{4}{5}\right)}.
  4. Calculate denominator: Now we calculate the denominator: 1−(45)=55−45=151 - \left(\frac{4}{5}\right) = \frac{5}{5} - \frac{4}{5} = \frac{1}{5}.
  5. Calculate sum: Next, we calculate the sum: S=(−8)/(15)S = (-8) / (\frac{1}{5}).\newlineTo divide by a fraction, we multiply by its reciprocal: S=(−8)×(51)=−40S = (-8) \times (\frac{5}{1}) = -40.

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