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Find the sum of the infinite geometric series.\newline−9−185−3625−72125+-9 - \frac{18}{5} - \frac{36}{25} - \frac{72}{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−9−185−3625−72125+-9 - \frac{18}{5} - \frac{36}{25} - \frac{72}{125} + \newlineWrite your answer as an integer or a fraction in simplest form.\newline______
  1. Identify Terms and Formula: 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. The first term of the series is −9-9, and we can find the common ratio by dividing the second term by the first term. r=−18/5−9=−18/5−1/9=1845=25r = \frac{-18/5}{-9} = \frac{-18/5}{-1/9} = \frac{18}{45} = \frac{2}{5}
  2. Calculate Common Ratio: Now that we have the first term a=−9a = -9 and the common ratio r=25r = \frac{2}{5}, we can use the formula for the sum of an infinite geometric series:\newlineS=a1−rS = \frac{a}{1 - r}\newlineS=−91−25S = \frac{-9}{1 - \frac{2}{5}}
  3. Apply Formula for Sum: We need to simplify the denominator of the fraction:\newline1−25=55−25=351 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5}\newlineSo the sum SS becomes:\newlineS=−9(35)S = -\frac{9}{\left(\frac{3}{5}\right)}
  4. Simplify Denominator: To divide by a fraction, we multiply by its reciprocal:\newlineS=−9×(53)S = -9 \times \left(\frac{5}{3}\right)\newlineS=−453S = -\frac{45}{3}\newlineS=−15S = -15

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