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Find the sum of the infinite geometric series.\newline−3−65−1225−24125+-3 - \frac{6}{5} - \frac{12}{25} - \frac{24}{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−3−65−1225−24125+-3 - \frac{6}{5} - \frac{12}{25} - \frac{24}{125} + \newlineWrite your answer as an integer or a fraction in simplest form.\newline______
  1. Identify terms and ratio: Identify the first term aa and the common ratio rr of the geometric series.\newlineThe first term aa is −3-3.\newlineThe common ratio rr can be found by dividing the second term by the first term: r=(−6/5)/(−3)=2/5r = (-6/5) / (-3) = 2/5.
  2. Check convergence: Determine if the series is convergent.\newlineA geometric series converges if the absolute value of the common ratio ∣r∣\lvert r \rvert is less than 11.\newlineIn this case, ∣25∣=0.4\lvert \frac{2}{5} \rvert = 0.4, which is less than 11, so the series is convergent.
  3. Use sum formula: Use the formula for the sum of an infinite geometric series.\newlineThe sum SS of an infinite geometric series is given by S=a1−rS = \frac{a}{1 - r}, where aa is the first term and rr is the common ratio.
  4. Calculate sum: Calculate the sum using the values of aa and rr.S=−31−(25)S = \frac{-3}{1 - \left(\frac{2}{5}\right)}S=−355−25S = \frac{-3}{\frac{5}{5} - \frac{2}{5}}S=−335S = \frac{-3}{\frac{3}{5}}S=(−3)⋅(53)S = (-3) \cdot \left(\frac{5}{3}\right)S=−5S = -5

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