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Find the sum of the infinite geometric series.\newline−10−152−458−13532+-10 - \frac{15}{2} - \frac{45}{8} - \frac{135}{32} + \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−10−152−458−13532+-10 - \frac{15}{2} - \frac{45}{8} - \frac{135}{32} + \newlineWrite your answer as an integer or a fraction in simplest form.\newline______
  1. Identify Terms: First, we need to identify the first term aa and the common ratio rr of the geometric series.\newlineThe first term aa is −10-10.\newlineTo find the common ratio rr, we divide the second term by the first term.\newliner=−15/2−10=34r = \frac{-15/2}{-10} = \frac{3}{4}
  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=a(1−r)S = \frac{a}{(1 - r)}, provided that ∣r∣<1|r| < 1. In this case, ∣34∣<1|\frac{3}{4}| < 1, so we can use the formula.
  3. Use Sum Formula: Let's calculate the sum using the formula:\newlineS=a1−r=−101−34S = \frac{a}{1 - r} = \frac{-10}{1 - \frac{3}{4}}
  4. Simplify Denominator: Simplify the denominator: 1−34=141 - \frac{3}{4} = \frac{1}{4}
  5. Divide First Term: Now, divide the first term by the simplified denominator:\newlineS=−10(14)=−10×(41)=−40S = -\frac{10}{\left(\frac{1}{4}\right)} = -10 \times \left(\frac{4}{1}\right) = -40

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