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Find the sum of the infinite geometric series.\newline−3−1−13−19+-3 - 1 - \frac{1}{3} - \frac{1}{9} + \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−3−1−13−19+-3 - 1 - \frac{1}{3} - \frac{1}{9} + \newlineWrite your answer as an integer or a fraction in simplest form.\newline______
  1. Identify terms and ratio: Identify the first term a1a_1 and the common ratio rr of the geometric series.a1=−3a_1 = -3To find the common ratio, divide the second term by the first term:r=a2a1=−1−3=13r = \frac{a_2}{a_1} = \frac{-1}{-3} = \frac{1}{3}
  2. Use sum formula: Use the formula for the sum of an infinite geometric series, which is S=a11−rS = \frac{a_1}{1 - r}, where SS is the sum, a1a_1 is the first term, and rr is the common ratio.\newlineHere, a1=−3a_1 = -3 and r=13r = \frac{1}{3}.
  3. Substitute values: Substitute the values of a1a_1 and rr into the formula to find the sum SS.\newlineS=−31−(13)S = \frac{-3}{1 - (\frac{1}{3})}
  4. Simplify denominator: Simplify the denominator of the fraction. 1−(13)=33−13=231 - \left(\frac{1}{3}\right) = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}
  5. Calculate sum: Now, calculate the sum SS with the simplified denominator.S=−323S = \frac{-3}{\frac{2}{3}}
  6. Multiply numerator: Multiply the numerator by the reciprocal of the denominator to find the sum.\newlineS=−3×(32)S = -3 \times \left(\frac{3}{2}\right)\newlineS=−92S = -\frac{9}{2}

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