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An infinite geometric series has first term -3/2 and sums to twice the common ratio.

Find the sum of all possible values for the common ratio.

If the common ratio r lies between −1 to 1 , we can have the sum of an infinite geometric series.

That is, the sum exits for \(| r |<1\) .

The sum S of an infinite geometric series with \( -1 is given by the formula, \(S=\frac{a_1}{1-r}\)

Let \(a_1 = -\frac{3}{2}\)

Let \( r =\,?\)

\(\begin{array}{|rcll|} \hline S = \frac{a_1}{1-r} &=& 2r \quad & | \quad a_1 = -\frac{3}{2} \\ \frac{-\frac{3}{2}}{1-r} &=& 2r \\ -\frac{3}{2} \cdot \frac{1}{1-r} &=& 2r \quad & | \quad : 2 \\ -\frac{3}{4} \cdot \frac{1}{1-r} &=& r \\ \frac{3}{4} \cdot \frac{1}{r-1} &=& r \quad & | \quad \cdot (r-1) \\ \frac{3}{4} &=& r \cdot (r-1) \quad & | \quad - \frac{3}{4} \\ r \cdot (r-1) - \frac{3}{4} &=& 0 \\ r^2-r- \frac{3}{4} &=& 0 \\\\ r &=& \frac{1\pm\sqrt{1-4\cdot (- \frac{3}{4}) } }{2} \\ r &=& \frac{1\pm\sqrt{1+3 } }{2} \\ r &=& \frac{1\pm 2 }{2} \\\\ r_1 &=& \frac{1 + 2 }{2} \\ r_1 &=& \frac{3}{2} \quad & | \quad \text{ no common ratio, because } r > 1 \\\\ r_2 &=& \frac{1 - 2 }{2} \\ r_2 &=& -\frac{1}{2} \\ \hline \end{array}\)

the sum of all possible values for the common ratio is \(-\frac12\)