+2115 Let
a + ar + ar^2 + ar^3 + \dotsb
be an infinite geometric series. The sum of the series is $4.$ The sum of the cubes of all the terms is $10.$ Find the common ratio.
+394
\(a+ar+a{r}^{2}+a{r}^{3} + \dots=4\\{a}^{3}+{a}^{3}{r}^{3}+{a}^{3}{r}^{6}+{a}^{3}{r}^{9}+\dots=10\\\text{Using the sum of infinite geometric series formula}:\\\begin{cases} \frac{a}{1-r}=4 \\ \frac{{a}^{3}}{1-{r}^{3}}=10 \end{cases} \\\text{Isolate a:}\\ \begin{cases} a=4(1-r) \\ {a}^{3}=10(1-{r}^{3}) \end{cases} \\\text{Substitute:}\\ 64{(1-r)}^{3}=10(1-{r}^{3})\\ 54{r}^{3}-192{r}^{2}+192r-54=0\\ (r-1)(9{r}^{2}-23r+9)=0\\ r=1, r=\frac{23+\sqrt{205}}{18}, r=\frac{23-\sqrt{205}}{18}\\ \text{take the range} -1 <1 \;\text{therefore}\; r=\frac{23-\sqrt{205}}{18} \)
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