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I can't find a way to get an answer for this. I have tried using the formula for the sum to infinity and dividing it by the sum to 4 terms but i can't get it to work.

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    $\begingroup$ Is it the sum of the $first$ 4 terms $a_0+a_{1}+a_{2}+a_{3}$ which is equal to 15, or a certain sum $a_k+a_{k+1}+a_{k+2}+a_{k+3}$ with $k$ unknown a priori? $\endgroup$ – Jean Marie Apr 3 '16 at 22:20
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    $\begingroup$ Are the terms necessarily real, or can they be complex? $\endgroup$ – Brian Tung Apr 3 '16 at 22:22
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Let the sequence have initial term $a_1$ and common ratio $r$. Then $a_k = a_1r^{k - 1}$. The sum of the first n terms of the geometric series is $$\sum_{k = 1}^{n} a_1r^{k - 1} = a_1(1 + r + r^2 + \cdots + r^{n - 1}) = a_1 \frac{1 - r^n}{1 - r}$$ provided that $r \neq 1$. If $r = 1$, then the series would not converge unless $a_1 = 0$, which cannot be the case here since the sum of the series is not equal to zero. Since the sum of the first four terms is $15$, we have $$a_1 \frac{1 - r^4}{1 - r} = 15 \tag{1}$$ If the series converges, then its limit is $$\sum_{k = 1}^{\infty} a_1r^{k - 1} = \frac{a_1}{1 - r}$$ Since the series has sum $16$, we have $$\frac{a_1}{1 - r} = 16 \tag{2}$$ Dividing equation 1 by equation 2 yields $$1 - r^4 = \frac{15}{16}$$ Solving for $r$ yields \begin{align*} 1 - \frac{15}{16} & = r^4\\ \frac{1}{16} & = r^4\\ \pm \frac{1}{2} & = r \end{align*}

Check: Substituting $r = 1/2$ into equation 1 yields \begin{align*} a_1 \cdot \frac{1 - \frac{1}{16}}{1 - \frac{1}{2}} & = 15\\ a_1 \cdot \frac{\frac{15}{16}}{\frac{1}{2}} & = 15\\ a_1 \cdot \frac{15}{8} & = 15\\ a_1 & = 8 \end{align*} Substituting $a_1 = 8$ and $r = 1/2$ into equation 2 yields $$\frac{a_1}{1 - r} = \frac{8}{\frac{1}{2}} = 16$$

If $r = -1/2$, then \begin{align*} a_1 \cdot \frac{1 - \frac{1}{16}}{1 + \frac{1}{2}} & = 15\\ a_1 \cdot \frac{\frac{15}{16}}{\frac{3}{2}} & = 15\\ a_1 \cdot \frac{15}{16} \cdot \frac{2}{3} & = 15\\ a_1 \cdot \frac{5}{8} & = 15\\ a_1 & = 24 \end{align*} Substituting $a_1 = 24$ and $r = -1/2$ into equation 2 yields $$\frac{24}{1 + \frac{1}{2}} = \frac{24}{\frac{3}{2}} = 24 \cdot \frac{2}{3} = 16$$ Thus, both $r = 1/2$ and $r = -1/2$ satisfy the given conditions.

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Why doesn't it work?

$a_1\frac{1-r^4}{1-r} = 15$ and $a_1\frac{1}{1-r} = 16$, so $1-r^4 = \frac{15}{16}$, giving $r^4 = 1 - \frac{15}{16} = \frac{1}{16}$. So $r = \pm\frac{1}{2}$.

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You have $a + ax + ax^2 + ax^3 = 15$ and $\frac{a}{1-x} = 16$ (taking $|x| < 1$. This gives $a = 16 (1-x)$. Substitute for $a$, giving $16 (1-x) (1 + x + x^2 + x^3) = 15$. You can then find $1 - x^4 = \frac{15}{16}$ giving $x^4 = \frac{1}{16}$ and so $x = \frac{1}{2}$ so that $a = 8$.

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    $\begingroup$ OK, $x \pm \frac{1}{2}$ so $a = 8$ or $a = 24$ $\endgroup$ – jim Apr 3 '16 at 22:12

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