<s>[INST] If the sum to 4 terms of a geometric progression is 15 and the sum to infinity is 16 find the possible values of the common ratio. 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.
 [/INST] 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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