1. There is not a constant ratio; the series is not geometric.
2. There is a constant ratio; the series is geometric. [latex]a_1=248.6[/latex] and [latex]r=\dfrac{99.44}{248.6}=0.4[/latex], so the sum exists. Substitute [latex]a_1=248.6[/latex] and [latex]r=0.4[/latex] into the formula and simplify to find the sum.
   [latex]\begin{align} \\ &S=\frac{a_1}{1-r} \\[1.5mm] &S=\frac{248.6}{1-0.4}=\frac{1243}{3} \\ \text{ }\end{align}[/latex]
3. The formula is exponential, so the series is geometric with [latex]r=-\frac{1}{3}[/latex]. Find [latex]a_1[/latex] by substituting [latex]k=1[/latex] into the given explicit formula.
   [latex]\begin{align} \\ a_1=4\text{,}374\cdot\left(-\frac{1}{3}\right)^{1-1}=4\text{,}374 \\ \text{ }\end{align}[/latex]

   Substitute [latex]4\text{,}374[/latex] and [latex]r=-\frac{1}{3}[/latex] into the formula, and simplify to find the sum.

   [latex]\begin{align}\\&S=\frac{a_1}{1-r} \\[1.5mm] &S=\frac{4\text{,}374}{1-\left(-\frac{1}{3}\right)}=3\text{,}280.5 \\ \text{ }\end{align}[/latex]
4. The formula is exponential, so the series is geometric, but [latex]r>1[/latex]. The sum does not exist.

[latex]0.\overline{3}=0.3+0.03+0.003+\dots[/latex]

Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to [latex]0.1[/latex] in the second term, and the second term multiplied to [latex]0.1[/latex] in the third term.

[latex]\begin{align}0.\overline{3}&=0.3+0.3\cdot(0.1)+0.3\cdot(0.01)+0.3\cdot(0.001)+\dots \\ &=0.3+0.3\cdot(0.1)+0.3\cdot(0.1)^2+0.3\cdot(0.1)^3+\dots\end{align}[/latex]

Notice the pattern; we multiply each consecutive term by a common ratio of [latex]0.1[/latex] starting with the first term of [latex]0.3[/latex]. So, substituting into our formula for an infinite geometric sum, we have

[latex]S=\dfrac{a_1}{1-r} =\dfrac{0.3}{1-0.1} =\dfrac{0.3}{0.9} =\dfrac{1}{3}[/latex]