This problem can be modeled by an arithmetic series with [latex]{a}_{1}=\frac{1}{2}[/latex] and [latex]d=\frac{1}{4}[/latex]. We are looking for the total number of miles walked after [latex]8[/latex] weeks, so we know that [latex]n=8[/latex], and we are looking for [latex]{S}_{8}[/latex]. To find [latex]{a}_{8}[/latex], we can use the explicit formula for an arithmetic sequence.

[latex]\begin{align} {a}_{n}&={a}_{1}+d\left(n - 1\right) \\ {a}_{8}&=\dfrac{1}{2}+\dfrac{1}{4}\left(8 - 1\right)=\dfrac{9}{4} \end{align}[/latex]

We can now use the formula for arithmetic series.

[latex]\begin{align} {S}_{n}&=\dfrac{n\left({a}_{1}+{a}_{n}\right)}{2} \\ {S}_{8}&=\dfrac{8\left(\frac{1}{2}+\frac{9}{4}\right)}{2}=11 \end{align}[/latex]

She will have walked a total of [latex]11[/latex] miles.

1. Identify the arithmetic series:
   • First term: [latex]a_1 = 1000[/latex]
   • Common difference: [latex]d = 200[/latex]
   • Number of terms: [latex]n = 14[/latex]
2. Find the last term: [latex]a_n = a_1 + (n-1)d = 1000 + (14-1)200 = 3600[/latex]
3. Use the formula for the sum of an arithmetic series: [latex]S_n = \frac{n}{2}(a_1 + a_n)[/latex]
4. Substitute the values:
   [latex]\begin{array}{rcl} S_{14} &=& \frac{14}{2}(1000 + 3600) \\&=& 7 \cdot 4600 \\ &=& 32,200 \end{array}[/latex]

Therefore, the parent will have saved [latex]$32,200[/latex]for their child’s college education.

1. Identify the arithmetic series:
   • First term: [latex]a_1 = 100[/latex]
   • Common difference: [latex]d = 5[/latex]
   • Number of terms: [latex]n = 30[/latex]
2. Find the last term: [latex]a_n = a_1 + (n-1)d = 100 + (30-1)5 = 245[/latex]
3. Use the formula for the sum of an arithmetic series: [latex]S_n = \frac{n}{2}(a_1 + a_n)[/latex]
4. Substitute the values:
   [latex]\begin{array}{rcl} S_{30} &=& \frac{30}{2}(100 + 245) \\ &=& 15 \cdot 345 \\ &=& 5,175 \end{array}[/latex]

Therefore, there are [latex]5,175[/latex] seats in total in the stadium.

1. Identify the geometric series:
   • Initial term: [latex]a = 5000[/latex]
   • Common ratio: [latex]r = 0.30[/latex]
   • Number of terms: [latex]n = 10[/latex]
2. Use the formula for the sum of a geometric series: [latex]{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}[/latex], where [latex]S_n[/latex] is the sum of the series
3. Substitute the values:
   [latex]\begin{array}{rcl} S_{10} &=&  \dfrac{5000(1-(0.30)^{10})}{1-0.30} \\  &\approx& 7,142 \end{array}[/latex]

Therefore, approximately [latex]7,142[/latex] people will have seen the product advertisement after [latex]10[/latex] days.

1. Identify the geometric series:
   • Initial term (amount decaying in first year): [latex]a = 100 \cdot 0.15 = 15[/latex] grams
   • Common ratio (rate of decay each subsequent year): [latex]r = 0.85[/latex]
   • Number of terms: [latex]n = 20[/latex]
2. Use the formula for the sum of a geometric series: [latex]{S}_{n}=\dfrac{{a}_{1}\left(1-{r}^{n}\right)}{1-r}[/latex]
3. Substitute the values:
   [latex]\begin{array}{rcl} S_{20} &=& \frac{15(1-(0.85)^{20})}{1-0.85} \\  &\approx& 96.12 \end{array}[/latex]

Therefore, approximately [latex]96.12[/latex] grams of the radioactive material will have decayed after [latex]20[/latex] years.