Detailed Answer

a) Both terms can be written as:

\[ 18 = u_1 + 4d\]

\[ 30 = u_1 + 8d\]

By subtracting two equations from one another we get:

\[ 12 = 4d \implies d = 3\]

b) By plugging d = 3 into the first equation from part (a) we get:

\[ 18 = u_1 + 4 * 3\]

\[ u_1 = 6\]

c)

\[ u_n= 6+3(n-1)=3+3n \]

Detailed Answer

a)

\[ u_5= u_1*r^{(5-1)}=15* (\frac{1}{3})^4=0.185\]

b)

\[ 15*(\frac{1}{3})^{(n-1)} < 1\]

\[ 15*(\frac{1}{3})^{(n-1)} - 1 < 0\]

This can be plugged into the GDC as a function and solved for its zero, like it's done on the figure below:

Since this is an inequality, we're looking for a value either greater than or smaller than 3.46. In this case, since we are raising to the power of \( n \) and are looking for a value smaller than 0, it has to be \( n > 3.46 \), as the fraction will become smaller and smaller with higher values of \( n \).

\[ n > 3.46\]

So the fourth term will be the first one with a value lower than 1.

c)

\[ S_7 = \frac{u_1(1-r^7)}{1-r} = \frac{15(1-\frac{1}{3}^7)}{1-\frac{1}{3}} = 22.5\]

Detailed Answer

a) By dividing consecutive terms of the series by the previous term we get that:

\[ \frac{20}{80}= \frac{1}{4}\]

\[ \frac{5}{20}= \frac{1}{4}\]

\[ \frac{5/4}{5}= \frac{1}{4}\]

As we can see, all common ratios are the same meaning that it is a geometric series.

b)

\[ S_{10} = \frac{u_1(1-r^{10})}{1-r} = \frac{80(1-\frac{1}{4}^{10})}{1-\frac{1}{4}} = 106.67\]

Detailed Answer

a) Knownig that March has 31 days we can use the formula:

\[ 72 = 12 + d(31 - 1)\]

\[ d = 2\]

b)

\[ S_n= \frac{n}{2}*(u_1+ u_n)\]

\[ S_{31}= \frac{31}{2}*(12 + 72)\]

\[ S_{31}= 1302\]

c) Using the formula for percentage error, we get that:

\[ ε= |\frac{v_a - v_e}{v_e}| * 100\% = |\frac{1200 - 1302}{1302}| * 100\% = 7.83\%\]

Detailed Answer

a)

\[ u_n = 25000 * 1.15^{n-1}\]

b) Since the end of 2015 is \( u_1 \), then the end of 2018 has to be \( u_4 \):

\[ u_4 = 25000 * 1.15^3 = 38021.88\]

c)

\[25000 * 1.15^{n-1} = 700000\]

This can be either solved in a GDC or using logarithms. To solve it in GDC it just needs to be slightly rearranged, input as a function and solved for zero:

\[25000 * 1.15^{n-1} - 70000= 0\]

\[n = 8.37\]

Hence, it will take 9 years for it to reach 70000.

d) Since the question asks for year 7, we first have to calculate the population in year 7 using our series:

\[ u_7 = 25000 * 1.15^6 = 57826.52\]

Now, using the formula for percentage error:

\[ ε= |\frac{v_a - v_e}{v_e}| * 100\% = |\frac{600000 - 57826.52}{57826.52}| * 100\% = 3.76\%\]

Detailed Answer

a) The swings will follow a gemoetric sequence with u₁ = 3, and r = 0.9. Then to get the fourth swing:

\[3 * 0.9^{4-1}= 2.19\]

b) We know that the distance has to fall below 0.5 for the dad to give a push, so:

\[3 * 0.9^{n-1}= 0.5\]

\[n = 18.005986\]

So, it will take 19 swings for dad to have to give a push.

Detailed Answer

a) Knowing that the car is worth $20,250 after two years from the initial value of $25,000, we have that:

\[25000 * (1-r)^2 = 20250\]

\[r = 0.1\]

b) Knowing that the starting value is $30,000 and the depreciation rate is 15%, we have that:

\[30000 * 0.85^5 = 13311.16\]

c) Using our answers from part (a) and (b)

\[30000 * 0.85^n < 25000 * 0.9^n\]

\[0 < 25000 * 0.9^n - 30000 * 0.85^n\]

So now it can be plugged as a function into the GDC and solved for zero:

We know that it's an inequality, so it \( n \) can either be greater than or smaller than 3.19. Since we know that \( 0.9^n \) will be raising faster than \( 0.85^n \), it has to be that for greater values of \( n \) the entire RHS will be positive. Thus:

\[n > 3.19\]

Detailed Answer

a) The first couple of terms of this sequence are: 7, 11, 15, ...

Thus, it can be clearly seen that this is an arithmetic sequence.

b) By looking at the answer from part (a) it can be easily found that the common difference is 11 - 7 = 4

c)

\[\sum_{n=1}^{5} (4n + 3)\]

d) A formula for the sum of an arithmetic series can be used:

\[ S_n= \frac{n}{2}*(2u_1+ d(n-1))\]

\[ S_{10}= \frac{10}{2}*(2 * 7+ 4(10-1)) = 5 * (14 + 36) = 250\]

Detailed Answer

a) This is an arithmetic sequence with a common difference of 25, so it can be easily found that u₁ is 225, and:

\[u_n = 225 + 25(n-1)\]

\[u_{25} = 225 + 25(25-1) = 825\]

b) It can be found with the formula:

\[ S_{15}= \frac{15}{2}*(2 * 225+ 25(15-1)) = 7.5 * (450 + 350) = 6000\]

c) Using the formula from part (a)

\[225 + 25(n-1) > 1000\]

Now it can be rearranged to find the equation which can be plugged directly into GDC:

\[225 + 25(n-1) - 1000 > 0\]

\[n > 32\]

So, the first term at which the value is greater than 1000 is the 33rd term.

d) By looking at this sequence it can be seen that this is a geometric sequence with r = 3, so:

\[9 = w_1 * 3^{3-1}\]

\[w_1 = 1\]

e) To answer this question we first need to find the general formula for this sequence:

\[w_n = w_1 * r^{n-1}\]

\[w_n = 1 * 3^{n-1}\]

\[w_n = 3^{n-1}\]

So the sigma notation will look as follows:

\[\sum_{n=1}^{4} (3^{n-1})\]

f) Now, let's compare the general formulas for those two sequences:

\[3^{n-1} > 225 + 25(n-1)\]

Let's rearrange it again to be able to plug it into GDC:

\[3^{n-1} -225 - 25n + 25> 0\]

\[n > 6.35\]

So, the first term will be term number 7.

Detailed Answer

a) Using the formula for the future value we get:

\[\text{FV} = \text{PV} \left(1 + \frac{r}{100}\right)^n\]

Since the interest is compounded annually, the formula becomes:

\[\text{FV} = 12000 \left(1 + \frac{4.75}{100}\right)^n = 12000 * 1.0475^n\]

b) By plugging the values into the formula from part (a):

\[\text{FV} = 12000 * 1.0475^2 = 13167.08\]

\[\text{FV} = 12000 * 1.0475^5 = 15133.92\]

\[\text{FV} = 12000 * 1.0475^{10} = 19086.29\]

c) Since we know that after 5 years Lisa will have $15,133.92 (calculated in part b), we can see that this value is greater than $15,000. So, Lisa will have enough money to put down her deposit.

Detailed Answer

a) Formula for the future value with depreciation is as follows:

\[\text{FV} = \text{PV} \left(1 - \frac{r}{100k}\right)^{kn}\]

Since it depreciates yearly, we have that \(k = 1\), so:

\[\text{FV} = 32500 \left(1 - \frac{10}{100}\right)^8 = 32500 * 0.9^8 = 13990.18\]

b) By plugging in the values to the same formula and using the GDC:

\[9200 = 18000 \left(1 - \frac{r}{100}\right)^5\]

Solving for \(r\), we get:

\[r \approx 12.56\%\]

Detailed Answer

a) Using the TVM Solver on GDC:

Converting 43.85 to years:

\[ \frac{43.85}{12} \approx 3.65 \]

So, it will take 3 full years plus a couple of months:

\[ 43.85 - 12 * 3 = 7.85 \]

Hence, it will take 3 years and 8 months.

b) The total amount paid is equal to:

\[ 43.85 * 1500 = $65775 \]

c) The table for Tom will look as follows:

So, the difference in the total money paid equals:

\[ 65775- 43 * 1500 = $1275 \]

Detailed Answer

a) Using the formula for future value:

\[ \text{FV} = \text{PV} \left(1 - \frac{r}{100}\right)^n \]

For the Honda motorcycle:

\[ \text{FV} = 20000 \left(1 - \frac{4}{100}\right)^6 \approx 15655.16 \]

b) The difference in values after 3 years:

\[ 20000 \left(1 - \frac{4}{100}\right)^3 - 25000 \left(1 - \frac{9}{100}\right)^3 \approx -1144.56 \]

c) Using a calculator:

\[ 20000 \left(1 - \frac{4}{100}\right)^n > 25000 \left(1 - \frac{9}{100}\right)^n \]

Solving for \( n \), we find:

\[ n \approx 5 \text{ years} \]

Detailed Answer

a) Using the formula for future value:

\[ \text{FV} = \text{PV} \left(1 + \frac{r}{100k}\right)^{kn} \]

\[ 20000 = 10000 \left(1 + \frac{5.5}{100 * 4}\right)^{4n} \]

Solving for \( n \), we find:

\[ n \approx 12.69 \text{ years} \]

b) For the high-growth stock option:

\[ 20000 = 10000 \left(1 + \frac{r}{100}\right)^6 \]

Solving for \( r \), we find:

\[ r \approx 12.25\% \]

Detailed Answer

a) To find the deposit:

\[ D = 20\% * 10000 = 2000 \]

b) The total cost of the loan is:

\[ \text{Total Cost} = \text{Deposit} + \text{Monthly Payments} * \text{Number of Payments} \]

\[ \text{Total Cost} = 2000 + 1100 * 10 = 11200 \]

c) By following the given equations:

\[ x + 11y = 8000 \]

d) Solving for \( x \) and \( y \):

Currently, there are 2 equations and 2 unknowns:

\[ (1) \ x + 4y = 4500 \]

\[ (2) \ x + 11y = 8000 \]

Solving for \( x \) in the first equation:

\[ x = 4500 - 4y \]

Now we can plug it into the second equation:

\[ (2) \ 4500-4y+11y=8000 \]

\[ y = 500 \]

By plugging the \( y \) back into the first equation:

\[ x=4500-4*500=2500 \]

e) Using the obtained values for \( x \) and \( y \):

\[ 2500+500*n=10000 \implies n = 15 \]

It will take 15 months to pay back the loan.

f) Number of months to repay the bank loan:

So, it will take Adam approximately 12 months to pay back his loan.

g) The total amount paid for the bank loan is:

\[ 900 * 11.37 = 10233 \]

Detailed Answer

a) Using the formula for future value:

\[ FV = PV \left(1 + \frac{r}{100 * k}\right)^{kn} \] where \( PV = 40000 \), \( r = 5.5 \), \( k = 12 \), and \( n = 5 \):

\[ FV = 40000 \left(1 + \frac{5.5}{100 * 12}\right)^{12 * 5} = 52628.15 \]

b) Using the formula for future value again to find \( n \) when \( FV = 60000 \):

\[ 60000 = 40000 \left(1 + \frac{5.5}{100 * n}\right)^{12 * n} \]

Solving it using a financial calculator or software:

\[ n \approx 7.39 \]

Detailed Answer

a) Using the formula for future value with quarterly compounding:

\[ FV = PV \left(1 + \frac{r}{100 * k}\right)^{kn} \] where \( PV = 12000 \), \( r = 5 \), \( k = 4 \) (quarterly), and \( n = 3 \):

\[ FV = 12000 \left(1 + \frac{5}{100 * 4}\right)^{12} = 13929.05 \]

b) The higher the frequency of compounding the higher will be the final value, keeping everything else constant. Thus, a yearly compound rate will result in a lower final value as compared to quarterly compounding.

c) Using the formula for future value with yearly compounding:

\[ FV = PV \left(1 + \frac{r}{100}\right)^{n} \] where \( PV = 12000 \), \( r = 5 \), and \( n = 3 \):

\[ FV = 12000 \left(1 + \frac{5}{100}\right)^{3} = 13891.5 \]

Detailed Answer

a) First, we have to find the value of the BMW after 5 years:

\[ \text{FV} = \text{PV} \left(1 - \frac{r}{100}\right)^n \] where \( \text{PV} = 35000 \), \( r = 8 \), and \( n = 5 \):

\[ \text{FV} = 35000 \left(1 - \frac{8}{100}\right)^5 = 23067.85 \]

So, the difference will be:

\[ 35000 - 23067.85 = 11932.15 \]

b) We have to equalize two equations for future value with one unknown (Mercedes’ price):

\[ 35000 \left(1 - \frac{8}{100}\right)^6 = P \left(1 - \frac{5}{100}\right)^6 \]

Solving for \( P \):

\[ P = \frac{35000 \left(1 - \frac{8}{100}\right)^6}{\left(1 - \frac{5}{100}\right)^6} = 28870.44 \]