# Examples of Calculations with Series

Example 1

You take a loan of $\text{}\text{}150\phantom{\rule{0.17em}{0ex}}000\text{}$ today, and plan to pay it back over 10 years with annual installments. The first installment is in one year, and the interest rate is $\text{}6\text{}\phantom{\rule{0.17em}{0ex}}\text{%}$. Find the size of each annual installment by using future values.

We solved this by using present values when we talked about annuity loans, but here we will use future values instead.

The installments can be expressed as the geometric series

$\begin{array}{llll}\hfill x& +x\cdot 1.06+x\cdot 1.0{6}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & +\cdots +x\cdot 1.0{6}^{9}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $x+x\cdot 1.06+x\cdot 1.0{6}^{2}+\cdots +x\cdot 1.0{6}^{9}.$

You can also visualize the geometric series with a timeline:

Notice that the series has 10 terms. This corresponds to one installment each year for 10 years. Here we have that ${a}_{1}=x$, $k=1.06$ and $n=10$. It’s important to remember that the sum ${S}_{10}$ is not equal to $150\phantom{\rule{0.17em}{0ex}}000$, but to $150\phantom{\rule{0.17em}{0ex}}000\cdot 1.0{6}^{10}$. This is because what you end up paying back isn’t just what you loaned, it’s the sum of all the installments. You use the formula for the sum of a geometric series and solve for $x$: $\begin{array}{llll}\hfill {S}_{n}& ={a}_{1}\frac{{k}^{n}-1}{k-1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 150\phantom{\rule{0.17em}{0ex}}000\cdot 1.0{6}^{10}& =x\cdot \frac{1.0{6}^{10}-1}{1.06-1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 150\phantom{\rule{0.17em}{0ex}}000\cdot 1.0{6}^{10}& \approx x\cdot 13.18\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\frac{150\phantom{\rule{0.17em}{0ex}}000\cdot 1.0{6}^{10}}{13.18}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& \approx 20\phantom{\rule{0.17em}{0ex}}381\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This means you have to pay around \$$20\phantom{\rule{0.17em}{0ex}}381$ each year to repay the loan of \$$150\phantom{\rule{0.17em}{0ex}}000$ in $10$ years. By adding the ten installments to each other, you can see that you have paid much more than \$$150\phantom{\rule{0.17em}{0ex}}000$. You can check if you want to!

If you compare this to the solution we got when we used the present value, you will see that they differ slightly. This is due to rounding of the numbers at different points during the calculations, which alters the results a little bit.

Example 2

At the start of a year, Lucy is considering taking up a loan of $\text{}\text{}10\phantom{\rule{0.17em}{0ex}}000\text{}$ to invest in a fund. The loan is an annuity loan, and to repay the loan, she’s paying $\text{}\text{}1627.45\text{}$ back to the bank at the end of each year for 10 years. The first installment is after 1 year.

Part 1 Show that the annual interest is $\text{}10\text{}\phantom{\rule{0.17em}{0ex}}\text{%}$.

We know that each installment is \$$1627.45$, which means that Lucy needs to pay this installment every year for 10 years.

You can express the installments as the geometric series

$\begin{array}{llll}\hfill & \frac{1627.45}{{\left(1+i\right)}^{1}}+\frac{1627.45}{{\left(1+i\right)}^{2}}+\cdots +\frac{1627.45}{{\left(1+i\right)}^{10}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{1em}{0ex}}=10\phantom{\rule{0.17em}{0ex}}000,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 $\frac{1627.45}{{\left(1+i\right)}^{1}}+\frac{1627.45}{{\left(1+i\right)}^{2}}+\cdots +\frac{1627.45}{{\left(1+i\right)}^{10}}=10\phantom{\rule{0.17em}{0ex}}000,$

Where the first part ${a}_{1}$ and the quotient $k$ is
 ${a}_{1}=\frac{1627.45}{1+i}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}k=\frac{1}{1+i}.$

 ${a}_{1}=\frac{1627.45}{1+i}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}k=\frac{1}{1+i}.$

There are two different methods that can be used to find the answer to this question. One is to use the formula for the sum of a geometric series
 ${S}_{10}=\frac{1627.45}{1+i}\cdot \genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\phantom{\rule{-0.17em}{0ex}}{\left(\frac{1}{1+i}\right)}^{10}-1\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\frac{1}{1+i}-1\phantom{\rule{0.17em}{0ex}}},$

letting $i=0.10$ and using a calculator to calculate what we get. The expression becomes

 ${S}_{10}=\frac{1627.45}{1+0.10}\cdot \genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\phantom{\rule{-0.17em}{0ex}}{\left(\frac{1}{1+0.10}\right)}^{10}-1\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\frac{1}{1+0.10}-1\phantom{\rule{0.17em}{0ex}}},$

which turns out to be the same as the size of the loan. That means you have shown that the annual interest is $10$ %, because when $i=0.10$, the sum of the first 10 terms of the series is equal to the loan.

The second method is to set up an equation where the sum is equal to the size of the loan, and then solve the equation for the interest $i$. The equation would look like this:

 $\frac{1627.45}{1+i}\cdot \genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\phantom{\rule{-0.17em}{0ex}}{\left(\frac{1}{1+i}\right)}^{10}-1\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\frac{1}{1+i}-1\phantom{\rule{0.17em}{0ex}}}=10\phantom{\rule{0.17em}{0ex}}000$

The easiest way to solve this is by using a digital tool, like `CAS` in `GeoGebra`. That gives us two solutions,

 ${i}_{1}=-7.8\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{i}_{2}=0.1,$

but as the interest can’t be negative, we know that only ${i}_{2}=0.10$ can be a valid answer. That is also the value we wanted to show that the interest had, which means we have shown that the interest is $10$ %.

The bank claims that if a fund has an annual return of $\text{}12\text{}\phantom{\rule{0.17em}{0ex}}\text{%}$, then Lucy will gain a solid profit by investing in that fund.

Part 2 – Determine the value of Lucy’s money at the end of the $10{}^{\text{th}}$ year in the fund.

Lucy has invested all \$$10\phantom{\rule{0.17em}{0ex}}000$ she loaned in the fund. With a guaranteed annual profit of $12$ % over a period of 10 years, you can use the formula for future value to find how much her money will be worth then. The formula is

 ${K}_{n}={K}_{0}\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{n},$

where ${K}_{0}=10\phantom{\rule{0.17em}{0ex}}000$, $p=12$ and $n=10$. That means the value of the money in the fund is

$\begin{array}{llll}\hfill {K}_{10}& =10\phantom{\rule{0.17em}{0ex}}000\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{12}{100}\right)}^{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 31\phantom{\rule{0.17em}{0ex}}058.48\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 ${K}_{10}=10\phantom{\rule{0.17em}{0ex}}000\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{12}{100}\right)}^{10}\approx 31\phantom{\rule{0.17em}{0ex}}058.48$

after $10$ years, which means the value of Lucy’s money in the fund is \$$31\phantom{\rule{0.17em}{0ex}}058.48$ after $10$ years.

Lucy’s net profit after 10 years is the difference between what she has paid for the loan and the value of her money in the fund.

Part 3 – Show that her net profit after 10 years will be $\text{}\text{}5121.05\text{}$.

You know that the future value of Lucy’s money in the fund is approximately \$$31\phantom{\rule{0.17em}{0ex}}058.48$, but what’s the future value of the loan after $10$ years? You know that the present value of the loan is \$$10\phantom{\rule{0.17em}{0ex}}000$, with an interest of $10$ %. You can use the formula again to find the future value of the loan:

 ${K}_{n}={K}_{0}\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{n},$

Where ${K}_{0}=10\phantom{\rule{0.17em}{0ex}}000$, $p=10$ and $n=10$. That makes the expression for the future value into

$\begin{array}{llll}\hfill {K}_{10}& =10\phantom{\rule{0.17em}{0ex}}000\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{10}{100}\right)}^{10}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 25\phantom{\rule{0.17em}{0ex}}937.42,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

 ${K}_{10}=10\phantom{\rule{0.17em}{0ex}}000\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{10}{100}\right)}^{10}\approx 25\phantom{\rule{0.17em}{0ex}}937.42,$

which means the future value of the loan in $10$ years is approximately \$$25\phantom{\rule{0.17em}{0ex}}937.42$.

Lucy’s net profit is then

 $\text{}31\phantom{\rule{0.17em}{0ex}}058.48-\text{}25\phantom{\rule{0.17em}{0ex}}937.42=\text{}5121.05.$

Instead of taking up this loan to invest in a fund, Lucy considers saving them in the bank. At the end of each year, she will deposit $\text{}\text{}1627.45\text{}$ into an account with a fixed annual interest. She will make the first deposit in a year.

Part 4 – What does the interest on Lucy’s savings account need to be for her to have as much money in the bank after 10 years as in part 2?

The value of Lucy’s money in the fund at the end of the $10{}^{\text{th}}$ year was found to be \$$31\phantom{\rule{0.17em}{0ex}}058.48$ in part 2.

Lucy’s savings can be expressed as the geometric series

$\begin{array}{llll}\hfill 1627.45& +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ⋮\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{9},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill 1627.45& +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{1}+1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & +\cdots +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{9},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

where you can see that the first term of the series ${a}_{1}$ and the quotient $k$ are
 ${a}_{1}=1627.45\phantom{\rule{0.33em}{0ex}}\text{and}\phantom{\rule{0.33em}{0ex}}k=\phantom{\rule{-0.17em}{0ex}}\left(1+\frac{p}{100}\right).$

 ${a}_{1}=1627.45\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}k=\phantom{\rule{-0.17em}{0ex}}\left(1+\frac{p}{100}\right).$

The problem can be expressed as the equation
$\begin{array}{lll}\hfill 31\phantom{\rule{0.17em}{0ex}}058.48=& \phantom{\rule{2em}{0ex}}& \hfill \\ \hfill 1627.45& +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ⋮\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{9},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill 31\phantom{\rule{0.17em}{0ex}}058.48=1627.45& +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{1}+1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & +\cdots +1627.45\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{9},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

or by using the formula for the sum of a geometric series:
 $31\phantom{\rule{0.17em}{0ex}}058.48=1627.45\genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\phantom{\rule{-0.17em}{0ex}}{\left(1+\frac{p}{100}\right)}^{10}-1\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\phantom{\rule{-0.17em}{0ex}}\left(1+\frac{p}{100}\right)-1\phantom{\rule{0.17em}{0ex}}}$

To solve this equation, it’s best to use a digital tool like `CAS` in `GeoGebra`. That gives you

 $p=13.73\phantom{\rule{0.17em}{0ex}}\text{,}$

which means the interest on the savings account needs to be at least $13.73$ % for Lucy to have the same profit from it after $10$ years as she would get from investing in the fund, provided she deposits \$$1627.45$ into the account each year.

Example 3

When you go shopping, certain items might have payment plans. This is the same as loaning money from the store in the form of the item. Due to the interest, this is always more expensive than paying the whole sum straight away.

Lindsay Lohan buys a laptop. She can either pay $\text{}\text{}1599\text{}$ straight away, or she can pay $\text{}\text{}69.9\text{}$ each month for 36 months. The first installment would be due one month after the laptop was purchased. Lindsay decide to go for the payment plan to pay for the laptop. Find the monthly and the annual interest she will have to pay.

Let the monthly interest be $x$, and find the present value of the installments. The present value gives you the geometric series

 $\frac{69,9}{x}+\frac{69,9}{{x}^{2}}+\cdots +\frac{69,9}{{x}^{36}}.$

This geometric series can also be expressed as a timeline:

The sum of this geometric series is the amount that Lindsay could have chosen to pay up front, $1599$. You can see that ${a}_{1}=\frac{69,9}{x}$ and $k=\frac{1}{x}$. You insert this into the formula for the sum of a geometric series and get $\begin{array}{llll}\hfill {S}_{n}& ={a}_{1}\cdot \frac{{k}^{3}6-1}{k-1},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 1599& =\frac{699}{x}\cdot \genfrac{}{}{1.0pt}{}{\phantom{\rule{0.17em}{0ex}}\frac{1}{{x}^{n}}-1\phantom{\rule{0.17em}{0ex}}}{\phantom{\rule{0.17em}{0ex}}\frac{1}{x}.-1\phantom{\rule{0.17em}{0ex}}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You can solve this equation with a digital tool like `CAS` in `GeoGebra`, which will give you that $x=-0.9$ and $x=1.03$. As the interest can’t be negative, the answer has to be $x=1.03$. This corresponds to an interest of $3$ % per month. The annual interest will then be

 $1.0{3}^{12}=1.43,$

which corresponds to an annual interest of $43$ % per year.

This goes to show that it’s always cheaper to pay up front. Lindsay would have saved a lot of money by paying for the laptop that way.