Arithmetic and Geometric Sequences (Economics)
Sequences
A sequence is a list of numbers which are written in a particular order. For example, $1, 3, 5, 7, 9$ is a sequence of odd numbers.
A finite sequence is a sequence which ends. The sequence has a known final value. For example $1, 3, 5, \dotso, 19$. is a finite sequence whose end value is $19$.
An infinite sequence is a sequence in which the terms go on forever, for example $2, 5, 8, \dotso$.
Note: The 'three dots' notation stands in for missing terms. If the dots are followed by a final number, the sequence is finite. If the dots have nothing after them, the sequence is infinite.
The Summation Operator
The Summation Operator, $\sum$, is used to denote the sum of a sequence. We call the sum of the terms in a sequence a series.
The starting index is written underneath and the final index above, and the sequence to be summed is written on the right. A series does not have to be the sum of all the terms in a sequence. For example, \[\sum^{5}_{i=1}{2^i}\] represents the sum of the first five terms in the sequence of powers of $2$.
Rules of the Summation Operator
Denote $a$ and $b$ as constants. Some useful properties of the summation operator are:
Constant Rule
\[\sum^{n}_{i=1}{a}=na\]
Constant Multiple Rule
\[\sum^{n}_{i=1}{a{x_i}}=a\sum^{n}_{i=1}{x_i}\]
The Sum of Sequences Rule
\[\sum^{n}_{i=1}{({x_i}+{y_i})}=\sum^{n}_{i=1}{x_i}+\sum^{n}_{i=1}{y_i}\]
Worked Examples
Worked Example
$\sum^{10}_{i=1}{a_i}=25$ and $\sum^{10}_{i=1}{b_i}=10$. Evaluate the following series using the above rules:
a) $\sum^{10}_{i=1}{(a_i+b_i)}$
b) $\sum^{10}_{i=1}{(4+b_i)}$
c) $\sum^{10}_{i=1}{(a_i-3b_i)}$
Solution
a) Using the Sum of Sequences Rule, we have: \[\sum^{10}_{i=1}{(a_i+b_i)}=\sum^{10}_{i=1}{a_i}+\sum^{10}_{i=1}{b_i}\] We know that $\sum^{10}_{i=1}{a_i}=25$ and $\sum^{10}_{i=1}{b_i}=10$ so: \begin{align} \sum^{10}_{i=1}{(a_i+b_i)}&=25+10\\ &=35 \end{align}
b) Using the Sum of Sequences Rule, we have: \[\sum^{10}_{i=1}{(4+b_i)}=\sum^{10}_{i=1}{4}+\sum^{10}_{i=1}{b_i}\] Using the Constant Rule we have: \begin{align} \sum^{10}_{i=1}{4}&=10\times 4\\ &=40 \end{align} and we know that $\sum^{10}_{i=1}{b_i}=10$ so: \begin{align} \sum^{10}_{i=1}{(4+b_i)}&=40+10\\ &=50 \end{align}
c) Using the Sum of Sequences Rule, we have: \[\sum^{10}_{i=1}{(a_i-3b_i)}=\sum^{10}_{i=1}{a_i}-\sum^{10}_{i=1}{3 b_i}\] Using the Constant Multiple Rule: \begin{align} \sum^{10}_{i=1}{3b_i}&=3\sum^{10}_{i=1}{b_i}\\ &=30 \end{align} and we know that $\sum^{10}_{i=1}{a_i}=25$ so: \begin{align} \sum^{10}_{i=1}{(a_i-3b_i)}&=25-30\\ &=-5 \end{align}
Arithmetic sequence
An arithmetic sequence (or arithmetic progression) is any sequence where each new term is obtained by adding a constant number to the preceding term. This constant number is referred to as the common difference. For example, $10, 20, 30, 40$, is an arithmetic progression increasing by $10$, or $-4, -3, -2, -1$ is an arithmetic progression decreasing by $1$ with each term.
An arithmetic progression can be expressed as $A, A+d, A+2d, A+3d, \dotso$, where $A$ denotes the first term and $d$ denotes the common difference between each term.
The $n^{\text{th} }$ term is given by \[A+(n-1)d\] The sum of the first $n$ terms of an arithmetic progression is given by: \[\sum^{n}_{i=1}{A+(i-1)d}\] A quicker formula for working out the sum of the first $n$ terms of an arithmetic progression is: \[nA+\dfrac{dn(n-1)}{2}\]
Worked Examples
Example 1
Write down the first five terms of the arithmetic progression with first term $8$ and common difference $7$.
Solution
We have $A=8$ and $d=7$.
The form of an arithmetic progression is $A, A+d, A+2d, A+3d, A+4d$, so using these values of $A$ and $d$ the first five terms are:
\[8, \; 8 + 7, \; 8 + 2\times 7, \; 8+3 \times 7, \; 8 + 4\times 7\]
\[8, 15, 22, 29, 36\]
Example 2
Find the $17^{\text{th} }$ term of the arithmetic progression with first term $5$ and common difference $2$.
Solution
We have $A=5$, $d=2$ and $n=17$.
The nth term is given by $A+(n-1)d$, so the $17^{\text{th} }$ term is:
\begin{align} &A+ (17-1)d\\ &= 5 + 16\times 2\\ &= 37 \end{align}
Example 3
Find the sum of the first fifty terms of the sequence: $1, 3, 5, 7, 9, \dotso$.
Solution
We have $A=1$, $d=2$ and $n=50$, and are asked to calculate \[\sum^{50}_{n=1} 1 + 2(i-1)\]
Using the formula, the sum of the first $50$ terms of the sequence is: \begin{align} 50&\times 1+\dfrac{2\times 50\times(50-1)}{2}\\ &=50+\dfrac{4900}{2}\\ &=2500 \end{align}
Geometric Sequence
A geometric sequence (or geometric progression), is a sequence where each new term is obtained by multiplying the preceding term by a constant number. This constant number is called the common ratio. A geometric progression can be expressed as $A, AR, AR^2, AR^3, \dotso$, where $A$ denotes the first term and $R$ is the common ratio between terms. The $n^{\text{th} }$ term is given by \[AR^{n-1}\]
The sum of the first $n$ terms of a geometric progression, denoted by $s$, is given by: \[s= \sum^n_{i=1} AR^{i-1}\] A quicker formula for working out the sum of the first $n$ terms of geometric progression is: \[s=A\dfrac{1-{R^n}}{1-R}\] Or equivalently (multiplying the numerator and denominator by $-1$): \[s=A\dfrac{{R^n}-1}{R-1}\]
A Special Case of the Geometric Progression
When the absolute value of the common ratio $R$ is less than $1$ $\left(\left|R\right|\lt1\right)$, as the number of terms in the sequence $n$ tends to infinity ($n\rightarrow \infty$) the sum of the series tends to \[\dfrac{A}{1-R}\]
Using limit notation, we have \[\displaystyle{\lim_{ n\to \infty} \left(A\dfrac{1-{R^n}}{1-R}\right)= \dfrac{A}{1-R}}\] so the formula for the sum to infinity of a geometric progression with first term $A$ and common ratio $R$ such that $\left(\left|R\right|\lt1\right)$ is: \[s=\dfrac{A}{1-R}\]
And we say that the sum of the sequence converges to $\dfrac{A}{1-R}$ as $n$ tends to infinity. If the absolute value of the common ratio is greater than $1$, $\left(\left|R\right|\gt1\right)$, then the terms in the sequence get larger and larger and so does the sum. In this case we say that the sum of the sequence diverges (tends to positive or negative infinity).
Worked Examples
Example 1
Write down the first five terms of the geometric progression which has first term $1$ and common ratio $\dfrac{1}{2}$.
Solution
We have $A=1$ and $R=\dfrac{1}{2}$.
The general form of a geometric progression is $A, AR, AR^2, AR^3, AR^4, \dotso$, so using these values of $A$ and $R$ the first five terms are:
\[1, \; 1\times \dfrac{1}{2}, \; 1\times \left(\dfrac{1}{2}\right)^2, \; 1\times \left(\dfrac{1}{2}\right)^3, \; 1\times \left(\dfrac{1}{2}\right)^4\]
\[1, \dfrac{1}{2}, \dfrac{1}{4}, \dfrac{1}{8},\dfrac{1}{16}\]
As we can see, the terms are getting smaller and smaller.
Example 2
Find the $10^{\text{th} }$ term of the geometric progression with first term $3$ and common ratio $2$.
Solution
We have $A=3$, $R=2$ and $n=10$.
Using the formula for the $n^{\text{th} }$ term, the $10^{\text{th} }$ term is
\begin{align} &3\times 2^{10-1}\\ &=3\times 2^{9}\\ &= 1536 \end{align}
Example 3
How many terms are in the geometric progression $2, 4, 8, \dotso, 128$?
Solution
We have $A=2$ and $r=2$. We also know that the $10^{\text{th} }$ term is $128$.
Using the formula for the $n^{\text{th} }$ term we can work out the value of $n$:
\begin{align} 128 &= 2\times 2^{n-1} \\ 64 &= 2^{n-1} \\ 2^6 &= 2^{n-1} \\ 6&= n-1 \\ n&=7 \end{align}
An alternative method would be to use logarithms:
\begin{align} 128 &= 2\times 2^{n-1}\\ 64 &= 2^{n-1} \\ \log_{10} 64 &= \log_{10}(2^{n-1})\\ \log_{10} 64 &= (n-1)\log_{10} 2\\ \frac{\log_{10} 64}{\log_{10} 2}&= n-1\\ 6&=n-1\\ n&=7 \end{align} So there are $7$ terms in the given geometric progression.
Example 4
Find the sum of the first five terms of the geometric progression $8, -4, 2, \dotso$.
Solution
We have $A=8$, $R=-\dfrac{1}{2}$ and $n=5$. We are asked to calculate \[\sum^5_{i=1} 8 \times \left(-\frac{1}{2}\right)^{i-1}.\]
Using the formula for the sum of a geometric progression we have: \begin{align} s&=8\times \dfrac{\left(\left(-\frac{1}{2}\right)^5 -1 \right)}{\left(-\frac{1}{2}-1\right)}\\ &=8\times \dfrac{\left(-\frac{1}{32}-1\right)}{\left(-\frac{3}{2}\right)}\\ &=8\times \dfrac{\left(-\frac{33}{32}\right)}{\left(-\frac{3}{2}\right)} \\ &=8\times \dfrac{11}{16} \\ &=5.5 \end{align}
Example 5
Find the sum to infinity of the geometric progression $1, \dfrac{1}{3}, \dfrac{1}{9}, \dfrac{1}{27},\dotso$.
Solution
We have $A=1$ and $R=\dfrac{1}{3}$.
As $-1\lt R\lt 1$, the series converges and we can use the formula $s = \dfrac{A}{1-R}$:
\begin{align} s&=\frac{1}{ 1-\frac{1}{3} }\\ &=\frac{1}{\; \frac{2}{3}\;}\\ &=\frac{3}{2} \end{align}
Arithmetic or Geometric?
If the sequence has a common difference, it is arithmetic; if it has a common ratio, it is geometric. We can therefore determine whether a sequence is arithmetic or geometric by working out whether adjacent terms differ by a common difference, or a common ratio.
Arithmetic?
Recall that a common difference is a number which is added to each term to get the next term in the sequence and that the $n$th term of an arithmetic sequence is given by: \[A+(n-1)d\]
Subtracting the $n$th term from the $(n+1)$th term gives: \begin{align} A+((n+1)-1)d-A+(n-1)d&=nd-(n-1)d\\ &=d \end{align}
Similarly, subtracting the $(n+1)$th term from the $(n+2)$th term gives: \begin{align} A+((n+2)-1)d-A+((n+1)-1)d&=(n+1)d-nd\\ &=d \end{align}
There is a common difference of $d$ between any $2$ terms in the sequence. One way to check if the sequence is arithmetic is therefore to subtract any term in the sequence from the term following it, and then pick another (different) term, and subtract this from the term following it. If the result of the subtractions is the same, then we have found the common difference and the sequence is arithmetic.
Geometric?
Recall that a common ratio is a number which is multiplied by each term to get the next term in the sequence and that the $n$th term of an geometric sequence is given by: \[AR^{(n-1)}\] Dividing the $(n+1)$th term by the $n$th term gives: \begin{align} \dfrac{AR^n}{AR^{(n-1)}}&=R^{(n-(n-1))}\\ &=R \end{align}
Similarly, dividing the $(n+2)$th term by the $(n+1)$th term gives: \begin{align} \dfrac{AR^{(n+1)}}{AR^n}&=R^{((n+1)-n)}\\ &=R \end{align}
There is a common ratio of $R$ between any $2$ terms in the sequence. One way to check if the sequence is geometric is therefore to divide any term in the sequence by the term preceding it, and then pick another (different) term, and divide this by the term preceding it. If the result is the same, then we have found the common ratio and the sequence is geometric.
Worked Examples
Determine whether the following sequences are arithmetic or geometric or neither:
a) $3, 21, 147, 1029,...$
b) $1125, 225, 45, 9,...$
c) $-13, 32, 69, 104, 137,...$
Solution
a) We will first test whether this sequence is arithmetic. Subtracting the first term from the second gives: \[21-3=18\] and subtracting the second term from the third gives: \[147-21=126\] Since $18\neq 126$ this sequence cannot be arithmetic so we will now test if it is geometric. Dividing the third term by the second term gives: \[\frac{147}{21}=7\] and dividing the fourth term by the third gives: \[\frac{1029}{147}=7\] Since these are the same there is a common ratio of $7$ between terms and we have a geometric sequence.
b) We will first test whether this sequence is arithmetic. Subtracting the third term from the fourth gives: \[9-45=-36\] and subtracting the second term from the third gives: \[45-225=-180\] Since $-36\neq -180$ this sequence is not arithmetic so we will now test if it is geometric. Dividing the second term by the first gives: \[\frac{225}{1125}=0.2\] and dividing the third term by the second gives: \[\frac{45}{225}=0.2\] Since these are the same this is a geometric sequence with a common ratio of $0.2$.
c) We will first test whether this sequence is arithmetic. Subtracting the second term from the third gives: \[69-32=37\] and subtracting the fourth term from the fifth gives: \[137-104=33\] Since $37\neq 34$ this sequence is not arithmetic so we will now test if it is geometric. Dividing the second term by the first gives: \[\frac{32}{-13}=-2.4615...\] and dividing the fifth term by the fourth gives: \[\frac{137}{104}=1.3173...\] so the sequence is neither arithmetic nor geometric.
Simple and Compound Interest
Simple Interest
Suppose you deposit $£A$ into a bank. We refer to $£A$ as the principal balance. You are paid $15\%$ interest on your deposit at the end of each year (per annum). At the end of the first year you will have a total of: \[£(A+0.15\times A)=£A(1+0.15)\] With simple interest, the key assumption is that you withdraw the interest from the bank as soon as it is paid and deposit it into a separate bank account. This means that the interest paid each year is only ever paid on the principal balance. At the end of the second year you will therefore have a total of: \[£(A+0.15A+0.15A)=£(A(1+2\times 0.15))\] At the end of the third year you will have: \[£(A+0.15A+0.15A +0.15A)=£(A(1+3\times 0.15))\] and so on.
We can see that the total balance-the principal balance and the interest earned each year up to the present year-forms an arithmetic progression with first term $A$ and common difference $0.15A$.
In general, for an interest rate of $r$ (or $(100\times r)\%$) per annum and principal balance of $£A$, the total balance will form an arithmetic progression with first term $A$ and common difference $d=r\times A$. Using the above formula for the $n^{\text{th} }$ term of an arithmetic progression, the total balance at the end of $n$ years is: \[£(A+nd)\]
Compound Interest
The difference between simple interest and compound interest is that with compound interest it is assumed that you do not withdraw the interest earned from the bank each time it is paid. At the end of each year, the interest will therefore be paid on the total balance earned so far (the principal sum and interest earned each year up to the present year).
If the principal sum is $£A$ and the interest rate is $r$ (or $(100\times r)\%$) per annum the total balance at the end of the first year will be: \[£A(1+r)\] At the end of the second year it will be: \[£A(1+r)^2\] At the end of the third year it will be: \[£A(1+r)^3\] and so on.
We can see that the total balance forms a geometric progression with first term $A$ and common ratio $(1+r)$. Using the above formula for the $n^{\text{th} }$ term of a geometric progression with $n-1$ replaced by $n$, the total balance at the end of $n$ years is: \[£A(1+r)^n\]
Note: The total balance at the beginning of the $n$th year is given by: \[£A(1+r)^{n-1}\]
Now suppose that the bank pays you interest on your deposit $t$ times per year. For example interest might be paid monthly, in which case we would have $t=12$. Assuming that the interest earned is not withdrawn after each payment, the total balance after $n$ years is given by: \[A\left(1+\frac{r}{t}\right)^{tn}\] where, as before, $r$ is the interest rate and $A$ is the principal sum.
Worked Examples
Example 1
James deposits $£2,000$ into a bank which pays an annual interest rate of $4\%$. He withdraws and spends the interest earned every time it is paid. Calculate:
a) James' total balance after $5$ years, and
b) The interest that he receives after $5$ years
Solution
a) As James withdraws the interest earned each year, his total balance forms an arithmetic sequence with first term $A=2,000$ and common difference $d=0.04\times 2000=80$. We want to calculate his total balance after $5$ years. Using the formula with $n=5$, we have: \[\text{Total balance }=£(2,000+5\times 80)=£2,400\]
b) The interest that James receives over $5$ years is equal to the difference between his total balance after $5$ years and his principal sum. We know that James' principal sum is $£2,000$ and we found in part a) that his total balance after $5$ years is $£2,400$. The interest received over $5$ years is thus: \[£2,400-£2,000=£400\]
Example 2
Now suppose that James finds out that his wife is expecting a baby girl and has been told that by the time she is $18$, he will have spent $£10,000$ on clothing for his daughter. In addition, James would like to buy a car for his daughter on her $18$th birthday worth $£5,000$. To make sure he has enough money, James decides he will deposit a sum of money into an account which pays an annual interest rate of $25\%$, and that he will withdraw the interest earned each year to pay for his daughter's clothes. How much does James need to put into the account to make sure that his total balance after $18$ years will be enough to cover the clothing costs and the cost of the car? How much will he have to spend on clothing each year?
Solution
Here we have $n=18$ and $d=0.25A$ where $A$ is the (unknown) amount of money that James needs to deposit. We want the total balance after $18$ years to be equal to $£15,000$. Using the formula with $n=18$, we have: \begin{align} 15,000&=A(1+18\times 0.25)\\ &=4.5A \end{align} Rearranging and solving this equation for $A$ gives: \begin{align} A&=\dfrac{15,000}{4.5} &=3333.333 \end{align} So James needs to deposit $£3333.34$ in order to have a total balance of at least $£15,000$ by his daughter's $18^{\text{th} }$ birthday.
The amount that James will have to spend on clothing each year is equal to the interest earned annually: $r\times A$. Here $A=3333.34$ and $r=0.25$ so we have:
$0.25\times 3333.333=833.34$ (to $2$d.p.) That is, James will have $£833.34$ to spend on clothing for his daughter each year.
Example 3
After discovering that his wife is in fact having triplets, James takes up a part time job to pay for their combined clothing costs and decides that he will enough money into a bank account to pay for a car worth $3,000$ for each of them on their $18$th birthday. Assuming that the interest rate is the same as before ($25\%$) and that James does not withdraw the interest payment each year, how much does James now need to deposit into the bank account?
Solution
As James leaves the interested earned each year in the account, his total balance forms a geometric sequence with (unknown) first term $A$ and common ratio $1+0.25=1.04$. Since each car costs $£3,000$, the combined cost is $3\times 3000=9000$. This is the total balance that James must have after $18$ years. Using the formula with $n=18$, we have: \begin{align} 9,000&=A(1.25^{18})\\ &=55.51115123A \end{align} Rearranging and solving this equation for $A$ gives: \begin{align} A&=\dfrac{9,000}{55.51115123}\\ &= 162.129... \end{align} So James needs to deposit $£162.13$ in order to have a total balance of at least $£9,000$ by his children's $18$th birthday.
Video Examples
Example 1
Hayley Bishop finds the fifteenth term and the sum of the first twenty terms of the arithmetic progression $2,6,10,14,\dotso$.
Example 2
Hayley Bishop finds the ninth term and the sum of the first five terms of the geometric progression $3,1,\frac{1}{3},\frac{1}{9}\dotso$.
Example 3
Hayley Bishop finds the sum to infinity of the geometric progression $3,1,\frac{1}{3},\frac{1}{9}\dotso$.
Test Yourself
Test yourself: Arithmetic sequences
Test yourself: Geometric sequences
