Sequences and Series
In this topic we shall study sequences and series, and their properties. Firstly we define the terms sequence and series.
Sequence: Any mathematical progression of numbers, following a pattern. (e.g. and )
Series: The sum of a finite or infinite sequence of terms. (e.g. and )
In general, sequences follow some rule. The rule is called the general term. For the sequence the general term is where represents the numbers . The general term, or rule may be called the th term.
Example 1Contents Write down the first three terms for the sequence given by the rule . Solution 1We have that,
Hence we have that the first three terms are given by and respectively. 
Note: The expression is in fact a short hand convention for . This is because a sequence is in fact simply the range of a function with a domain equal to the natural numbers. That is, the general rule, is simply a function which converts natural numbers into a sequence.
Arithmetic Sequences
A special family of sequences called arithmetic sequences (or arithmetic progressions abbreviated to AP), are of particular interest to us. In an arithmetic progression, ensuing terms follow the pattern of differing by the same amount. For example, consider the arithmetic progression all the adjacent terms of which differ by .
Example 2Show that forms an arithmetic progression. Solution 2For three successive terms to form an arithmetic progression, we simply require that the difference between the third and second term be equal to the difference between the second and first term. That is, each successive pair, differ by the same amount. So we have that,
And
Hence as the difference is constant, then it follows that the terms form an arithmetic progression. 
As a convention, we denote the common difference between successive terms of an AP by . That is,
We shall now find the general term describing the th term of any arithmetic progression. Suppose that the first term of an arithmetic progression is and that the common difference is given by . Then we have the following table.
By observation of the general pattern, we can observe that the th term of an arithmetic sequence is given by;
where is the first term, is the common difference and is the term number.
We shall now consider some examples to illustrate use of this formula.
Example 3Find of . Solution 3So we have that and . Hence we have that the general term is given by,
So, we have that,

Example 4Find the term number that is equal to in the sequence . Solution 4Firstly, we find the general term. We have that and . Hence we have that
So we must set the general term to and solve the resulting equation in to find the term number.
Hence we have that the rd term is equal to . 
Example 5The rd term of an AP is and the st term is . Find the first term and the common difference of this sequence. Solution 5We firstly have that the general term of an AP is given by . We have that,
So we have,
And
Now, gives us,
Now, substituting this back into gives,
Thus we have that the first term is and the common difference is for this sequence. 
Arithmetic Series
We now shall look at arithmetic series (the summation of an arithmetic progression). We denote the sum up to terms for any series as .
Now, suppose we are adding terms from an arithmetic progression to obtain a sum. That is,
Now, we consider written in the reverse order with terms lining up vertically with written in correct order.
Now, adding the two series with terms lining vertically being added gives,
Now, on the RHS there exists terms. So we have,
which is the formula for the summation of an AP. We may further simplify the formula;
Now, since there exists terms in the series, we thus have that where is the last term. Hence we have that,
where is the first term, is the number of terms in the series, is the last term and is the common difference.
Note: The second form of the summation formula for an AP is only used if the last term is known. Otherwise we simply use the standard form which is the first.
Example 6What is the sum of the first 80 terms of the sequence ? Solution 6We have that , and . We are required to calculate .
Thus the sum of the first terms is . 
Now, in some questions, students are given the summation formula, and are required to find the general term of the sequence being summed. This not only applies to AP’s but to all series. The method used to find the general term given the summation formula for terms is,
Now, this formula is obviously true since,
is obtained by substituting whenever occurs in the formula for .
Consider the below example to illustrate the method used throughout such questions.
Example 7Find the general term for the sequence being summed if the formula for the sum of the first terms of this sequence is given by , and hence find the rd term of the sequence. Solution 7
Now,
Hence we have that,
That is, the general term is and the rd term is given by . 
Sigma Notation
Here we introduce a notation for the summation of a series. The Greek letter (Sigma) is used to denote the summing of terms. With the summation formula, we simply place the values of from the lowest value (shown at the bottom) to the highest value (shown at the top), into the formula, incrementing the number by , and then simply sum the terms together. Examples are the best way to introduce this notation.
Example 8Find the value of
Solution 8

Geometric Sequences
Another special sequence of numbers is called a geometric sequence or geometric progression (GP).
These sequences have the property that each succeeding term is a constant multiple of the previous term. This constant multiple is called the common ratio and is denoted (for ratio). An example of such a series is where the common ratio is .
For a sequence of numbers to be a geometric progression, the ratio of any two adjacent terms (succeeding term over preceding term) must be equal to the common ratio. In other words,
For all natural numbers .
Example 9Show that the sequence is a geometric progression. Solution 9For the sequence to be a geometric progression, we must simply show that ordered adjacent terms share a common ratio.
Hence as the ratio of the third term to the second is equal to the ratio of the second term to the first, then it follows that is a GP. 
We shall now find the th term of a geometric sequence. Suppose that a geometric sequence has first term and common ratio . We have that,
It is obvious from that pattern that the th term of a geometric sequence is,
where is the term number, is the first term and is the common ratio.
Consider the below examples to illustrate the use of the formula.
Example 10Find of the following GP;
Solution 10We have that and . Hence we have that,
and consequently,

Example 11Find the number of terms in the following GP;
. Solution 11So we have , . Hence we have that,
Now, we have that the last term of the sequence is . Equating this t and solving the resultant equation for gives;
Now, we must take logarithms of both sides to solve for .
Now, we have that,
By the rule . Hence we have that,
Hence we have that,
Hence there are terms in the sequence. 
Example 12In a certain geometric progression and . Find the general term. Solution 12Assuming the general form for the th term, we have that;
Now,
And
Hence we have two equations;
Now gives,
Substituting back into gives,
Hence we have that the general term is given by,

Geometric Series
If we are to sum a geometric progression, we obtain a geometric series. We shall denote the sum to terms as . Hence we have that,
To obtain a summation formula for geometric series, we must firstly perform the operation, .
That is,
Upon cancelation of terms. Now, dividing through by and factoring gives,
Provided that . By removing a factor of from top and bottom we obtain,
with .
Obviously both formulae are equivalent and it doesn’t matter which formula one uses, however, the use of the first is recommended when and the use of the second is recommended when to avoid difficulties with the negatives.
Example 13Find the sum to 12 terms of the geometric sequence . Solution 13We have that and that . Also we are given that . Hence we have,

Limiting Sum of a Geometric Series
In the case where we have that (i.e ) we have the following situation.
As , since . Hence we have that,
That is, the sum to an infinite number of terms, of a geometric series with common ratio between and is a finite value. We denote this by . Hence we have the formula,
when .
Note: If then the limiting sum of the geometric series does not exist and the formula does not hold.
We shall now examine some of the uses of this formula.
Example 14Does the limiting sum of the geometric series exist? If so, find its value. Solution 14Firstly, the only condition required for the limiting sum to exist is that . We have that,
Hence the limiting sum does exist. Now, we have that and . Applying the formula for infinite geometric series gives,

Questions involving limiting sums may at times involve expressing repeating decimals as fractions. The technique for such questions is quite straightforward. Observe the example below.
Example 15Express as a fraction, using geometric series. Solution 15Firstly we write as a decimal.
Now, we have that this repeating decimal can be represented in terms of fractions as
One can see that the bracketed segment is a geometric series with and . Hence we have that,
That is,

Exam Style Questions
Here we consider three exam style questions, typical of assessments and exams.
Example 16The sum of the first five terms of an arithmetic series is four times the fourth term. Also, the sum of the fifth and sixth terms is 65. Find the sum of the first 15 terms of the series. Solution 16So firstly, the written information must be translated into symbolic mathematics. We have that,
and
Now, we know that since the sequence is an AP, and also that,
Using these formulae gives,
and
Hence we have the equations,
Now, substituting into gives,
and hence we have that
Now, we are required to find .
Hence the sum of the first 15 terms is . 
Example 17Find the number which added to each of and will give a set of three numbers in geometric progression. Solution 17So, let this number be . So we have that must form a geometric progression. So we require that,
Since ordered adjacent terms must have the same common ratio.
Upon cross multiplying.
That is,
Hence the number is . 
Example 18a) Show that
b) Hence find value of,
Solution 18a) The series forms an Arithmetic Series, with and . Now, to find the number of terms, we simply need to find the th term of the sequence and equate this to the last term, upon which we may solve for and find the term number. So we have that,
So, equating this to gives,
We can now use the formula for summing an AP. In this case we shall use the last term form of the formula.
b) Now, to sum the series,
We firstly notice that this can be broken up into the series from (a) and another geometric series.
Now, there are 16 terms for the series from (a) and hence there are 16 terms for the geometric series as well. Now, we can see that for the geometric series, and . Hence we have that the sum is equal to,
Using (a) and the summation formula for a GP. Simplifying gives,
Hence we have that,

Series Applications
A practical application of arithmetic and geometric series arises within financial mathematics. Within the Advanced Mathematics course there are 2 main types of questions related to financial mathematics; Superannuation and Time Payments.
Superannuation: A fixed amount is invested each period over a number of years.
Time Payment: A loan is taken, and the repayments are made until the total amount owing is zero. Interest is charged on the amount owing periodically.
It is not necessary to know the details of these financial terms to be able to solve the mathematical problems correctly. More importantly, students should be able to convert a worded problem into symbolic mathematics.
Other applications also arise including physical problems (construction problems), compound interest and wage increase.
Example 19A pile of sand dumped at the start of a long road has to be distributed in truckloads at metre intervals along the road. If there are truckloads of sand in the pile, calculate how far in total the truck has to travel in order to deliver the sand. Solution 19Observe the diagram below depicting the situation. For every truckload the truck must travel back and forth along the same distance. This doubles every distance travelled. It is obvious that the total distance travelled is an arithmetic series, since each time the truck goes to the next point to dump the sand, the distance constantly increases by m from the previous point. Now, the first distance travelled is since the pile is put m from where the big pile is. This becomes our value of . The common difference between distances travelled is m since every time the truck travels back and forth, an extra 1600m is added onto the previous distance travelled. So, as there are 15 piles, then we have that the total distance is given by,
Hence the total distance travelled by the truck is km. 
Example 20Find the accumulated amount after years, if dollars is compounded annually at per annum. Solution 20So, let be the amount accumulated after years. We have that
Since after at the start, no interest is to be paid on the principle. Now, after the first year, interest is paid on the principle and added to the total. So we have that is equal to,
Now, after 2 years, interest is paid on the total amount accumulated and then added to the total. That is,
We can now see a general pattern here, and may extrapolate the accumulated amount after years as:
By comparing the changing factors in and . Hence the total accumulated amount after years is

The above example formulates the compound interest formula. The next question is an example of a time payment question.
Example 21A family loans to buy a house. The interest rate is at a constant 9% p.a., and the family wishes to repay the money over 20 years. Find the monthly repayments required for this to occur. Solution 21We firstly note that after 20 years there occurs years. Letting be the amount owing after months, we require . Also, the interest rate is in terms of years. We require the calculation to be in terms of months. Hence the interest rate is now .
We note that,
And after one month we have interest added on to the total (so we multiply the total amount owing by where is the interest rate), after which the monthly repayment made. That is,
Now, after the second month, interest is charged on the previous months final balance, after which another monthly repayment is made.
Upon substituting the expression for in terms of into the expression. Now, we have that after the third month, interest is charged on the previous months balance, after which another monthly repayment is made.
From here we can see a general pattern for . Extrapolating this gives,
Now, using the summation formula for a GP on the second expression gives,
Now, we require . Using this, and solving for gives,
i.e. $ Hence the monthly repayments are $. 
The next question is an example of a superannuation problem.
Example 22Nam’s father paid $ into an account on the day Nam was born. After that, he paid $ into the account on Nam’s birthday until Nam’s 18th birthday (He does not pay on Nam’s 18th birthday). If the account accrued interest at per annum compounded monthly, calculate how much Nam would receive on his 18th birthday. Solution 22Firstly, we let be the amount accumulated after years. is obviously true. Now, after the first year, this amount is compounded monthly at per annum. The monthly interest rate is and the total number of periods in this time is . Also, at the end of the first year, $ is added into the account. Applying the compound interest formula gives and adding $ gives,
Now, at the end of the second year, per annum interest is paid on the accumulated amount, and an extra $ is added into the account. We have;
Substituting the previous expression for in terms of gives;
We can now extrapolate the th term as,
Now the expression contained within brackets is obviously a geometric series with , and the number of terms being . Hence we have that,
Now, to find the amount accumulated after the th birthday, we simply substitute into the above expression for .
= $
Now, since Nam’s father does not pay on the 18^{th} birthday, and the above formula assumes he does, we must subtract $ only. We do not need to consider any interest to subtract since the $ is added at the time of the 18th birthday according to the above formula, thus not allowing it to accrue any interest of its own.
$ – $ = $
Hence we have that Nam will obtain dollars on his th birthday. 
It can be seen that both time payment and superannuation problems can be solved using the same techniques. Simply formulate an expression for by writing the first few terms down, use the geometric series formulae to simplify the expression, and calculate what is required.
Note: It is important that you do not round off numbers early (especially the modified interest rate), as this will lead to very large differences later on in the calculation. Try to make use of the memory function of your calculator.