SEQUENCES AND SERIES

Sequences

What is a sequence?

Here are a few lists of numbers:

3, 5, 7, …
21, 16, 11, 6, …
1, 2, 4, 8, …

Ordered lists of numbers like these are called sequences. Each number in a sequence is called a term.

Sequences usually have patterns that allow us to predict what the next term might be. For example, in the sequence 3, 5, 7, …, you always add two to get the next term.

The three dots that come at the end indicate that the sequence can be extended, even though we only see a few terms. We can do so by using the pattern.
For example, the fourth term of the sequence should be 9, the fifth term should be 11, etc.

There are two types of sequences: arithmetic sequences and geometric sequences. We will discuss what they are and how they are applied in the real world.

Arithmetic Sequences/Progressions (A.P.)

An arithmetic sequence is a sequence that involves adding or subtracting a number to each term to get the next term.

In an arithmetic sequence, the difference between consecutive terms is always the same. This difference is called the common difference (d) of the sequence. For example, the common difference of 10, 21, 32, 43, … is 11.

Suppose we want to write a formula to find a certain term in the arithmetic progression 3, 8, 13, … We already know the common difference is +5 and the first term is 3. Therefore, our formula is: $n^{th}\> term = 3+5(n-1)$ The last part of the equation has an $n -1$ instead of $n$ because the first term in an arithmetic sequence does not have the common difference added, therefore the last term will have the common difference added $n-1$ times.

The general form of an arithmetic sequence is $a, \,\,a+d, \,\,\,a+2d,\,…\,\,\, a+(n-1)d$ where $a$ is the first term and $d$ is the common difference.

The $n^{th}$ term of an arithmetic sequence is given by $a + (n-1)d$ . This is because the nth term is the first term plus the common difference added $n-1$ times.

Arithmetic sequences show up where values are changed incrementally. For example, when training AI neural networks, the weight of each connection between neurons is changed by a constant amount to determine the best settings for the neural network.

Geometric Sequences/Progressions (G.P.)

In geometric sequences, the ratio between consecutive terms is always the same. We call that ratio the common ratio.

For example, in the sequence 1, 2, 4, 8, 16, … each term is obtained by multiplying the previous term by 2. Therefore, 2 is the common ratio of this geometric sequence.

Generally, the $n^{th}$ term of a geometric sequence with $a$ as the first term and $r$ as the common ratio is $ar^{n-1}$ . Once again, $r$ is raised to the $(n-1)^{th}$ power because in the first term, $r$ is raised to the power of zero, so each successive term is multiplied by one less $r$ . For example, the 5th term is equal to the first term multiplied by $r$ four times. The sequence is $a, ar, \,ar^2, \,ar^3$ …

Geometric sequences have plenty of real world applications. One which you may already know is the calculation of compound interest. They also come into play in something as simple as bouncing a ball, as the ball loses a certain percentage of its height after every bounce. They are evident even in the natural world, for example in food chains whereby there is a proportional decrease in energy as it is transferred from one level of feeding to the next.

Series

A series is the sum of the terms in an arithmetic or geometric sequence. The expression $S_n$ is used to represent the sum of a certain number of terms of a sequence, where $n$ is the number of terms added.

The concept of a series is one of the most powerful ideas in mathematics and has a wide range of applications.

Like sequences, there are two types of series: arithmetic series and geometric series.

Arithmetic Series

It is obtained by adding the terms of an arithmetic sequence.

I’m going to throw the formula at you, but don’t run away yet. You will understand it once I break it down for you. The sum of the first n terms of an arithmetic series is obtained by the equation: $S_n=\frac{n}{2}[2a+(n-1)d]$ Where $n$ is the number of terms, $a$ is the first term and $d$ is the common difference.

Explain please!

While this may seem like a random equation, it is obtained in a very simple way. We can represent the sum of terms in the A.P as: $S_n=a+(a+d)+(a+2d)+\dots+[a+(n-2)d]+[a+(n-1)d]$

Now, we rewrite this backwards, starting with the last term and ending with the first term. $S_n=[a+(n-1)d]+[a+(n-2)d]+\dots+(a+d)+a$ If we add the 2 equations, we notice an interesting pattern. Each pair of terms adds up to $[2a+(n-1)d]$ . $S_n+S_n=2S_n=[2a+(n-1)d]+[2a+(n-1)d]+\dots+[2a+(n-1)d]$ The expression $[2a+(n-1)d]$ repeats $n$ times, as there are $n$ terms in the series. Thus, we have $2S_n=n[2a+(n-1)d]$ Dividing both sides by 2, $S_n=\frac{n}{2}[2a+(n-1)d]$ And thus we have our equation!

Arithmetic series provide a useful framework for understanding and calculating cumulative sums in various scenarios. The applications don’t always have to be complex. For example, when saving a certain amount of money every month, the total amount forms an arithmetic series. Each deposit adds a constant amount to the total. So the next time you’re counting up your savings, maybe use the formula for the sum of an arithmetic series!

Geometric Series

A geometric series is obtained by adding the terms of a geometric sequence.

The general formula is: $S_n=\frac{a(r^n-1)}{r-1}$

Where does this formula come from?

The sum of the geometric progression is expressed as:

S_n=a+ar+ar^2+…+ar^{n-1}

Multiplying it by the common ratio r:

rS_n=ar+ar^2+ar^3+…+ar^{n-1}+ar^n

Subtracting the first equation from the second one:

\begin{align*} rS_n=\quad\quad\>&\cancel{ar+ar^2+ar^3+…+ar^{n-1}}+ar^n \\ \underline{-S_n=-a-}&\underline{\cancel{ar-ar^2-ar^3-…-ar^{n-1}}} \end{align*}

rS_n-S_n=ar^n-a

We then factor out the

S_n

on the left side of the equation and the

a

on the right side.

S_n(r-1)=a(r^n-1)

Dividing both sides by

(r-1)

\boxed{ S_n=\frac{a(r^n-1)} {r-1}}

Alternatively, one can also use the formula:

S_n=\frac{a(1-r^n)}{1-r}

Prove it!

We get this formula in a very similar way to the way whe derived the first formula. The sum of the geometric progression is expressed as: $S_n=a+ar+ar^2+…+ar^{n-1}$ Multiplying it by the common ratio r: $rS_n=ar+ar^2+ar^3+…+ar^{n-1}+ar^n$ Subtracting the second equation from the first one:

\begin{align*}S_n=a\,\,&\cancel{+ar+ar^2+ar^3+…+ar^{n-1}} \\ \underline{-rS_n=\quad}&\underline{\cancel{-ar-ar^2-ar^3-…-ar^{n-1}}-ar^n}_ㅤ \end{align*}

$S_n-rS_n=a-ar^n$ We then factor out the $S_n$ on the left side of the equation and the $a$ on the right side. $S_n(1-r)=a(1-r^n)$ Dividing both sides by $(1-r)$ : $\boxed{S_n=\frac{a(1-r^n)} {1-r}}$

The first formula is preferrable when $|r| >1$ , while the second one is preferrable when $|r|<1$ . This is simply to avoid dealing with negative numbers in our calculations. However, either formula can be used in any case.

Geometric series have a variety of uses, ranging across various fields such as physics, medicine, and even music. They can be used to calculate the amount of medicine in a person’s body, if you know the dosing schedule and amount and how quickly the drug decays in the body. Geometric series also provide an introduction to infinite series, which serve a diverse array of purposes in the real world.

Example 1

The first term of an Arithmetic Progression is equal to the first term of a Geometric Progression. The second term of the AP is equal to the fourth term of the GP while the tenth term of the AP is equal to the seventh term of the GP.

a) Find the value of $r$ that satisfies the progressions.

Let us start with what we know, then look for an insight that will lead to the solution. We have that the first terms of both progressions are equal ( $a+0d=ar^0$ ), the second term of the AP is equal to the fourth term of the GP ( $a+d=ar^3$ ), and the 10th term of the AP is equal to the 7th term of the GP ( $a+9d=ar^6$ ).

Do you notice anything about the terms we are given in the geometric progression? The powers of r are multiples of 3! ( $ar^0, ar^3, ar^6$ ) Therefore, we can get an equality from these. We find that: $\frac{ar^6}{ar^3}=\frac{ar^3}{ar^0}$ Now, substituting for the values in the AP: $\frac{a+9d}{a+d}=\frac{a+d}{a}$ $a(a+9d)=(a+d)(a+d)$ $a^2+9ad=a^2+2ad+d^2$ $d^2=7ad$ $d=7a$

Now, it may seem like we have hit a wall. We want $r$ , and all we have is an equation in terms of $d$ and $a$ . But what if we could express one side of the equation to include $r$ ?
We are told that the second term of the AP is equal to the fourth term of the GP. Thus we have: $a+d=ar^3$ From this, we see that $d=ar^3-a$ .

And we can finally solve the problem. Substituting $d=ar^3-a$ into the equation: $ar^3-a=7a$ $8a=ar^3$ $r^3=8$ $r=2$

b) Given that the tenth term of the GP is 5120, find the values of a and d.

The 10th term of the GP can be expressed as $ar^9$ . Thus: $ar^9 =5120$ $a(2^9)=5120$ $512a=5120$ $a=10$

To find d, we use the equation that we found earlier, $d=ar^3-a$ . $d=10(2^3)-10$ $d=80-10$ $d=70$

c) Calculate the sum of the first 20 terms of the AP.

We use the formula $S_n=\frac{n}{2}[2a+(n-1)d]$ to find the sum of an AP. Substituting the values of $n$ , $a$ and $d$ : $\begin{align*} S_{20}&=\frac{20}{2}[2\times10+(20-1)70] \\ &=10[20+1330]\\&=13500 \end{align*}$

Example 2

The first term of a Geometric Progression is 2. The common ratio of the GP is also 2. The product of the last 2 terms of the GP is 512. Determine the number of terms in the GP.

Generally, the last term of a GP is $ar^{n-1}$ . If we substitute our values of $a$ and $r$ , our last term is $2(2^{n-1})$ . The second last term is equal to the last term divided by the common ratio, 2. This evaluates to $2(2^{n-2})$ .

The product of the last 2 terms is 512, hence: $2(2^{n-2})\times2(2^{n-1})=512$ $2^{n-1}\times2^n=512$ $2^{2n-1}=2^9$ $2n-1=9$ $2n=10$ $n=5$

Sequences and series are not just mathematical concepts—they’re powerful tools applied across various fields. From predicting financial trends to modeling natural processes, these mathematical patterns offer valuable insights into real-world phenomena. Hopefully, you now have a deeper understanding of the order in our daily lives.