# Arithmetic & Geometric Progression

Contents

## What is Progression in Maths?

In our daily lives, we observe different patterns in nature. You might have observed the V-formation that different kinds of birds including geese, ducks and other migratory birds adopt during their flight, the magnificent pattern in which the petals of a rose or sunflower are arranged, the mesmerizing geometric art formed by the swinging pendulum, the positions of twinkling stars in a night sky, and the list goes on.

Mathematicians try to express such patterns in form of numbers, symbols and formula. They come up with different terminologies and define these fundamental concepts. Progression (or sequence) is an example of one such jargon.

Definition: In arithmetic (branch of mathematics), a set or list of ordered elements is called a progression (or a sequence). Elements can be anything such as letters, numbers, words etc. These elements are ordered based on some rule. For example, the following list represents a progression (or a sequence).

100, 10, 1, 0.1, 0.01            (1)

The underlying rule governing this list is quite evident i.e., list of five elements starting from 100 where each next element is one tenth of the current element in magnitude. It should be noted that the order of elements is also an important aspect of a progression. For example, if we reverse the order of the progression stated above i.e., “0.01, 0.1, 1, 10, 100”, then we get a new progression.

It should also be noted here that the elements in a progression can be infinite as well. You might be thinking how it is possible. Well, it is simple. We do not need to write down all the elements of the list. All we need is to define the rule and mention first few terms of the progression. The following is an example.

-1, -2, -3, -4….            (2)

Looking at the example, we can immediately come to the conclusion that the next terms will be -5, -6, -7 and so on. This is how we define an infinite progression with only few elements of that sequence.

Note that we can also represent the progression in the form of a statement. For example, the progression given above can be asserted as “A progression of all negative integer numbers starting from -1 and expanding towards negative infinity”.

There is one thing to remember here, however. It is often not possible to deduce a single rule from an infinite progression with its first few elements mentioned only. As an example, consider the following progression.

3, 5, 7, ….           (3)

What do you think is the rule that governs this progression? “Sequence of all positive odd numbers starting from 3”? Yes, that is right, but not completely. We can state a different rule as “Sequence of all positive prime numbers starting from 3”. In other words, a list of few elements can represent more than a single progression!

Exercise: As an exercise, try to come up with two examples of finite and two examples of infinite progressions. Also state the governing rule behind the progressions.

There are several widely known types of progression in mathematics. Of course, each one of them is based on a unique rule. They include:

• Arithmetic Progression
• Geometric Progression
• Harmonic Progression
• Fibonacci Progression

In this article, we will explore in detail two of the above-mentioned types of progressions (Arithmetic and Geometric Progressions).

## Arithmetic Progression

The fascinating aspect about a progression is that it is based on a single rule. With the help of a very simple rule, a uniquely creative progression can be constructed. One such example is the arithmetic progression. The rule is straightforward: “Consecutive elements of the progression differ only by a constant number.” That constant number is intuitively called the common difference. Common difference is usually represented by the letter ‘d’. Following is the example of an arithmetic progression.

1, 3, 5, 7, 9, 11, 13…           (4)

As you can observe that each next number is two more than the one preceding it i.e., 3 is two more than 1, 5 is two more than 3, 7 is two more than 5 and so on. It is easy to realize that the common difference for this arithmetic progression is 2 i.e., d=2. Note that the common difference can be negative as well as positive.

Exercise: As an exercise try to construct an arithmetic progression starting from 1 with a common difference of -2. (Hint: Each term will be two less than the one preceding it)

Notation: For the progression (4), we can say that ‘1’ is the first term, ‘3’ is the second term, ‘5’ is the third term and so on. It would be great if we could express the arithmetic progression in terms of some generalized expression. Luckily, the mathematicians are sane enough to make our lives easier. The equation that is used to communicate any term (i.e., $1^{st}$, $2^{nd}$, $5^{th}$, $100^{th}$ term) of the arithmetic progression is called the $N^{th}$ term. It is denoted as ‘$a_{n}$’. The subscript ‘n’ represents the position of the term. The first term of the progression is also of great significance. It is also called the initial term and is represented as ‘$a_{1}$’. The entire arithmetic progression can be developed based on the information of the first term and the common difference of an arithmetic progression.

General Form and nth term of Arithmetic Progression: Let us try to build the general form for the arithmetic progression. Let us say that the first term of an arithmetic progression is ‘a1’. The common difference of the arithmetic progression is ‘d’. The second term of the arithmetic progression will be equal to preceding term plus the common difference (i.e., $a_{1}$$a_{2}=a_{1}+d$’).

Similarly, the third term of the progression will be $a_{3}=a_{2}+d$. Now here comes a mathematical trick. Note that as ‘$a_{2}=a_{1}+d$’, therefore, the third term can be rewritten as, $a_{3}= (a_{1}+d) + d = a_{1} + 2d$. Extending this idea, the fourth term can be found as, $a_{4}= a_{3} + d= a_{1} + 3d$. Can you observe a pattern here? You may realize that we can rewrite the arithmetic progression in a relatively generalized form as:

$a_{1}, \,a_{1}+d,\,a_{1}+2d,\,a_{1}+3d,\,....$                     (5)

Looking at the above expression, the expression for the nth term can be registered as:

$a_{n}= a_{1} + (n-1)d$                     (6)

Congratulations! We have successfully derived the formula for the general term of an arithmetic progression.

It may be noted that using the above relation we can find the $a_{n-1}$ term as:

$a_{n-1}= a_{1} + (n-1-1)d=a_{1} + (n-2)d$                     (7)

Using the above two equations, it can be realized that the common difference can be expressed as:

$d= a_{n}-a_{n-1}$                     (8)

Note that the above relation makes sense because the common difference is actually the difference between the two consecutive terms of an arithmetic progression.

Let us solve two examples. Suppose that we are given the common difference ‘d’ of the arithmetic progression, equal to -5. The first term is known to be equal to 300. We have been asked to construct the arithmetic progression using this information. Using the relation of the nth term, the arithmetic progression can be written as 300, 295, 290, 285, 280….

As the second example, consider that we are given the arithmetic progression 4, 2.5, 1, -0.5 …. We have been asked to find the common difference, the first term and the nth term of the given progression. How do we approach the problem? To find the common difference, we observe the difference between any two consecutive terms of the progression. The difference comes out to be -1.5, therefore, the common difference ‘d’ = -1.5. The first term is 4, evidently. Finally, given that we have figured out the first term as well as the common difference, the nth term of the progression comes out to be: $a_{n}=4+(n-1)(-1.5) = 4-1.5(n-1)$

Exercise: Find which among the following progressions qualify as the arithmetic progressions. Find the common difference for the ones that are arithmetic progressions.

• 10, 20, 40, 80….
• 1,2,4,5,7,8…
• 35, 36, 37, 38 ….
• 1/3, 2/3, 1, 4/3, ….

Find the respective common difference, the first term and nth term of each of the following arithmetic progressions.

• 0.1, 0.9, 1.7, 2.5….
• -10, -9.5, -9.0, -8.5….
• 1400, 1550, 1700, 1850….
• 1, -1, -3, -5 …..

Sum of n-terms of an AP: It is sometimes valuable to know the sum of a finite portion of an arithmetic progression. In other words, we are often interested in finding the sum of first n-terms of an arithmetic progression. There is a rather fascinating story associated with this topic. The story is about an elementary school student who later became a famous Mathematician. His name was Carl Friedrich Gauss.

Carl Friedrich Gauss was a German mathematician. One day in his elementary school class the teacher asked the young Gauss to find the sum of first 100 natural numbers (1-100). It was expected to be a laborious and time-consuming task. Young Gauss quickly figured out the answer to the problem, that was 5050. The teacher was already astonished with the swift answer made by the young lad, while he asked him to explain how he did it. Gauss explained his simple yet elegant technique. His explanation goes like this. Let $S_{100}$ be sum of first 100 natural numbers.

$S_{100}=1 + 2 + 3 + ... + 98 + 99 + 100$

It can be rewritten again as:

$S_{100}=100 + 99 + 98 + ...+ 3 + 2 + 1$

Gauss argued that above two equations can be summed and we get,

$2S_{100}= (100 + 1) + (99 + 2) + (98 + 3) ... (1 + 100)$

$2S_{100}= 101 + 101 + 101 + ... + 101$

Since there were 100 terms in each equation,

$2S_{100}= 100 x 101$

$S_{100}= 5050$

Amazing, isn’t it?

This method can be extended to find the $S_{n}$ (sum of first ‘n’ terms of an AP). The final result comes out to be:

$S_{n}= \frac{n}{2} (a_{1}+a_{n})$

You are highly encouraged to pull this one off yourself and derive it. Try to play around with the formula of nth term of the arithmetic progression and use similar technique mentioned above.

## Geometric Progression

A geometric progression is a list of ordered numbers where each following number is obtained by multiplying the preceding number by a non-zero real number. The non-zero real number is called as the common ratio. The common ratio is represented by the letter ‘r’. Following is the example of a geometric progression.

2, 4, 8, 16, 32…                 (9)

It can be realized that each term is actually twice the term preceding it. Therefore, the common ratio ‘r’ is 2. Note that the common ratio ‘r’ can be either positive or negative but should not be unity (since for r=1 there will be a sequence of same repeating numbers and it will be a boring progression). As with the case of the arithmetic progression, the geometric progression can be finite and infinite although there is a huge difference when it comes to the sum of the terms of the infinite Arithmetic progression and infinite Geometric progression (it will be discussed later).

It is noteworthy that if the common ratio is greater than 1, then it will be a progression increasing in magnitude. On the other hand, if the common ratio is less than 1, then it will be a decreasing progression.

Geometric progressions have a number of applications throughout engineering, mathematics, physics, economics, computer science and even the biology.

Exercise: As an exercise try to develop a geometric progression using the common ratio ‘r’ equal to -2. The first term of the progression should be 1. (Hint: Start with the first term. Multiply it with -2 to get the second term. Then multiply the second term with -2 to get the third term and so on).

Notation: As far as the notation is concerned, the same notation is used for the geometric progression. The symbol ‘$a_{n}$’ is usually used to represent the nth term of the geometric progression.

General Form and the nth term: Now let us derive the general form for a geometric progression. It is known that the common ratio ‘r’ is the ratio between each element and the preceding one. Therefore, if we represent the first element of a geometric progression as $a_{1}$ then we can deduce that the next term ‘$a_{2}$’ will be equal to ‘$ra_{1}$’. Similarly, the third term ‘$a_{3}$’ would be equal to ‘$ra_{2}$’.’$a_{2}$’ can be replaced by ‘$ra_{1}$’. Thus, ‘$a_{3}$’ will be equal to ‘$r^{2}a_{1}$’ and so on. A very noticeable pattern thus appears. In this way, it is easy to deduce that the general form of a geometric progression would be:

$a_{1}, \,ra_{1}, \,r^{2}a_{1}, \,r^{3}a_{1}...$           (10)

From the above expression, the nth term can be written as:

$a_{n}=r^{n-1}a_{1}$           (11)

Let us solve two examples relating to the geometric progression (as we did for AP). Suppose that we are given the common ratio ‘r’ (of a geometric progression) equal to 1/3. The first term of the geometric progression is also known. It is equal to 270. It is asked to construct the geometric progression using the aforementioned information. To solve this example, we refer to the general form of the geometric progression. After plugging in the value of ‘r’ and the first term i.e., $a_{1}$, we can get the geometric progression as 270, 90, 30, 10, 10/3,… .

As the second example, consider that we are given the geometric progression 1, 1.1, 1.21, 1.331… . We have been asked to find the first term ‘$a_{1}$’, the common ratio ‘r’ and the nth term ‘$a_{n}$’ of the given progression. The first term is simply equal to 1. To find the common ratio of the given geometric progression, we can find the ratio between any element and the one preceding it. Let us take third number i.e., 1.21 and divide it by the preceding number i.e., 1.1. We get ‘r’ equal to 1.21/1.1 = 1.1. Finally, to find the nth term of the progression, we refer to the relation $a_{n}=r^{n-1}a_{1}$. Plugging in the value of ‘r’ equal to 1.1 and ‘$a_{1}$’ equal to 1, we get $a_{n}=1.1^{n-1}(1)=1.1^{n-1}$.

Exercise:

1. Find the value of $99^{th}$ term of the above given geometric progression.
2. Find which of the following progressions are geometric. (Hint: Find the ratio between each pair of the numbers. If the ratios come out to be equal then the progression is said to be a GP)
1. 1, 1, 2, 4, 8, 16, …
2. 0.1, 0.5, 2.5, 12.5, …
3. $e, \,-e^{2}, \,e^{3}, \,-e^{4}...$
4. 100, 10, 100, 10, 100, …
3. Construct a geometric progression given that its first term is 15 and the ratio between its second and third term is 2 (Note the order of numbers of the given ratio i.e., $a_{2}/a_{3}=2$. Be careful about the fact that common ratio is defined with succeeding element as numerator).

More about the common ratio ‘r’: There are some interesting details related to the common ratio ‘r’.

The magnitude of the common ratio determines the nature of a geometric progression. If ‘r’ is greater than 1 then each succeeding number is larger than the one preceding it. Therefore, as a whole, the geometric progression expands towards infinity. If ‘r’ is less than 1 but greater than 0, then each succeeding number is smaller in magnitude than the one preceding it. In this case, the progression seems to shrink as ‘n’ increases.

Now consider a special progression below,

3, 3, 3, 3, …                                      (12)

Can you find its common ratio? Well, by applying the usual procedure of obtaining the common ratio, we see that its common ratio comes out to be unity i.e., 1.

What if ‘r’ is less than 0 i.e., ‘r’ is negative? In that case, we get a progression whose consecutive numbers switch between being positive and negative. The following progression is one such example.

2, -4, 8, -16…                                  (13)

The value of ‘r’ is -2 for the progression stated above. Three cases appear when r < 0.

• When ‘r’ is between -1 and 0, the GP seems to be shrinking (the absolute magnitudes of numbers get smaller as ‘n’ increases). The sequence, 10, -1, 0.1, -0.01, 0.001… is an example.
• When ‘r’ is less than -1, the GP seems to be expanding (the absolute magnitudes of numbers get larger as ‘n’ increases). The sequence, 10, -100, 1000, -10000… is an example.
• When ‘r’ is equal to -1, the GP seems to be constant (only alternating between positive and negative values). The sequence, 10, -10, 10, -10, 10… is an example.

Here is a question. What happens when ‘r’ is equal to 0? You might be tempted to say the progression would appear something like below.

$a_{1}$, 0, 0, 0, 0…

But remember that ‘r’ is the ratio between the consecutive numbers of a GP. And 0/0 is not defined. Therefore, ‘r’ cannot be equal to 0.

Sum of n-terms of a GP: The sum of the first n-terms of a geometric progression can be evaluated by using the following formula.

$S_{n}= \frac{a_{1}(1-r^{n})}{(1-r)}$                               (14)

Here is an interesting fact about the sum of GP that this formula shows. You should be able to recall that the sum of all terms of an infinite arithmetic progression cannot converge to a single value. But in this case, we see that if 0<r<1 and as n approaches infinity, the term $r^{n}$ approaches zero in the above expression and thus, the sum of the terms of infinite GP converges to a constant!

Let us look at an example. Suppose we are given a geometric progression {3, 6, 12, 24 …}. The sum of first 12 terms of the progression would be $S_{12}= \frac{3(1-2^{12})}{(1-2)}=12,285$. Similarly, the sum of first 6 terms would be $S_{6}= \frac{3(1-2^{6})}{(1-2)}=189$.

Note that if we were asked to find the sum of terms from position 7 to 12, then it can simply be evaluated as the difference of $S_{12}$ and $S_{6}$

Exercise:

1. Find the sum of first 10 terms of the geometric progression 1, 2, 4, 8, 16 …
2. Find the sum of
• First 10 terms
• First 100 terms
• Terms from position 11 to 100.
• All terms

of the geometric progression 1, 1/2, 1/4, 1/8 …

Sources: