# Arithmetic Progression

A progression is another term for
sequence
. Therefore, Arithmetic Progressions (also known as Arithmetic Sequences)
are special sequences defined by the property that the difference between any two
consecutive terms of the sequence are constant. Whereas the rule for regular sequences
is that the difference between consecutive terms has to have some kind of relationship,
it may or may not be constant. The relationship may not even be of a ‘difference’
nature but may be of a ‘multiplicative’ nature. This is not so for Arithmetic sequences,
the relationship has to be of a difference nature only. This common difference is
denoted by the letter d.

As an example, take a look at the sequence below;

for the sequence below to be an arithmetic progression, the difference between consecutive
terms has to be some constant d such that:

since that relationship must hold true, we can move further and say that

the same applies to a4

the same also applies to the last term an where byan-1
is the second last term

So now let us take a replace the terms in the sequence with their corresponding
terms that we have obtained

From the above new sequence, we should observe that to get the next term, you always
add the first term to the product of the common difference and one less than the
position of the next term.

i.e. if the next term is denoted by the letter n, then

The above relationship should always hold true as the relationship between consecutive
terms in an arithmetic progression. You can check that it is true by substituting
for the values of the sequence we’ve been working with.

This is what makes arithmetic progressions such special sequences, all you need
to know is the first term and the common difference d and then you’ll be
able to find any term in the sequence without having to find its preceding term.

For example, given that the common difference in an arithmetic progression is 5
and the first term is 3, find the 10th and 25th terms of the sequence.

solution:

The common relationship between all terms in an arithmetic progression is given
by

## Arithmetic Series

Since there exist Arithmetic Sequences, Arithmetic Series also exist and are the
sums of the terms in arithmetic sequences.

We have already seen that a series is given in the following form

where Sn is the notation for the sum of a series.

We have also established that the above terms can also be written in terms of the
first term and the common difference

but the above can also be written in terms of the last term as follows

If we were to add the two sums we would obtain the following

The above is obtained after some tedious addition and collecting like terms but
it turns out as a nice simple expression

So we now have an expression for the sum in terms of only the first and last terms
as

but we can simplify this further since we already established a way of expressing
the any term in terms of the first term and the common difference d

so now we have two options of formula to use to find the sum of an arithmetic progression.

Remembering from the section on Series, the notation for summation:

## Examples of Arithmetic Progression

### Example 1

Find the sum of the Arithmetic Progression below given that the total number of
terms is 15

Step 1

Since we haven’t been told what the common difference textbf{d} is, we need to
find that first

Step 2

Next to find the sum, all we have to do is substitute in the formula since we already
have the first and last terms

Step 3

Step 4

### Example 2

Given that the following is an Arithmetic Progression with 20 terms, find its sum

Step 1

Like in the previous example, we’re not given the common difference, and in this
case we don’t even have the last term. We could choose to find both or we could
just find the difference and use only this and the first term to find the sum.

Step 2

The formula below only utilizes the first term and the common difference

Step 3

Substituting:

Step 4

## Quiz on Arithmetic Progression

1. How many gifts do you receive in total in the song “Twelve Days of Christmas”?

The first step is to list out the sequence for the 12 days:

• On the first day, you receive 1 gift
• On the second day you receive 1 + 2 = 3 gifts
• On the third day you receive 1 + 2 + 3 = 6
• On the fourth day you receive 1 + 2 + 3 + 4 = 6
• And so on, until you reach day 12

The sequence for the gifts received on each day becomes:

To find the total number of gifts, just change the sequence into a series and find
its sum

364
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