Arithmetic Progressions

Introduction

Arithmetic Progression (AP) is basically a series of numbers that are increasing/progressing by a fixed common difference.

Examples

Example 1: The natural numbers follow an arithmetic progression :

$$ AP: 1, 2, 3, 4, 5, 6, … $$

Here, one is added to a number to get the next number in the series.

$$ 1 = 1\\ 1+1 = 2\\ 2+1 = 3\\ 3+1 = 4\\ 4+1 = 5\\ 5+1 = 6\\ … $$

Here, the first term (denoted by a ) is 1. The common difference (denoted by d ) is also 1 here.
$$ a=1\\ d=1\\ \therefore AP = 1, 2, 3, 4 … $$

Example 2: A taxi service runs only 100 kilometers. Its initial minimum fee is ₹50 and ₹20 is added for each kilometer. $$ a=50\\ d=10\\ \therefore AP=50, 60, 70 … 1050 $$

This is an example of a finite AP. It means that this AP has a last term (denoted by l ), and it stops there.

$$ \implies l=1050 $$

The first example we looked at was an infinite AP, as natural numners go on till infinity.

1. General Form of an AP

Let’s express APs in a general form applicable to all APs.

Remember that:
First term → a
Common difference → d

Let’s write an AP using just ‘a’ and ‘d’

$$ \text{1st term} = a\\ \text{2nd term} = a + d\\ \text{3rd term} = a + 2d\\ \text{4th term} = a + 3d\\ $$

That’s how it works.

2. Testing for AP

To know if a series forms an AP. We find the difference between consecutive terms.

Let’s assume a series,

$$ A1, A2, A3, A4 … A99, A100 $$

If this is an AP, the difference between terms should be constant, right?

So, we check if the difference is the same. If the series is an AP,

$$ A2 - A1 = A3 - A2 = A4 - A3 … = A100 - A99 $$

If the differences are not the same, then it’s not an AP.

3. Term between two terms in an AP

Let’s assume an arithmetic progression,

$$ 3, 6, \text{\textunderscore\textunderscore} , 12 … $$

In this series, one number is missing between two numbers. To find that number, we just have to take the arithmetic mean of the two numbers beside it.

In this case

$$ \frac{6+12}2 = 9 $$

4. n th term of an AP

Let’s assume an AP with $$ a=2\\ \text{and}\\ d=2\\ \implies \text{AP= } 2, 4, 6, 8, 10, 12, 14 … $$

In this AP, the first term is 2 and the 4th term is 8. Let’s write the same AP in its general form.

$$ 2, 2 + 2, 2 + 4, 2 + 6, 2 + 8 … $$

Here the 4th term is 2 + 6, which is a + 3d.

Which can be written as: $$ a + (4-1)d $$

This works with all terms, so we use a general formula for it.

For an AP, the nth term is : $$ \tag{1} a+(n-1)d $$

If a question comes to check if a number is a part of an AP, then you just have to use this formula backwards. If you get a natural number, then it’s probably a part of the AP. For eg:-

Example: Check whether 300 is a term of the AP: 100, 150, 200 …

Here, $$ a=100\\ d=50 $$

If we substitute the a and the d in formula (1) and simply equate it to 300: $$ 100 + (n-1)50 = 300 $$ After equating we get $$ n = 5 $$

We get n to be a natural number. This means that 300 is a part of the AP and is the 5th term in it.

Example: Find the total number of two-digit even numbers.

The first two digit number, 10, is even. So we can start an AP from 10 to the last two-digit even number, which is 98.

$$ a = 10\\ l = 98\\ d = 2\\ \text{(}d=2 \because \text{even numbers are multiples of two)} $$

Using formula (1) $$ 98 = 10+(n-1)2 $$

Do the calculation, you will get….. $$ n = 45 $$

5. Sum of first n terms of an AP

Lets consider the AP: $$ 2, 4, 6, 8, 10 … $$

We can write this as the sum of each term, $$ 2+4+6+8+10…….. $$

How do we find, say, the sum of the first five terms of this AP?

For this, we can use the formula:
$$ S_n=\frac{n}2[2a+(n-1)d] $$
where n is the number of terms whose sums is to be calculated.

Note: To find the nth term of an AP. Just find the sum of n terms, and subtract sum of (n-1) terms from it.

6 Sum of first n natural numbers

As we said before, natural numbers form an AP with
$$ a=1 d=1 $$
So, to find the sum of first n natural numbers we just have to substitute a and d.
$$ S_n=\frac{n}2[2a+(n-1)d]\\ S_n=\frac{n}2[2(1)+(n-1)(1)]\\ S_n=\frac{n}2[2+n-1]\\ \implies S_n=\frac{n}2[n+1]\\ $$
Therefore, the formula to find the first n natural numbers is:
$$ S_n=\frac{n}2[n+1]\\ $$