# Arithmetic Progressions

An **arithmetic progression** (AP), also called an arithmetic sequence, is a sequence of numbers which differ from each other by a *common difference*. For example, the sequence $2, 4, 6, 8, \dots$ is an arithmetic sequence with the common difference $2$.

We can find the common difference of an AP by finding the difference between any two adjacent terms.

The following sequence is an AP with common difference 5 and initial term 0:

$\LARGE \color{#3D99F6}{0} \underbrace{\quad \quad }_{5} \color{#D61F06}{5} \underbrace{\quad \quad }_{5} \color{#20A900}{10} \underbrace{\quad \quad }_{5} \color{cyan}{15} \underbrace{\quad \quad }_{5} \color{orangered}{20} \underbrace{\quad \quad }_{5} \color{grey}{25}$

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## Describing Arithmetic Progressions

**Important terminology**

**Initial term:**In an arithmetic progression, the first number in the series is called the "initial term."**Common difference:**The value by which consecutive terms increase or decrease is called the "common difference."

**Recursive Formula**

We can describe an arithmetic sequence with a recursive formula, which specifies how each term relates to the one before. Since in an arithmetic sequence, each term is given by the previous term with the common difference added, we can write a recursive description as follows:

$\text{Term} = \text{Previous term} + \text{Common Difference.}$

More concisely, with the common difference $d$, we have:

$a_n=a_{n-1}+d.$

**Explicit Formula**

While the recursive formula above allows us to describe the relationship between terms of the sequence, it is often helpful to be able to write an explicit description of the terms in the sequence, which would allow us to find any term.

If we know the initial term, the following terms are related to it by repeated addition of the common difference. Thus, the explicit formula is

$\text{Term} = \text{Initial Term} + \text{Common Difference} \times \text{Number of steps from the initial term}.$

We can write this with common difference $d$, as:

$a_n = a_1 + d(n-1).$

What is the sequence described by $a_n = 2 + 4(n-1)$?

The sequence is $2, 6, 10, 14, \dots$.

We can see from the explicit formula that the initial term is 2 and the common difference is 4.

What is the explicit formula for the arithmetic progression $3, 6, 9, 12, \dots$?

Using the form given above, we have an initial term, $a_1=3$, and a common difference, $d$, of 3. Thus, $a_n = 3 + 3(n-1)$.

Note that we can simplify this expression to $a_n=3+3n-3=3n$.

## Sums of Arithmetic Progressions

**Sum of terms:** The sum of the first $n$ terms of an AP with initial term $a$ and common difference $d$ is given by

$S_n=\frac n2 \big[ 2a+(n-1)d\big] \qquad \text{or} \qquad S_n = \dfrac{n}{2} \big[T_1 + T_n\big] \qquad \text{or} \qquad S_n = n \times (\text{middle term}).$

What is the sum of the first 100 positive integers?

Let us assume that the sum of the first 100 positive integers is $S$, then$S=1+2+3+ \cdots +98+99+100.$

Now writing this expression in reverse order gives

$S=100+99+98+ \cdots +3+2+1.$

On adding the above two values, we get

$\begin{aligned} 2S&=(1+100)+(2+99)+(3+98)+\cdots +(98+3)+(99+2)+(100+1)\\ &=(101)+(101)+(101)+ \cdots + (101)+(101)+(101)\\ &=101 \times 100\\ \\ \Rightarrow S&=\dfrac{101 \times 100}{2}\\&=101 \times 50\\&=5050.\ _\square \end{aligned}$

From the example above, notice that the sums of corresponding terms in $S=1+2+3+ \cdots +98+99+100$ and $S=100+99+98+ \cdots +3+2+1$ are all the same, i.e.

$1+100=2+99=3+98= \cdots =50+51=51+50=\cdots =98+3=99+2=100+1.$

The sum of the terms of an AP can be found manually by adding all the terms, but this can be a very tedious process. Based on the above property possessed by an AP, there is a generalized formula for the sum of an AP.

For an arithmetic progression with initial term $a_1$ and common difference $d$, the sum of the first $n$ terms is

$S_n = \frac{n}{2}\big[2a_1+(n-1)d\big].$

What is the sum of the first 50 odd positive integers?

We recognize that this is an arithmetic sequence with common difference $2$ and initial term $1.$ We can then use the formula to obtain$S_n = \frac{1}{2}n\big(2a_1+(n-1)d\big) \implies S_{50} =25\times(2+49\times2) = 2500. \ _\square$

Note:We can generalize that the sum of the first $n$ odd positive integers is$\begin{aligned} S_n &= \frac{ a_1 + a_n } { 2} \times n \\ &= \frac{ 1 + \big(1 + (n-1)\cdot 2\big)} { 2} \times n \\ &= n^2. \end{aligned}$

## Properties of Arithmetic Progressions

**Increasing/Decreasing Sequence:**

If the common difference is positive, i.e. $d>0,$ then the arithmetic progression is an

*increasing*sequence and satisfies the condition $a_{n-1}<a_n:$ $a_1 < a_2 < a_3 < \cdots .$ For example, the arithmetic progression with initial term $2$ and common difference $3,$ i.e. $2,5,8,11,14, \dots,$ is an increasing sequence.If the common difference is negative, i.e. $d<0,$ then the arithmetic progression is a

*decreasing*sequence and satisfies the condition $a_{n-1}>a_n:$ $a_1 > a_2 > a_3 > \cdots .$ As an example, the arithmetic progression with initial term $1$ and common difference $-3,$ i.e. $1, -2, -5, -8, -11, \dots,$ is a decreasing sequence.

**Other properties:**

- If $a,b,c$ are in AP, then $2b = a + c.$
- If each term of an AP is increased, decreased, multiplied, or divided by a constant non-zero number, then the resulting sequence is also in AP.
- If the $n^\text{th}$ term of any sequence is of the form $an + b$, then the sequence is in AP where the common difference is $a$.

**Geometrical Interpretations:**

If $a_n$ represents the $n^{\text{th}}$ term of a the sequence $1, 2, 3, 4, 5, \dots$, corresponding points can be drawn $(n,a_n)$ in the Cartesian coordinate system as follows:

All the points $(n,a_n)$ are collinear, meaning a straight line can be formed by joining all the points. The result looks like this:
$$

*Slope of line:* The slope of the line is equal to the common difference of the AP, i.e. $d$. In the above case, the slope of the line $1$ is the same as its common difference.

## Problem Solving

$100$ potatoes, but can only carry one potato at a time. The potatoes are in a line in front of you, with the first potato $1$ meter away and each subsequent potato is located an additional one meter away.

You are standing next to a bucket and are tasked with collectingHow much distance would you cover while performing this task?

## See Also

**Cite as:**Arithmetic Progressions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/arithmetic-progressions/