# Arithmetic Progressions

An **arithmetic progression** (AP) is a sequence of numbers satisfying the condition that the difference between any two consecutive numbers is constant. Observe that the page numbers written at the bottom of the book below follow the pattern of 47, 45, 43, etc. In this sequence, each term decreases by two. Similar patterns, where sequences increase or decrease by a fixed constant, can convey information about a data set.

Consider the book described above. If there is a series of pages with the pattern 39, 37, 35, 27, 25, what might you hypothesize?Given that the established arithmetic progression is broken, you might hypothesize that pages are missing from the book.

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## Arithmetic Progressions Terminology and Examples

**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".

**General term:** The \(n^{\text{th}}\) term \(a_n\) of an AP with first term \(a\) and common difference \(d\) is given by \[a_n=a+(n-1)d. \]

Note:It is sometimes easier to compute values in an arithmetic progression based on terms in the middle of the arithmetic progression rather than the initial term. Given an arithmetic progression with \(k^{\text{th}}\) term \(a_k\) and common difference \(d,\) the term \(a_n\) can be expressed in terms of \(a_k\) as follows: \[ a_n = a_k + (n-k)d. \]

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

\[\LARGE \color{blue}{0} \underbrace{\quad \quad }_{5} \color{red}{5} \underbrace{\quad \quad }_{5} \color{green}{10} \underbrace{\quad \quad }_{5} \color{cyan}{15} \underbrace{\quad \quad }_{5} \color{orangered}{20} \underbrace{\quad \quad }_{5} \color{grey}{25} \]

Given an arithmetic progression with initial term 4 and common difference 8, what is the 10\(^\text{th}\) term?If the arithmetic sequence is given by \(a_1, a_2, a_3, \ldots\), then \(a_1=4\) and the common difference is \(d=8.\) By the formula above, we have\[ a_{10}= 4 + (10-1) \times 8 = 4 + 72 = 76 .\]

Thus, the 10\(^\text{th}\) term in the arithmetic progression is 76.

## Given an arithmetic progression with 35\(^\text{th}\) term 15 and common difference 2, what is the 32\(^\text{nd}\) term?

Observe that we are not given the initial term of the arithmetic progression, but we are given the 35\(^\text{th}\) term. Then by the formula above with \(k=35,\) we have\[a_{32}=a_{35}+(32-35) \times 2 = 15+(-3) \times 2 = 9. \]

Alternatively, since the \(35^\text{th}\) term is \(15\) and the common difference is \(2,\) the \(34^\text{th}\) term is \(15-2=13,\) the \(33^\text{rd}\) term is \(13-2 =11, \) and the \(32^\text{nd}\) term is \(11-2=9. \)

**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.

Consider an arithmetic progression whose first term and common difference are both 100. If the \(n^\text{th}\) term of this progression is equal to \(100!\), find \(n\)

\[\]**Notation**: \(!\) denotes the factorial notation. For example, \(8! = 1\times2\times3\times\cdots\times8 \).

## Sum of Terms in 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

\[S_n=\frac n2 [ 2a+(n-1)d]\]

## There is a famous story about the mathematician Carl Friedrich Gauss:

Gauss was 9 years old, and was sitting in his math class. He was a genius even at this young age, and was therefore often bored in class and would goof off and get into trouble. One day his teacher wanted to punish him for goofing off, and told him that if he was so smart, why didn't he go sit in the corner and add up all the integers from 1 to 100. Gauss went and sat in the corner, but didn't pick up his pencil.

Teacher:“Carl! Why aren't you working? I suppose you've figured it out already, haven't you?”

Gauss:"Yes. It is 5050."The teacher didn't believe him and spent the next ten minutes or so adding everything up by hand, only to find that Gauss was right!

Gauss could solve the problem in the blink of an eye, as he recognized a pattern in the terms.

## 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

\[\begin{align} 2S&=(1+100)+(2+99)+(3+98)+\cdots +(98+3)+(99+2)+(100+1)\\ &=(101)+(101)+(101)+ \cdots + (101)+(101)+(101)\\ &=101 \times 100\\ \\ \Rightarrow S&=\dfrac{101 \times 100}{2}\\&=101 \times 50\\&=5050. \end{align}\]

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 \[\]

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

This theorem can be proven using the property of an AP that the sum of the corresponding terms from the beginning and the end is constant. That is, for an AP with first term \(a_1\) and last term \(a_n\), the following condition holds:

\[a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\cdots=a_n+a_1.\]

Suppose we wanted to add the first \(n\) terms of an arithmetic progression. Then we would first have

\[S_n = a_1 + (a_1 + d) + (a_1 + 2d) + \ldots + (a_n-2d) + (a_n - d) + a_n .\]

Writing this expression in reverse gives

\[S_n = a_n + (a_n - d ) + (a_n - 2d) + \ldots + (a_1+2d) + (a_1 + d) + a_1 .\]

Adding the two expressions, we have

\[\begin{align} 2S_n &= (a_1 + a_n) + (a_1+a_n) + (a_1+a_n) + \ldots + (a_1+a_n) + (a_1+a_n) \\ &= n(a_1+a_n) \\ \\ \Rightarrow S_n &=\frac{n}{2}\left(a_1+a_n\right) \\ &= \frac{n}{2}[a_1+\left(a_1+(n-1)d\right)] \qquad (\text{since }a_n=a_1+(n-1)d) \\ &= \frac{n}{2}[2a_1+(n-1)d]. \ _\square \end{align}\]

## 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 \[S_n = \frac{1}{2}n\left(2a_1+(n-1)d\right) \Rightarrow S_{50} =25\times(2+49\times2) = 2500. \ _\square\]

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

The initial term of an arithmetic progression is \(1\). If the sum of its first \(9\) terms is

threetimes as large as the sum of its first \(5\) terms, what is the common difference of this progression ?Let us write down all the relevant information and solve for the required value:\[\begin{align} S_n & = \frac{n(2a_1 + (n-1)d )}{2} \\ S_9 &= \frac{9(2 + 8d)}{2} \\ S_5 &= \frac{5(2 + 4d)}{2}. \end{align}\]

Since \(S_9 = 3S_5,\) it follows that

\[\frac{9(2 + 8d)}{2} = \frac{15(2 + 4d)}{2}.\]

Solving for \(d,\) we get \(d=1.\)

Given \(p\) arithmetic progressions, each of which consisting of \(n\) terms, if their first terms are \(1,2,3,\ldots,p-1,p\) and common differences are \(1,3,5,7,\ldots,2p-3,2p-1,\) respectively, what is the sum of all the terms of all the arithmetic progressions?

Consider an arithmetic progression whose third term is 11 and seventh term is 27. What is the minimum number of terms of this arithmetic progression that is needed to yield the sum of all terms divisible by 5 as 2030?

## Properties of Arithmetic Progressions

The following are significant properties of an arithmetric progression:

If the terms of an AP \(a_1,a_2, \dots, a_n\) are increased or decreased by the same number \(k\), then the resulting sequences \(a_1 \pm k , a_2 \pm k , \dots , a_n \pm k\) are also in AP.

If the terms of an AP \(a_1,a_2, \dots, a_n\) are multiplied or divided by the same non-zero number \(k\), then the resulting sequence \(a_1 \cdot k , a_2 \cdot k , \dots , a_n \cdot k\) or \(a_1/ k , a_2 / k , \dots , a_n / k\) is also in AP.

If \(S_k\) represents the sum of the first \(k\) terms of an AP, then \(S_p=S_q\) implies \(S_{p+q}=0\) for two distinct natural numbers \(p\) and \(q\).

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### Geometrical Interpretations:

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### 1. Representation of an AP on the number line

When the terms of an AP are represented on the real number line, the displacement between any two consecutive terms is constant. For example:

\[\begin{array} &-6 &-4 &-2 &0 &2 &4 &6. \end{array}\]

Marked on the number line, these numbers look as follows:

### 2. Representation of an AP on the Cartesian plane

\[\begin{array} &1 &2 &3 &4 &5. \end{array}\]

If \(a_n\) represents the \(n^{\text{th}}\) term of an AP, 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.

When the points \((n,a_n)\) corresponding to any AP are drawn on the Cartesian coordinate system, the result is a straight line with slope \(d\).

### 3. Representation of the sum of an AP on the Cartesian plane

If \(S_n\) represents the sum of the first \(n\) terms of the above AP, then locating the points \((n, S_n)\) on the Cartesian plane will give the following:

These points lie on a parabola with vertex \(\left( -\frac 12 , -\frac 18 \right) \) and focus \(\left( -\frac 12, \frac 38 \right)\). The graph of the parabola is as follows:

When all the points \((n,S_n)\) corresponding to any AP on the Cartesian coordinate system are located, they all lie on a parabola with vertex \(\left[ - \left( \dfrac{2a-d}{2d} \right), - \left( \dfrac{(2a-d)^2}{8d} \right) \right] \) and focus \(\left[ - \left( \dfrac{2a-d}{2d} \right), \left( \dfrac{1}{2d} - \dfrac{(2a-d)^2}{8d} \right) \right] \).

Consider a series of equilateral triangles set up on the \(x\)-axis as in the figure above. The side lengths of the triangles are an arithmetic progression with initial term \(2\) and common difference \(3.\) If the top vertices of these triangles are colored red, what is the curve that the red dots sit on?

Let the Cartesian coordinates of the red dots be \( (x_n, y_n) \). Also, consider an arithmetic progression given by the general term \( a_n = 2 + 3 (n-1) \). Then it is clear that \( y_n,\) the height of the \(n^{\text{th}}\) equilateral triangle, can be obtained as follows:\[ y_n = a_n \frac{ \sqrt{3} } { 2} = \frac{ [ 2 + 3 (n-1) ] \sqrt{ 3} } { 2 } . \]

In order to find \(x_n\), find the sum of the side lengths of the previous triangles, and add half of \( a_n \) to it.

\[ x_n = \frac{a_n}{2} + \sum_{i=1} ^ { n-1 } a_n = \frac{2 + 3 (n-1) } { 2} + \frac{n-1}{2} [2 \times 2 + (n-2) 3 ] = \frac{3 n^2 -2 n +1 } { 2 }. \]

Hence, we see that \( (x_n, y_n) \) would lie on a parabola. In fact, by converting the Parametric Equations, we get

\[ x_n = \frac{ 2 y_n ^2 + 3 } { 9 }. \]

In a certain arithmetic progression, the following holds:

\[S_{1729}=S_{28},\]

where \(S_n\) denotes the sum of the first \(n\) terms. What is the \(879^{\text{th}}\) term of this arithmetic progression?

###### I made this problem while I was pondering over my friend Sandeep Bhardwaj's problem.

## Additional Problems with Arithmetic Progressions

Your salary increases by \($1000\) every three months. If your present salary is \($20000\) per month, what will be your salary (per month) after 4.5 years from now?

As the salary increases by a fixed amount regularly, we can consider this event of salary increase as an arithmetic progression with initial term \($20000\) and common difference \($1000.\) Since \(4.5\) years is \(54\) months, the salary after \(54\) months (after \(54\div3=18\) terms of AP) will be the \(19^{\text{th}}\) term of the AP. Using the formula for general term \(a_n=a+(n-1)d\), we get\[\begin{align} a_{19}& = 20000+(19-1)1000& \\ &=20000+18000& \\ &=38000. \end{align} \]

Hence your salary per month after \(4.5\) years from now will be \(\$ 38000. \)

You deposit \($100\) in a bank at the rate of \(8\)% p.a. Calculate the total interest accrued on the account at the end of the \(1^{\text{st}}, 2^{\text{nd}}\) and \(3^{\text{rd}},\) respectively, and also calculate the balance on the account at the end of the \(25^{\text{th}}\) year.

At the end of the first year, the total interest accrued on the account is \($100 \times 8\text{%}=$8\).

At the end of the second year, the total interest is \($100 \times 8\text{%}+$100 \times 8\text{%}=$8+$8=$16\).

Similarly, at the end of the third year, the total interest is \(3\times $100 \times 8\text{%}=$24\).So, after \(n\) years the balance (in dollars) on your account will be an arithmetic sequence with initial term \(a=108\) and common difference \(d=8:\)

\[a_n=a+(n-1)d=108+(n-1)\cdot 8.\]

Since \(n\) is given as \(25,\) the balance on the account (in dollars) at the end of the \(25^{\text{th}}\) year is

\[\begin{align} a_{24}&=108+(25−1)d\\ &=108+24 \times 8\\ &= 300. \end{align}\]

How much distance would you cover while performing this task?

## Find the number of terms which are common in the following two arithmetic progressions: \[\]

\[\begin{array} &&2,4,6,8,\ldots &\text{ up to } 100 \text{ terms} \\ &3,6,9,12,\ldots &\text{ up to } 80\text{ terms}. \end{array}\]

Let \(r\) terms are common in the given two arithmetic progressions. Then the sequence of the common terms is \(6,12,18,\ldots\) and the \(r^{\text{th}}\) term would be \(a_r=6+(r-1)6=6r. \qquad (1)\)Now, the \(100^{\text{th}}\) term of the sequence \(2,4,6,8,\ldots\) is \(2+(100-1)2=200.\)

Similarly, the \(80^{\text{th}}\) term of the sequence \(3,6,9,12,\ldots\) is \(3+(80-1)3=240.\)

Then from \((1)\) it must be true that\[6r \leq 200 \Rightarrow r \leq 33.\bar{3} \Rightarrow r=33.\]

Hence, there are \(33\) common terms in the given arithmetic progressions.

## See Also

**Cite as:**Arithmetic Progressions.

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