Arithmetic Progression

An arithmetic progression (also called an arithmetic sequence) is a sequence of numbers in which every term after the first is obtained by adding the same number to the previous term. This number is denoted by $d$ and is called the common difference of the progression.

Examples

• $1,\ 2,\ 3,\ 4,\ \ldots$ is an arithmetic progression with common difference ${d=1}.$
• $4,\ 7,\ 10,\ 13,\ \ldots$ is an arithmetic progression with common difference ${d=3},$ because ${7-4}={10-7}={13-10}={3}.$
• $20,\ 10,\ 0,\ -10,\ -20,\ \ldots$ is an arithmetic progression with common difference ${d=-10}.$
• $5,\ 5,\ 5,\ 5,\ \ldots$ is an arithmetic progression too, with common difference ${d=0}.$
• $1,\ 2,\ 4,\ 8,\ \ldots$ is not an arithmetic progression: the differences ${2-1=1}$ and ${4-2=2}$ are not equal. (Here every term is twice the previous one. Such a sequence is a geometric progression.)

To find the common difference, subtract any term from the term after it: ${d=a_2-a_1}={a_3-a_2}=\ldots$

On this page: Notation · The $n$th term · Properties · Sum of the first $n$ terms · Calculator · Solved problems · Common mistakes · Practice

Notation

• $a_1$ is the first term of the progression;
• $a_n$ is the $n$th term (the general term), and $n$ is its position;
• $d$ is the common difference: ${d=a_{n+1}-a_n}$ for every $n;$
• $S_n$ is the sum of the first $n$ terms: ${S_n}={a_1+a_2+\ldots+a_n}.$ A sum of terms of an arithmetic progression is called an arithmetic series.

If ${d\gt 0},$ every term is greater than the previous one and the progression is increasing. If ${d\lt 0},$ the progression is decreasing. If ${d=0},$ all terms are equal and the progression is constant.

Formula for the $n$th term

Start at $a_1$ and keep adding $d$:

${a_2=a_1+d},$   ${a_3}={a_2+d}={a_1+2d},$   ${a_4=a_1+3d},$   ${a_5=a_1+4d},$   $\ldots$

We reach $a_n$ after ${n-1}$ steps, so the common difference $d$ has been added to $a_1$ exactly ${n-1}$ times:

Example. Find the 20th term of the progression $4,\ 7,\ 10,\ 13,\ \ldots$

Here ${a_1=4}$ and ${d=3},$ so

${a_{20}}={a_1+19d}={4+19\cdot 3}={61}$

From $a_1$ to $a_{20}$ there are 19 steps, not 20. That is why the common difference is multiplied by ${n-1},$ not by $n.$

We can also start from any other term $a_k.$ There are ${n-k}$ steps from $a_k$ to $a_n$:

Example. In an arithmetic progression ${a_3=7}$ and ${a_8=22}.$ Find $a_1$ and $d.$

There are 5 steps from $a_3$ to $a_8,$ so

${d}={\frac{a_8-a_3}{8-3}}={\frac{22-7}{5}}={3},$   ${a_1}={a_3-2d}={7-6}={1}$

The progression is $1,\ 4,\ 7,\ 10,\ 13,\ \ldots$

The terms lie on a straight line

The formula for the general term can also be written as ${a_n}={a_1+(n-1)d}={dn+(a_1-d)}.$ This is a linear function of the position $n.$ If we plot the points ${(n,\ a_n)},$ they lie on one straight line, and the common difference $d$ is how much the line rises for each step to the right. For the progression $4,\ 7,\ 10,\ 13,\ \ldots$ we have ${a_n=3n+1}$:

The points ${(n,\ a_n)}$ of the progression ${a_n=3n+1}.$ Each point is ${d=3}$ higher than the one before.

Properties

Each term is the arithmetic mean of its neighbors

The previous term is ${a_{n-1}=a_n-d}$ and the next one is ${a_{n+1}=a_n+d}.$ When we add them, $d$ cancels: ${a_{n-1}+a_{n+1}=2a_n}.$ Therefore

Every term (except the first) is the arithmetic mean of the term before it and the term after it. This is where the name of the progression comes from. The converse is also true: if every term except the first is the arithmetic mean of its neighbors, the sequence is an arithmetic progression.

So three numbers $a,$ $b,$ $c$ (in this order) form an arithmetic progression exactly when ${2b=a+c}.$

In the same way, every term is the arithmetic mean of any two terms at the same distance from it: ${a_n=\frac{a_{n-k}+a_{n+k}}{2}}$ for ${k\lt n}.$

Terms equally far from the ends

Two terms that are equally far from the beginning and from the end always have the same sum:

The reason: when we move one place to the right at the front and one place to the left at the back, the first term grows by $d$ and the second one shrinks by $d.$ More generally, if ${m+k=p+q},$ then ${a_m+a_k=a_p+a_q}.$

Example. In the progression $1,\ 11,\ 21,\ 31,\ 41,\ 51$:

${1+51}={11+41}={21+31}={52}$

${11}={\frac{1+21}{2}},$   ${21}={\frac{11+31}{2}},$   ${31}={\frac{21+41}{2}}$

Sum of the first $n$ terms

The story goes that when Carl Friedrich Gauss was a schoolboy, his teacher told the class to add up the numbers from 1 to 100. Gauss answered almost at once: he added the first number to the last, the second to the second-to-last, and so on.

${1+100}={2+99}={3+98}=\ldots={50+51}={101}$

There are 50 pairs, so the sum is ${50\cdot 101=5{,}050}.$

The same idea works for every arithmetic progression. Write the sum twice, the second time backwards:

$$\begin{aligned}S_n&=a_1+a_2+\ldots+a_{n-1}+a_n\\S_n&=a_n+a_{n-1}+\ldots+a_2+a_1\end{aligned}$$

Add the two equations column by column. Each column holds two terms equally far from the ends, so every column adds up to ${a_1+a_n}.$ There are $n$ columns, therefore ${2S_n=(a_1+a_n)\,n}.$

The sum ${1+3+5+7+9}$ written twice: the blue and the orange columns form a ${5\times 10}$ rectangle, so ${2S_5=50}$ and ${S_5=25}.$

The first formula is handy when you know the last term; the second one when you know the common difference.

Example 1. ${1+2+3+\ldots+100}={\frac{(1+100)\cdot 100}{2}}={5{,}050}.$

Example 2. The sum of the first $n$ odd numbers is

${1+3+5+\ldots+(2n-1)}={\frac{\bigl(1+(2n-1)\bigr)\,n}{2}}={n^2}$

For example ${1+3+5+7+9}={25}={5^2},$ as the picture shows.

The other way round: from the sums to the terms. If we know the sums, the terms come from a subtraction: ${a_1=S_1},$ and for ${n\ge 2}$ we have ${a_n=S_n-S_{n-1}}.$

Arithmetic Progression Calculator

Enter the first term, the common difference and the number of terms. The answer and the working appear as you type.

Solved Problems

Problem 1. Is the sequence $1,\ 11,\ 21,\ 31,\ \ldots$ an arithmetic progression?

Solution: Yes. Every term is 10 more than the previous one, so it is an arithmetic progression with ${a_1=1}$ and ${d=10}.$ Its general term is ${a_n}={1+(n-1)\cdot 10}={10n-9}.$

Problem 2. Find the sum of the first 10 terms of the progression $1,\ 11,\ 21,\ 31,\ \ldots$

Solution: Here ${a_1=1},$ ${d=10}$ and ${n=10}$:

${S_{10}}={\frac{(2\cdot 1+9\cdot 10)\cdot 10}{2}}={92\cdot 5}={460}$

Check with the other formula: ${a_{10}}={1+9\cdot 10}={91}$ and ${S_{10}}={\frac{(1+91)\cdot 10}{2}}={460}.$

Problem 3. Is 100 a term of the progression $4,\ 7,\ 10,\ 13,\ \ldots?$ What about 200?

Solution: The general term is ${a_n}={4+(n-1)\cdot 3}={3n+1}.$ We look for a natural number $n$ such that

${3n+1=100},$   ${3n=99},$   ${n=33}$

So ${a_{33}=100}.$ For 200 we get ${3n=199},$ and 199 is not divisible by 3. The number 200 is not a term of the progression.

Problem 4. Insert three numbers between 2 and 14 so that the five numbers form an arithmetic progression.

Solution: We need a progression with ${a_1=2}$ and ${a_5=14}.$ There are 4 steps from $a_1$ to $a_5$:

${d}={\frac{14-2}{4}}={3}$

The numbers are $5,\ 8,\ 11,$ and the progression is $2,\ 5,\ 8,\ 11,\ 14.$

Problem 5. Find the sum of all two-digit numbers divisible by 3.

Solution: These numbers are $12,\ 15,\ 18,\ \ldots,\ 99,$ an arithmetic progression with ${a_1=12}$ and ${d=3}.$ We find how many there are from ${a_n=99}$:

${12+(n-1)\cdot 3=99},$   ${(n-1)\cdot 3=87},$   ${n-1=29},$   ${n=30}$

${S_{30}}={\frac{(12+99)\cdot 30}{2}}={111\cdot 15}={1{,}665}$

Problem 6. How many terms of the progression $3,\ 7,\ 11,\ \ldots$ must be added to get 210?

Solution: Here ${a_1=3}$ and ${d=4}$:

${S_n}={\frac{\bigl(2\cdot 3+(n-1)\cdot 4\bigr)\,n}{2}}={\frac{(4n+2)\,n}{2}}={2n^2+n}$

We solve ${2n^2+n=210},$ that is ${2n^2+n-210=0}$:

${D}={1+4\cdot 2\cdot 210}={1681}={41^2},$   ${n}={\frac{-1+41}{4}}={10}$

The other root, ${\frac{-1-41}{4}=-10.5},$ is not a natural number. Answer: 10 terms. Check: ${a_{10}=3+9\cdot 4=39}$ and ${S_{10}}={\frac{(3+39)\cdot 10}{2}}={210}.$

Problem 7. Three numbers form an arithmetic progression. Their sum is 15 and their product is 80. Find the numbers.

Solution: It is convenient to call the middle number $x$ and the other two ${x-d}$ and ${x+d}.$ Then

${(x-d)+x+(x+d)}={3x}={15},$   so ${x=5}$

${(5-d)\cdot 5\cdot (5+d)}={80},$   ${25-d^2=16},$   ${d^2=9},$   ${d=\pm 3}$

The numbers are $2,\ 5,\ 8$ (for ${d=3}$) or $8,\ 5,\ 2$ (for ${d=-3}$).

Problem 8. The sum of the first $n$ terms of a sequence is ${S_n=n^2+2n}$ for every $n.$ Prove that the sequence is an arithmetic progression and find its general term.

Solution: ${a_1=S_1=3}.$ For ${n\ge 2}$:

${a_n}={S_n-S_{n-1}}={n^2+2n-(n-1)^2-2(n-1)}={2n+1}$

The formula ${a_n=2n+1}$ also gives ${a_1=3}$ for ${n=1}.$ Since ${a_{n+1}-a_n=2}$ for every $n,$ the sequence is an arithmetic progression with ${a_1=3}$ and ${d=2}$: $3,\ 5,\ 7,\ 9,\ \ldots$

Problem 9. A hall has 15 rows. The first row has 20 seats, and every row has 2 seats more than the row in front of it. How many seats are in the last row, and how many seats are there in all?

Solution: The numbers of seats in the rows form an arithmetic progression with ${a_1=20},$ ${d=2}$ and ${n=15}$:

${a_{15}}={20+14\cdot 2}={48},$   ${S_{15}}={\frac{(20+48)\cdot 15}{2}}={510}$

The last row has 48 seats, and the hall has 510 seats.

Problem 10. The numbers ${\frac{1}{b+c}},$ ${\frac{1}{c+a}},$ ${\frac{1}{a+b}}$ form an arithmetic progression. Prove that the numbers $a^2,$ $b^2,$ $c^2$ form an arithmetic progression too.

Solution: The middle term is the arithmetic mean of its neighbors:

${\frac{2}{c+a}}={\frac{1}{b+c}+\frac{1}{a+b}}={\frac{a+2b+c}{(b+c)(a+b)}}$

The denominators are not 0 (otherwise the fractions would not exist), so we can cross-multiply:

${2(b+c)(a+b)}={(a+2b+c)(c+a)}$

The left side is ${2ab+2b^2+2ac+2bc}.$ The right side is ${(a+c)^2+2b(a+c)}={a^2+2ac+c^2+2ab+2bc}.$ After cancelling the equal terms we are left with

${2b^2=a^2+c^2}$

So $b^2$ is the arithmetic mean of $a^2$ and $c^2,$ which means that $a^2,$ $b^2,$ $c^2$ form an arithmetic progression.

Common Mistakes

• $nd$ instead of ${(n-1)d}.$ The term ${a_{10}}$ of $2,\ 5,\ 8,\ \ldots$ is ${2+9\cdot 3=29},$ not ${2+10\cdot 3=32}.$
• Miscounting the terms. From $a_k$ to $a_m$ inclusive there are ${m-k+1}$ terms, not ${m-k}.$ For example, the numbers from 12 to 99 in steps of 3 are ${\frac{99-12}{3}+1=30}.$
• The sign of the difference. For $10,\ 7,\ 4,\ \ldots$ the common difference is ${d=7-10=-3}.$ Always subtract the previous term from the next one, not the other way round.
• A sequence in two parts. The formulas only work if the difference is the same everywhere. In $1,\ 3,\ 5,\ 8,\ 11$ the differences are 2, 2, 3, 3, so this is not an arithmetic progression.

Practice Problems

Try them yourself, then open the answer.

1. Find ${a_{15}}$ of the progression $2,\ 9,\ 16,\ \ldots$

${d=7},$ ${a_{15}}={2+14\cdot 7}={100}$

2. ${a_1=10}$ and ${d=-3}.$ Find ${a_{12}}$ and ${S_{12}}.$

${a_{12}}={10+11\cdot(-3)}={-23},$   ${S_{12}}={\frac{(10-23)\cdot 12}{2}}={-78}$

3. ${a_5=17}$ and ${a_{12}=45}.$ Find $a_1$ and $d.$

${d}={\frac{45-17}{12-5}}={4},$   ${a_1}={a_5-4d}={17-16}={1}$

4. Which term of the progression $5,\ 9,\ 13,\ \ldots$ equals 101?

${a_n}={5+(n-1)\cdot 4}={4n+1},$   ${4n+1=101},$   ${n=25}.$ It is ${a_{25}}.$

5. For which $x$ do the numbers ${x+1},$ ${3x},$ ${4x+2}$ form an arithmetic progression?

${2\cdot 3x}={(x+1)+(4x+2)},$ that is ${6x=5x+3}$ and ${x=3}.$ The numbers are $4,\ 9,\ 14$ with common difference 5.

6. Find ${S_{20}}$ if ${a_1=3}$ and ${a_{20}=60}.$

${S_{20}}={\frac{(3+60)\cdot 20}{2}}={630}$

7. Find the sum of all natural numbers from 1 to 200 that are divisible by 7.

The numbers are $7,\ 14,\ \ldots,\ 196.$ Since ${196=7\cdot 28},$ there are 28 of them. ${S_{28}}={\frac{(7+196)\cdot 28}{2}}={203\cdot 14}={2{,}842}$

8. The angles of a triangle form an arithmetic progression. Find the middle angle.

The angles are ${\alpha-d},$ $\alpha,$ ${\alpha+d}$ and their sum is ${3\alpha=180^\circ}.$ The middle angle is ${\alpha=60^\circ},$ whatever the other two are.

9. ${S_5=40}$ and ${S_{10}=155}.$ Find $a_1$ and $d.$

${S_5}={\frac{(2a_1+4d)\cdot 5}{2}}={5a_1+10d}={40},$ that is ${a_1+2d=8}.$

${S_{10}}={\frac{(2a_1+9d)\cdot 10}{2}}={10a_1+45d}={155},$ that is ${2a_1+9d=31}.$

From the first equation ${a_1=8-2d},$ then ${16-4d+9d=31},$ ${d=3}$ and ${a_1=2}.$

10. A gardener plants 12 trees on the first day and every following day 4 more than on the day before. In how many days will the gardener plant 168 trees?

${S_n}={\frac{\bigl(2\cdot 12+(n-1)\cdot 4\bigr)\,n}{2}}={2n^2+10n}.$ From ${2n^2+10n=168}$ we get ${n^2+5n-84=0}$ and ${n=7}$ (the other root is $-12$). In 7 days.

More practice, with new numbers every time:

List of arithmetic progression problems

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See also: Geometric Progression