What is an Arithmetic Progression? Class 10 Maths Guide

The def­i­n­i­tion of an arith­metic pro­gres­sion#

An arith­metic pro­gres­sion is a sequence in which each term, except the first, is obtained by adding a fixed num­ber to the term just before it. That fixed num­ber is called the com­mon dif­fer­ence and is writ­ten $d$. The first term is usu­ally writ­ten $a$.

In sym­bols, a list $a_1, a_2, a_3, \ldots$ is an AP when

$$a_{k+1} - a_k = d \quad \text{for every } k = 1, 2, 3, \ldots$$

The com­mon dif­fer­ence can be pos­i­tive, neg­a­tive or zero, and it does not have to be a whole num­ber.

• If $d \gt 0$, the terms increase: 3, 7, 11, 15, … has $d = 4$.
• If $d \lt 0$, the terms decrease: 10, 7, 4, 1, … has $d = -3$.
• If $d = 0$, every term is the same: 6, 6, 6, 6, … is an AP with $d = 0$.

The gen­eral form of an AP#

Two num­bers describe the whole list#

If you know the first term $a$ and the com­mon dif­fer­ence $d$, you can write down every term. Start at $a$ and keep adding $d$:

$$a, \quad a + d, \quad a + 2d, \quad a + 3d, \quad \ldots$$

This is called the gen­eral form of an AP. Notice the pat­tern: the sec­ond term has one $d$ added, the third term has two, the fourth has three. Each term has one fewer $d$ than its posi­tion num­ber. This obser­va­tion leads straight to the for­mula for the $n$th term, $a_n = a + (n-1)d$, which is the sub­ject of the next les­son in this chap­ter.

Why the dif­fer­ence test works#

The def­i­n­i­tion says each term is the pre­vi­ous term plus $d$. Sub­tract­ing the pre­vi­ous term from both sides gives $a_{k+1} - a_k = d$. So a list is an AP exactly when all of its con­sec­u­tive dif­fer­ences are equal. If even one dif­fer­ence is dif­fer­ent from the oth­ers, the list is not an AP. This is the whole rea­son the test below works, and it is why check­ing only the first two terms is never enough.

Finite and infi­nite APs#

A finite AP has a last term. The two-digit mul­ti­ples of 5, namely 10, 15, 20, …, 95, form a finite AP with 18 terms. An infi­nite AP goes on for ever, such as the list of odd num­bers 1, 3, 5, 7, … . We show that a list con­tin­ues by writ­ing three dots at the end. For a finite AP, the last term is often writ­ten $l$.

How to test whether a list is an AP#

Use this method every time:

1. Write down the terms in the given order.
2. Work out every con­sec­u­tive dif­fer­ence: $a_2 - a_1$, $a_3 - a_2$, $a_4 - a_3$, and so on. Always sub­tract a term from the term after it.
3. If all the dif­fer­ences are equal, the list is an AP and that com­mon value is $d$.
4. If any two dif­fer­ences are unequal, the list is not an AP. Say which dif­fer­ences dis­agree.
5. If the list is an AP and the ques­tion asks for more terms, keep adding $d$ to the last term.

Here are some quick checks.

The last two rows are impor­tant non-exam­ples. In 2, 4, 8, 16 each term is mul­ti­plied by 2, so the steps keep grow­ing. In 1, 4, 9, 16 (the square num­bers) the dif­fer­ences increase by 2 each time, so again they are not equal.

Worked exam­ples#

Exam­ple 1: writ­ing an AP from a and d#

Write the first four terms of the AP with $a = -5$ and $d = 3$.

Start at $-5$ and add 3 each time:

$$\begin{aligned} a_1 &= -5 \\ a_2 &= -5 + 3 = -2 \\ a_3 &= -2 + 3 = 1 \\ a_4 &= 1 + 3 = 4 \end{aligned}$$

Answer: $-5, -2, 1, 4$.

Exam­ple 2: a frac­tional, neg­a­tive com­mon dif­fer­ence#

Write the first four terms of the AP with $\displaystyle a = \frac{3}{4}$ and $\displaystyle d = -\frac{1}{2}$.

Adding $\displaystyle -\frac{1}{2}$ is the same as sub­tract­ing $\displaystyle \frac{1}{2} = \frac{2}{4}$:

$$\displaystyle \frac{3}{4}, \quad \frac{3}{4} - \frac{2}{4} = \frac{1}{4}, \quad \frac{1}{4} - \frac{2}{4} = -\frac{1}{4}, \quad -\frac{1}{4} - \frac{2}{4} = -\frac{3}{4}$$

Answer: $\displaystyle \frac{3}{4}, \frac{1}{4}, -\frac{1}{4}, -\frac{3}{4}$.

Exam­ple 3: test­ing a list with dec­i­mals#

Is 12, 9.5, 7, 4.5, … an AP? If so, write the next two terms.

Find the con­sec­u­tive dif­fer­ences:

$$9.5 - 12 = -2.5, \qquad 7 - 9.5 = -2.5, \qquad 4.5 - 7 = -2.5$$

All three are equal, so the list is an AP with $d = -2.5$. The next two terms are $4.5 - 2.5 = 2$ and $2 - 2.5 = -0.5$.

Answer: Yes, $d = -2.5$; the next two terms are 2 and $-0.5$.

Exam­ple 4: a list that is not an AP#

Is 1, 3, 6, 10, … an AP?

The dif­fer­ences are $3 - 1 = 2$, $6 - 3 = 3$ and $10 - 6 = 4$. Since $2 \ne 3$, the dif­fer­ences are not all equal.

Answer: No. The gaps grow by one each time, so there is no com­mon dif­fer­ence.

Exam­ple 5: find­ing an unknown so that three terms form an AP#

For what value of $k$ are $k + 1$, $2k + 1$ and $4k - 1$ con­sec­u­tive terms of an AP?

Three num­bers are in AP when the mid­dle dif­fer­ence equals the first dif­fer­ence:

$$(2k + 1) - (k + 1) = (4k - 1) - (2k + 1)$$

Sim­plify each side:

$$\begin{aligned} k &= 2k - 2 \\ k &= 2 \end{aligned}$$

Check: with $k = 2$ the terms are 3, 5, 7, whose dif­fer­ences are 2 and 2.

Answer: $k = 2$.

Why this works: the con­di­tion $b - a = c - b$ can be rearranged to $2b = a + c$, so the mid­dle term of three num­bers in AP is always the aver­age of the other two.

Exam­ple 6: an AP hid­den in surds#

Show that $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$ is an AP and write its next term.

Sim­plify each surd by tak­ing out square fac­tors:

$$\sqrt{3}, \quad \sqrt{12} = 2\sqrt{3}, \quad \sqrt{27} = 3\sqrt{3}, \quad \sqrt{48} = 4\sqrt{3}$$

Now the dif­fer­ences are all $\sqrt{3}$, so the list is an AP with $d = \sqrt{3}$. The next term is $5\sqrt{3} = \sqrt{25 \times 3} = \sqrt{75}$.

Answer: It is an AP with $d = \sqrt{3}$; the next term is $\sqrt{75}$, that is, $5\sqrt{3}$.

Com­mon mis­takes#

• Sub­tract­ing the wrong way round. The com­mon dif­fer­ence is a later term minus the ear­lier one, so for 10, 7, 4 it is $-3$, not $3$.
• Check­ing only the first two terms. A list such as 1, 3, 6, 10 starts with a dif­fer­ence of 2 but is not an AP. Check every dif­fer­ence you are given.
• Think­ing $d$ must be a pos­i­tive whole num­ber. Neg­a­tive, zero and frac­tional dif­fer­ences are all allowed.
• Com­par­ing surds with­out sim­pli­fy­ing them first. $\sqrt{12} - \sqrt{3}$ is not $\sqrt{9}$; write $\sqrt{12} = 2\sqrt{3}$ and then sub­tract.
• Con­fus­ing an AP with a list that mul­ti­plies by a fixed num­ber, such as 2, 4, 8, 16. Mul­ti­ply­ing gives grow­ing gaps, so it is not an AP.
• For­get­ting the sign when adding a neg­a­tive frac­tion, for exam­ple writ­ing $\displaystyle \frac{1}{4} - \frac{1}{2} = \frac{1}{4}$ instead of $\displaystyle -\frac{1}{4}$.

Try these#

1. Write the first four terms of the AP with $a = 7$ and $d = -4$. Answer: $7, 3, -1, -5$.
2. Is $\displaystyle -1, \frac{1}{4}, \frac{3}{2}, \frac{11}{4}, \ldots$ an AP? If so, give $d$ and the next term. Answer: Yes, $\displaystyle d = \frac{5}{4}$; the next term is 4.
3. Is 2, 6, 18, … an AP? Answer: No; the dif­fer­ences are 4 and 12.
4. Find $x$ if $x$, 11 and $2x + 4$ are con­sec­u­tive terms of an AP. Answer: $x = 6$.
5. Show that $\sqrt{5}, \sqrt{20}, \sqrt{45}, \ldots$ is an AP and write its next term. Answer: $d = \sqrt{5}$; the next term is $\sqrt{80} = 4\sqrt{5}$.

Key terms#

Sequence

A list of num­bers writ­ten in a def­i­nite order.

Term

Each num­ber in a sequence; the term in posi­tion $n$ is writ­ten $a_n$.

Arith­metic pro­gres­sion (AP)

A sequence in which each term after the first is found by adding the same fixed num­ber to the pre­vi­ous term.

First term

The start­ing num­ber of an AP, usu­ally writ­ten $a$.

Com­mon dif­fer­ence

The fixed num­ber $d$ added at every step; it equals any term minus the term before it.

Gen­eral form

The way of writ­ing an AP as $a, a + d, a + 2d, a + 3d, \ldots$

Finite AP

An AP that has a last term, often writ­ten $l$.

Infi­nite AP

An AP that con­tin­ues with­out end, shown by three dots.

Com­mon ques­tions#

Can the com­mon dif­fer­ence of an AP be zero?#

Yes. A list such as 6, 6, 6, 6 is an AP with $d = 0$, because every term equals the pre­vi­ous term plus zero.

Is a list of only two num­bers an AP?#

Any two num­bers can be treated as the first two terms of an AP, since there is only one dif­fer­ence to com­pare. Ques­tions on test­ing an AP there­fore always give at least three terms.

How do I know whether to write d as a frac­tion or a dec­i­mal?#

Match the form of the ques­tion. If the terms are frac­tions, keep $d$ as a frac­tion; if they are dec­i­mals, a dec­i­mal is fine. Both are cor­rect if they are equal in value.

Why is the mid­dle of three AP terms their aver­age?#

If $a$, $b$ and $c$ are in AP, then $b - a = c - b$, which rearranges to $\displaystyle b = \frac{a + c}{2}$. This is a quick way to find an unknown mid­dle term.

Is the list of prime num­bers an AP?#

No. The primes 2, 3, 5, 7, 11 have dif­fer­ences 1, 2, 2, 4, which are not all equal.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.
3. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.