Class 10 Mathematics Chapter 5 Notes: Arithmetic Progressions

Tell whether a list of numbers is an arithmetic progression (AP), and give the reason
Write the first term $a$ , the common difference $d$ and the general form of any AP
Find any term with $a_n = a + (n-1)d$ , and find which term a given number is
Add the first $n$ terms in one step with $S_n = \frac{n}{2}[2a+(n-1)d]$ or $S_n = \frac{n}{2}(a+l)$
Turn word problems on salary, savings, prizes, production and rows of objects into an AP and solve them

arithmetic progression (AP)termcommon difference
general formfinite APinfinite APnth term / general term
last term lsum of first n termsarithmetic mean

1 Patterns in Everyday Life

Many lists of numbers around us follow a pattern. The book begins with patterns in nature: the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, and the spirals on a pineapple and on a pine cone. It then gives six patterns from day-to-day life. Look at
how each next number is made, because that decides everything in this chapter.

Situation (from the book)	The list	How the next number is made	Adds a fixed number?
Reena’s monthly salary: starts at ₹8000, annual increment (yearly rise) ₹500	8000, 8500, 9000, . . .	add 500	Yes
Rungs of a ladder: bottom rung 45 cm, each rung 2 cm shorter going up (8 rungs)	45, 43, 41, . . ., 31	add −2	Yes
Savings scheme: ₹8000 invested becomes $\frac{5}{4}$ times itself every 3 years (amounts after 3, 6, 9, 12 years)	10000, 12500, 15625, 19531.25	multiply by $\frac{5}{4}$	No
Unit squares in squares of side 1, 2, 3, . . . units	$1^2, 2^2, 3^2, \ldots$	squares of consecutive (one-after-another) numbers	No
Shakila’s money box for her daughter: ₹100 at age one, ₹50 more every year	100, 150, 200, 250, . . .	add 50	Yes
Pairs of rabbits at the start of months 1 to 6	1, 1, 2, 3, 5, 8	add the two previous numbers	No

This chapter studies only the lists marked “Yes”: lists where every next term comes from
adding the same fixed number.

2 What Is an Arithmetic Progression?

Each number in a list is called a term. Look at these five lists from the book:

List	Rule	Fixed number added
1, 2, 3, 4, . . .	each term is 1 more than the one before	1
100, 70, 40, 10, . . .	each term is 30 less than the one before	−30
−3, −2, −1, 0, . . .	add 1 to the term before	1
3, 3, 3, 3, . . .	add (or subtract) 0	0
−1.0, −1.5, −2.0, −2.5, . . .	add −0.5 (that is, subtract 0.5)	−0.5

Learn by heartDefinition 1

An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a
fixed number to the preceding term, except the first term.

(Preceding term = the term just before it. The first term has nothing before it, so it is simply given.)

Learn by heartDefinition 2

The fixed number added each time is called the common difference of the AP, written $d$ .
It can be positive, negative or zero.

We write the terms as $a_1$ (first term), $a_2$ (second term), . . ., $a_n$ ( $n$ th term). Since the same
$d$ is added every time, the gap between any two neighbouring terms is always $d$ (here $a_{n-1}$ is the term just before $a_n$ ):

a_2 – a_1 = a_3 – a_2 = \ldots = a_n – a_{n-1} = d

Learn by heartDefinition 3

The general form of an AP is

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

a

= first term ·

d

= common difference

2.1 Finite and infinite APs

Finite AP

Stops after a certain number of terms, so it has an end
Has a last term
e.g. cash prizes for Classes I to XII: 200, 250, 300, . . ., 750

Infinite AP

The terms never end (the list ends with “. . .”)
Has no last term
e.g. 1, 2, 3, 4, . . .

The book’s finite APs, with their common differences:

Finite AP	List	$d$
Heights of students in a queue (cm)	147, 148, 149, . . ., 157	1
Minimum temperatures of a week in January, lowest to highest (°C)	−3.1, −3.0, −2.9, . . ., −2.5	0.1
Balance after paying 5% of a ₹1000 loan every month (₹)	950, 900, 850, . . ., 50	−50
Cash prizes to toppers of Classes I to XII (₹)	200, 250, 300, . . ., 750	50
Total savings after each month, saving ₹50 a month for 10 months (₹)	50, 100, 150, . . ., 500	50

2.2 Writing an AP from a and d

To write down an AP you need both the first term $a$ and the common difference $d$ . Knowing only one
of them is not enough. Start at $a$ and keep adding $d$ :

$a$	$d$	The AP
6	3	6, 9, 12, 15, . . .
6	−3	6, 3, 0, −3, . . .
−7	−2	−7, −9, −11, −13, . . .
1.0	0.1	1.0, 1.1, 1.2, 1.3, . . .
0	$1\frac{1}{2}$	$0,\ 1\frac{1}{2},\ 3,\ 4\frac{1}{2},\ 6, \ldots$
2	0	2, 2, 2, 2, . . .

2.3 Test for an AP: checking whether a list is an AP

Now the other way round: you are given a list and must decide whether it is an AP. The first term is
simply the first number. For $d$ , use the fact that in an AP each term minus the term just before it is always
the same:

d = a_{k+1} – a_k

Here $a_k$ is any term (the $k$ th term) and $a_{k+1}$ is the term just after it.

Step 1Work out

a_2 – a_1

a_3 – a_2

a_4 – a_3

, . . .

↓

Step 2All the same? The list is an AP and that value is

d

↓

Step 3Any two different? The list is not an AP.

Tip: once you already know a list is an AP, one difference is enough to find $d$ . A list that fails the test:
1, 1, 2, 3, 5, . . . (its differences 0, 1, 1, 2 are not all the same, so it is not an AP).

Common Mistake: order of subtraction

Always subtract the earlier term from the later term ( $a_{k+1} – a_k$ ), even when the later
term is smaller. For 6, 3, 0, −3, . . . write $d = 3 – 6 = -3$ , not $6 – 3 = 3$ .

Solved Example: finding a and d

Q. For the AP $\frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}, \ldots$ write the first term $a$ and the common difference $d$ .

Solution: $a = \frac{3}{2}$ and $d = \frac{1}{2} – \frac{3}{2} =$ $-1$ .

Solved Example: AP or not?

Q. Which of these lists form an AP? If they do, write the next two terms.
(i) 4, 10, 16, 22, . . . (ii) 1, −1, −3, −5, . . . (iii) −2, 2, −2, 2, −2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Solution:
(i) $10-4 = 6$ , $16-10 = 6$ , $22-16 = 6$ . Same every time, so it is an AP with $d = 6$ . Next two terms: $22+6 = 28$ and $28+6 = 34$ .
(ii) $-1-1 = -2$ , $-3-(-1) = -2$ , $-5-(-3) = -2$ . It is an AP with $d = -2$ . Next two terms: $-5+(-2) = -7$ and $-7+(-2) = -9$ .
(iii) $2-(-2) = 4$ but $-2-2 = -4$ . The differences are not equal, so it is not an AP.
(iv) $1-1 = 0$ , $1-1 = 0$ , but $2-1 = 1$ . Not equal, so it is not an AP.

3 The nth Term of an AP

What is Reena’s monthly salary in her 5th year? Adding ₹500 year by year works, but what about the
25th year? Watch the pattern as we go:

Figure 2: Reena's monthly salary for her first five years: each bar is ₹500 taller than the one before. — **Figure 2**: Reena’s monthly salary for her first five years: each bar is ₹500 taller than the one before.

3rd year:

8000 + 500 + 500 = 8000 + 2 \times 500

= 8000 + (3-1)\times 500 = ₹9000

4th year:

8000 + 3 \times 500

= 8000 + (4-1)\times 500 = ₹9500

5th year:

8000 + 4 \times 500

= 8000 + (5-1)\times 500 = ₹10000

15th year:

8000 + 14 \times 500

= 8000 + (15-1)\times 500 = ₹15000

25th year:

8000 + 24 \times 500

= 8000 + (25-1)\times 500 = ₹20000

So each year’s salary = first salary + (year number − 1) × annual increment. The same thing happens in
every AP with first term $a$ and common difference $d$ :

a_2 = a + d = a + (2-1)d

one

d

added

a_3 = a_2 + d = a + 2d = a + (3-1)d

two

d

‘s added

a_4 = a_3 + d = a + 3d = a + (4-1)d

three

d

‘s added

a_n = a + (n-1)d

(n-1)

d

‘s added to reach the

n

th term

Formula: nth term

a_n = a + (n-1)d

a_n

is also called the general term of the AP. If the AP has

m

terms,

a_m

is the
last term, which is sometimes written

l

What the symbols mean

a

first termany number

d

common differencepositive, negative or 0

n

position of the term (1st, 2nd, . . .)positive integer

a_n

the

n

th term (general term)any number

l

last term of a finite APany number

3.1 Finding a term

Given

AP: 2, 7, 12, . . .

To find

the 10th term

Solution

$a = 2$ , $d = 7 – 2 = 5$ , $n = 10$ . So $a_{10} = 2 + (10-1)\times 5 = 2 + 45 =$ 47.

3.2 Which term is a given number?

Put the given number in place of $a_n$ and solve for $n$ .

Solved Example: which term is −81?

Q. Which term of the AP 21, 18, 15, . . . is −81? Is any term 0? Give a reason.

Solution: $a = 21$ , $d = 18 – 21 = -3$ .
$-81 = 21 + (n-1)(-3) = 21 – 3n + 3 = 24 – 3n \Rightarrow 3n = 24 + 81 = 105 \Rightarrow n = 35$ . So the 35th term is −81.
For 0: $21 + (n-1)(-3) = 0 \Rightarrow 21 = 3(n-1) \Rightarrow n – 1 = 7 \Rightarrow n = 8$ . So yes, the 8th term is 0.

Exam Tip: n must be a positive integer

$n$ counts positions (1st, 2nd, 3rd, . . .), so it can only be a positive integer. If solving gives
a fraction, a negative number or zero, the given number is not a term of the AP. Write this reason in the
answer, because it explains why the given number is not a term of the AP.

Solved Example: is 301 a term?

Q. Check whether 301 is a term of the list 5, 11, 17, 23, . . .

Solution: The differences $11-5$ , $17-11$ , $23-17$ are all 6, so it is an AP with $a = 5$ , $d = 6$ .
If 301 is the $n$ th term: $301 = 5 + (n-1)\times 6 = 5 + 6n – 6 = 6n – 1$ , so $6n = 302$ and $n = \frac{302}{6} = \frac{151}{3}$ .
This is not a positive integer, so 301 is not a term of the list.

Solved Example: how many numbers?

Q. How many two-digit numbers are divisible by 3?

Solution: They are 12, 15, 18, . . ., 99, an AP with $a = 12$ , $d = 3$ , $a_n = 99$ .
$99 = 12 + (n-1)\times 3 \Rightarrow 87 = (n-1)\times 3 \Rightarrow n – 1 = 29 \Rightarrow n = 30$ .
There are 30 such numbers.

3.3 Finding the AP from two conditions

If two terms are given, write each one using $a_n = a + (n-1)d$ . You get two equations in $a$ and $d$ ; subtract one from the other to find $d$ , then find $a$ .

Solved Example: two terms given

Q. Determine the AP whose 3rd term is 5 and 7th term is 9.

Solution: $a_3 = a + 2d = 5$ . . . (1) and $a_7 = a + 6d = 9$ . . . (2).
Subtracting (1) from (2): $4d = 4$ , so $d = 1$ . Then from (1), $a = 5 – 2 = 3$ .
The AP is 3, 4, 5, 6, 7, . . .

3.4 The kth term from the end

In a finite AP with $n$ terms, the $k$ th term from the end is the
$(n – k + 1)$ th term from the start. The “+1” is there because the last term itself is the 1st term from the end (so the 11th term from the end of 25 terms is the 15th, not the 14th).
Another way: start at the last term $l$ and step back $(k-1)$ times, so the $k$ th term from the end $= l – (k-1)d$ .

Figure 3: In an AP of 25 terms, the 11th term from the end is the 15th term from the start. — **Figure 3**: In an AP of 25 terms, the 11th term from the end is the 15th term from the start.

Solved Example: term from the end

Q. Find the 11th term from the last term (towards the first term) of the AP 10, 7, 4, . . ., −62.

Method 1 (count the terms): $a = 10$ , $d = 7 – 10 = -3$ , $l = -62$ .
$-62 = 10 + (n-1)(-3) \Rightarrow -72 = (n-1)(-3) \Rightarrow n – 1 = 24 \Rightarrow n = 25$ .
The 11th term from the end is the $(25 – 11 + 1) =$ 15th term (a common slip is to write the 14th):
$a_{15} = 10 + (15-1)(-3) = 10 – 42 =$ −32.

Method 2 (reverse the AP): Written backwards the AP is −62, −59, −56, . . ., so now $a = -62$ and
$d = +3$ . The required term is the 11th term of this new AP: $a_{11} = -62 + (11-1)\times 3 = -62 + 30 =$ −32.

3.5 Word problems on the nth term

Solved Example: simple interest

Q. ₹1000 is invested at 8% simple interest per year. Do the interests at the end of each year form an
AP? If so, find the interest at the end of 30 years.

Solution: Simple interest $= \frac{P \times R \times T}{100}$ ( $P$ = money invested, $R$ = rate per cent per year, $T$ = time in years). After 1, 2, 3 years:
$\frac{1000 \times 8 \times 1}{100} = 80$ , then 160, then 240, . . . Each is 80 more than the last, so it is an AP with
$a = 80$ , $d = 80$ . Then $a_{30} = 80 + (30-1)\times 80 = 80 + 2320 = 2400$ . Interest after 30 years = ₹2400.

Solved Example: rows in a flower bed

Q. A flower bed has 23 rose plants in the first row, 21 in the second, 19 in the third, and so on,
with 5 plants in the last row. How many rows are there?

Solution: 23, 21, 19, . . ., 5 is an AP (each row has 2 fewer plants) with $a = 23$ , $d = -2$ , $a_n = 5$ .
$5 = 23 + (n-1)(-2) \Rightarrow -18 = (n-1)(-2) \Rightarrow n – 1 = 9 \Rightarrow n = 10$ . There are 10 rows.

4 Sum of the First n Terms of an AP

Shakila puts ₹100 in her daughter’s money box on her first birthday, ₹150 on the second, ₹200 on the third,
and so on. How much is collected by the 21st birthday? Adding 21 numbers one by one is slow. A formula does it in
one line, and the idea behind it comes from the famous mathematician Gauss.

Gauss’s trick: add the list to itself reversed

When Gauss was just 10 years old, he was asked to add the whole numbers from 1 to 100. He wrote the sum forwards and
backwards and added the two lines. Every pair adds to 101, and there are 100 pairs:

$2S = 100 \times 101 \Rightarrow S = \frac{100 \times 101}{2} = 5050$

Figure 4: Writing 1 to 100 forwards and backwards: every column adds to 101, so 2S = 100 × 101. — **Figure 4**: Writing 1 to 100 forwards and backwards: every column adds to 101, so 2S = 100 × 101.

The same trick works for any AP $a,\ a+d,\ a+2d, \ldots$ whose $n$ th term is $a+(n-1)d$ :

S = a + (a+d) + \ldots + [a+(n-1)d]

(1) the sum written forwards

S = [a+(n-1)d] + [a+(n-2)d] + \ldots + a

(2) the same sum written backwards

2S = [2a+(n-1)d] + \ldots + [2a+(n-1)d]

add (1) and (2) term by term: every pair gives

2a+(n-1)d

, e.g. 2nd pair

(a+d) + [a+(n-2)d] = 2a+(n-1)d

2S = n[2a+(n-1)d]

there are

n

such pairs

S = \frac{n}{2}[2a+(n-1)d]

divide both sides by 2

Formula: sum of the first n terms

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

S_n

= sum of the first

n

terms (for example,

S_{20}

= sum of the first 20 terms).
Since

2a + (n-1)d = a + [a+(n-1)d] = a + a_n

, this is also

S_n = \frac{n}{2}(a + a_n)

Formula: when the last term is known

S_n = \frac{n}{2}(a + l)

l

= last term (

= a_n

when the AP has

n

terms). Use this form when the first and last terms
are given but

d

is not.

Given

Shakila’s money box: $a = 100$ , $d = 50$ , $n = 21$

To find

total money on the 21st birthday, $S_{21}$

Solution

$S_{21} = \frac{21}{2}\big[2 \times 100 + (21-1)\times 50\big] = \frac{21}{2}\big[200 + 1000\big] = \frac{21}{2}\times 1200 =$ ₹12600.

The formula for the sum of the first n terms involves four quantities S, a, d and n. If we know any three of them, we can find the fourth.

NCERT textbook, Section 5.4

Remark: nth term from the sums

a_n = S_n – S_{n-1}

The

n

th term is the sum of the first

n

terms minus the sum of the first

(n-1)

terms.
Useful when the question gives

S_n

as a formula in

n

, such as

S_n = 4n – n^2

. Also

a_1 = S_1

4.1 Solved examples on the sum

Solved Example: sum of 22 terms

Q. Find the sum of the first 22 terms of the AP 8, 3, −2, . . .

Solution: $a = 8$ , $d = 3 – 8 = -5$ , $n = 22$ .
$S_{22} = \frac{22}{2}\big[16 + 21(-5)\big] = 11(16 – 105) = 11(-89) =$ −979.

Solved Example: sum given, find a term

Q. The sum of the first 14 terms of an AP is 1050 and its first term is 10. Find the 20th term.

Solution: $1050 = \frac{14}{2}\big[2 \times 10 + (14-1)d\big] = 7(20 + 13d) = 140 + 91d \Rightarrow 91d = 910 \Rightarrow d = 10$ .
So $a_{20} = 10 + (20-1)\times 10 =$ 200.

Solved Example: two answers

Q. How many terms of the AP 24, 21, 18, . . . must be taken so that their sum is 78?

Solution: $a = 24$ , $d = -3$ , $S_n = 78$ .
$78 = \frac{n}{2}\big[48 + (n-1)(-3)\big] = \frac{n}{2}(51 – 3n)$
$\Rightarrow 156 = 51n – 3n^2 \Rightarrow 3n^2 – 51n + 156 = 0 \Rightarrow n^2 – 17n + 52 = 0$ (dividing by 3) $\Rightarrow (n-4)(n-13) = 0$ , so
$n = 4$ or $n = 13$ . Both answers are correct.

Why two answers? Here $a$ is positive and $d$ is negative, so the terms turn negative after a while.
The 5th to 13th terms (12, 9, . . ., −9, −12) add up to zero, so the sum of the first 4 terms and the sum of the
first 13 terms are both 78.

Common Mistake

When the sum formula gives a quadratic equation in $n$ (one with an $n^2$ term), it can have up to two solutions (roots). Do not throw one away out of habit.
Reject a solution only if it is not a positive integer, or if it is impossible in the story (for example, an answer that would give some row a negative number of plants). Both can be correct.

Solved Example: sum of the first n positive integers

Q. Find the sum of (i) the first 1000 positive integers (ii) the first $n$ positive integers.

Solution: (i) $1 + 2 + \ldots + 1000$ has $a = 1$ , $l = 1000$ , $n = 1000$ :
$S_{1000} = \frac{1000}{2}(1 + 1000) = 500 \times 1001 =$ 500500.
(ii) Here $a = 1$ and the last term is $n$ : $S_n = \frac{n}{2}(1 + n)$ , that is,

S_n = \frac{n(n+1)}{2}

Solved Example: list given by its nth term

Q. Find the sum of the first 24 terms of the list whose $n$ th term is $a_n = 3 + 2n$ .

Solution: $a_1 = 3 + 2 = 5$ , $a_2 = 3 + 4 = 7$ , $a_3 = 3 + 6 = 9$ , . . . The list 5, 7, 9, 11, . . .
has a constant difference 2, so it is an AP with $a = 5$ , $d = 2$ .
$S_{24} = \frac{24}{2}\big[2 \times 5 + (24-1)\times 2\big] = 12(10 + 46) =$ 672.

Solved Example: production of TV sets

Q. A manufacturer produced 600 TV sets in the third year and 700 in the seventh year. Production rises
by the same fixed number every year. Find (i) the production in the 1st year (ii) the production in the 10th
year (iii) the total production in the first 7 years.

Solution: Production increases by a fixed number, so the yearly figures form an AP.
$a_3 = a + 2d = 600$ and $a_7 = a + 6d = 700$ . Subtracting, $4d = 100$ , so $d = 25$ and $a = 550$ .
(i) 1st year: 550 sets.
(ii) $a_{10} = a + 9d = 550 + 225 =$ 775 sets.
(iii) $S_7 = \frac{7}{2}\big[2 \times 550 + (7-1)\times 25\big] = \frac{7}{2}(1100 + 150) = \frac{7}{2}\times 1250 =$ 4375 sets.

5 Arithmetic Mean

If three numbers $a, b, c$ are in AP (here $a$ , $b$ , $c$ are simply three numbers in a row), the middle one is
exactly halfway between the other two, because $b – a = c – b$ gives $2b = a + c$ .

Arithmetic mean

b = \frac{a+c}{2}

a, b, c

are in AP,

b

is called the arithmetic mean of

a

and

c

Solved Example: finding k

Q. For what value of $k$ are $k + 2$ , $4k – 6$ and $3k – 2$ three consecutive terms of an AP?

Solution: The middle term is the mean of the other two: $2(4k – 6) = (k + 2) + (3k – 2)$
$\Rightarrow 8k – 12 = 4k \Rightarrow k = 3$ . The terms are 5, 6, 7, and $6 – 5 = 7 – 6 = 1$ , so the answer
$k = 3$ checks out.

6 Solving Word Problems: One Method for All

Many questions in this chapter are word problems, such as the salary, savings, flower-bed and TV-production examples. The method stays the same: say why the numbers form
an AP, write $a$ and $d$ (or equations for them), pick the right formula, and answer in the words of the question.

What the problem gives	What it asks	Formula to use
A starting value and a fixed yearly/weekly increase (salary, savings)	the value in some year, or when a value is reached	$a_n = a + (n-1)d$
Two terms (like the 3rd and 7th years)	$a$ and $d$ , then another term or a total	two equations, then $a_n$ or $S_n$
Rows of objects (such as rose plants in a flower bed)	number of rows, or objects in the last row	$a_n = a + (n-1)d$ , or $S_n$ if the total is given
A total (money collected in a money box, total production)	the sum	$S_n = \frac{n}{2}[2a+(n-1)d]$ or $\frac{n}{2}(a+l)$
A real-life list (“Is it an AP, and why?”)	yes or no, with a reason	check whether $a_{k+1} – a_k$ stays the same

✓ Do this

Write one line on why the list is an AP
Write $a$ , $d$ , $n$ clearly before using a formula
Check that $n$ is a positive integer
End with a sentence: “So there are 10 rows.”

✗ Avoid this

Calling any growing list an AP: doubling and squares are not APs
Using 14 instead of 15 for “11th from the end of 25 terms”
Rejecting a valid second answer of a quadratic equation
Mixing up $a_n$ (one term) with $S_n$ (a total)

About Exercise 5.4

Your NCERT textbook marks its last exercise, Exercise 5.4, as Optional, with the footnote “These exercises are not from the examination
point of view.” Its questions use exactly the formulas above, so they are good extra practice.

Summary and quick revision: read this the night before the exam

An AP is a list where each term after the first is the previous term plus a fixed number $d$ (the common difference); $d$ can be positive, negative or zero
General form: $a,\ a+d,\ a+2d,\ a+3d, \ldots$ ; a finite AP has a last term $l$ , an infinite AP does not
A list is an AP if $a_{k+1} – a_k$ is the same for every $k$ ; always subtract the earlier term from the later one
$n$ th term: $a_n = a + (n-1)d$ , and $n$ must be a positive integer
$k$ th term from the end of an AP with $n$ terms = $(n – k + 1)$ th term, or reverse the AP
Sum: $S_n = \frac{n}{2}[2a+(n-1)d] = \frac{n}{2}(a+l)$ ; first $n$ positive integers: $\frac{n(n+1)}{2}$
$a_n = S_n – S_{n-1}$ ; a quadratic equation in $n$ can have two valid answers
If $a, b, c$ are in AP, then $b = \frac{a+c}{2}$ (arithmetic mean)

More for this chapter

Important Questions MCQ / Objective NCERT Solutions Short Notes