Arithmetic Progressions: Class 10 Maths · Chapter 5

Lists that grow by equal steps: any term, and the sum of many terms, in one line

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

70 min
61 practice questions
16 MCQs

What you will learn in Arithmetic Progressions

Decide whether a list of numbers is an arithmetic progression by checking that every difference between neighbouring terms is the same, and read off its first term and common difference
Write an AP from its first term and common difference, use the general form a, a + d, a + 2d, …, and tell finite APs from infinite ones
Derive the nth term a + (n − 1)d and use it to find a term, find the position of a term, or show that a number is not a term
Find an AP from two of its terms, count the terms of a finite AP and find a term counted from the end
Derive the sum of the first n terms, n/2 [2a + (n − 1)d] = n/2 (a + l), by writing the sum forwards and backwards
Use the nth term as the difference of two sums, and explain why a question about the number of terms can have two answers
Model everyday situations such as seats in rows, savings, instalments and simple interest as APs and solve them

Original companion text aligned to NCERT Mathematics Class 10, Chapter 5, pp. 49–72. Not affiliated with or endorsed by NCERT.

5.1 Introduction

Lay out matchsticks to make squares in a row, each new square sharing a side with the one before. One square needs 4 sticks, two squares need 7 and three need 10. Every new square costs exactly 3 more sticks. Lists that change by the same amount at every step are all around us: in fares, rents, savings plans, seating plans and loan instalments.

Not every list behaves like that. Here are five lists, each with its own rule:

Monthly rent of a shop (in ₹) in years 1, 2, 3 and 4, when it starts at ₹12,000 and rises by ₹600 every year: 12000, 12600, 13200, 13800, …
Matchsticks needed for 1, 2, 3 and 4 squares in a row: 4, 7, 10, 13, …
Height (in cm) a ball reaches after each bounce, when every bounce is half as high as the one before: 160, 80, 40, 20, …
Handshakes when 2, 3, 4 and 5 friends all shake hands with one another: 1, 3, 6, 10, …
The Virahānka numbers from Class 9, where each number is the sum of the two before it: 1, 2, 3, 5, 8, …

The rent and the matchsticks grow by adding a fixed number each time. The ball heights are multiplied by $\frac{1}{2}$ each time. In the handshake list the jumps themselves grow: 2, then 3, then 4. In this chapter we study only the first kind of list. We find a quick way to get any term, a quick way to add many terms at once, and we use both on everyday problems.

3/3Doubling gives jumps of 50, 100 and 200: they keep growing, so the jumps are not equal.

Fig. 5.1Adding a fixed number makes equal jumps; doubling does not.NCERT Fig. 5.1–5.3 · p. 49

Step 1 shows four groups of matchstick squares side by side: one, two, three and four squares in a row, labelled with their stick counts 4, 7, 10 and 13. Step 2 lines up the counts 4, 7, 10, 13 and draws an arrow labelled plus 3 between each pair, so every jump is the same. Step 3 adds a second list below, 50, 100, 200, 400, where each number is double the one before; its jumps are labelled plus 50, plus 100 and plus 200, which are not equal.

Quick check

1 of 2

Which list grows by adding the same number each time?

2, 6, 18, 54, …5, 9, 13, 17, …1, 4, 9, 16, …1, 3, 6, 10, …

Show answer

Answer (B) 5, 9, 13, 17, …

In 5, 9, 13, 17 every jump is 4. The list 2, 6, 18, 54 multiplies by 3; the squares 1, 4, 9, 16 have jumps 3, 5, 7; and 1, 3, 6, 10 has jumps 2, 3, 4.

5.2 Arithmetic Progressions

Look at these five lists of numbers:

2, 5, 8, 11, …
60, 45, 30, 15, …
−7, −5, −3, −1, …
9, 9, 9, 9, …
1.5, 1.25, 1, 0.75, …

We call each number in a list a term. In every one of these lists you get the next term by adding the same number to the term before it: add 3 in (1), add −15 in (2), add 2 in (3), add 0 in (4) and add −0.25 in (5). Lists like these have a name.

Definition

Arithmetic progression (AP)

A list of numbers where you get from any term to the next by adding the same fixed number (समांतर श्रेढ़ी).

Example5, 9, 13, 17, … (add 4 each time).

Definition

Common difference

The amount by which an AP changes from each term to the next, written $d .$ An AP may go up ( $d > 0$ ), go down ( $d < 0$ ) or stay level ( $d = 0$ ).

ExampleIn 60, 45, 30, 15, … the common difference is $d = - 15.$

We name the terms $a_{1}, a_{2}, a_{3}, \dots$ The small number is the position of the term, not a multiplier: $a_{7}$ is the 7th term. In an AP every difference "later term minus earlier term" is the same, so $a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = \dots = d .$ In short:

Common difference

d = a_{k + 1} - a_{k}

(5.1)

where

$a_{k}$: the $k$ th term
$a_{k + 1}$: the term just after it

APs turn up in daily life. Each list below stops after a certain term. Can you see why each one is an AP, and what its $d$ is?

Lengths (in cm) of the bars of a toy xylophone, each 1.5 cm shorter than the one before: 24, 22.5, 21, …, 12
What is still owed (in ₹) on an interest-free loan of ₹6000, repaid at ₹500 a month: 5500, 5000, 4500, …, 0
Lowest night temperatures (in °C) in Shimla over a December week, in rising order: −2.4, −2.1, −1.8, …, −0.6
Evening readings (in litres) of a water meter for one week, when a family uses 450 litres a day: 12350, 12800, …, 15050
House numbers on the odd side of a street: 1, 3, 5, …, 59

Start at $a$ and keep adding $d$ : you get $a,$ then $a + d,$ then $a + d + d,$ and so on. This is the general form of an AP:

General form of an AP

a, a + d, a + 2 d, \dots

(5.2)

Some APs stop and some never do. The everyday lists above each end at a definite number, such as 12 or 59, so each is a finite AP with a last term. The five lists at the start of this section carry on for ever: each is an infinite AP, with no last term.

Two APs can share a first term and still be different: 4, 9, 14, … and 4, −1, −6, … both begin with 4. Two APs can also share a common difference and still differ. An AP is pinned down only when you know both $a$ and $d,$ as the table shows.

Table 5.1The same first term with different common differences gives different APs.

$a$	$d$	First four terms
$4$	$5$	$4, 9, 14, 19$
$4$	$- 5$	$4, - 1, - 6, - 11$
$- 3$	$0.5$	$- 3, - 2.5, - 2, - 1.5$
$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}, \frac{2}{3}, 1, \frac{4}{3}$
$7$	$0$	$7, 7, 7, 7$

Often the job is reversed: you are handed a list and must decide whether it is an AP. Work out every difference "later term minus earlier term". For 7, 4, 1, −2, … we get $4 - 7 = - 3,$ $1 - 4 = - 3$ and $- 2 - 1 = - 3.$ All are equal, so it is an AP with $a = 7$ and $d = - 3.$ For 1, 4, 9, 16, … the differences are 3, 5 and 7, so it is not an AP. Fig. 5.2 shows what equal and unequal jumps look like.

1/4Equal jumps up: 2, 6, 10, 14, 18 is an AP with $d = 4.$

Fig. 5.2In an AP the bar tops climb, fall or stay level by the same jump every time.

Each step shows five bars for five terms, with the jump between neighbouring bar tops labelled underneath. Step 1: bars 2, 6, 10, 14, 18 rise by equal jumps of plus 4, and a dashed straight line passes through all the tops. Step 2: bars 18, 14, 10, 6, 2 fall by equal jumps of minus 4, again on a straight line. Step 3: five bars all equal to 6, jumps of 0, on a flat line. Step 4: bars 1, 2, 4, 7, 11 with jumps plus 1, plus 2, plus 3, plus 4, drawn dashed; the tops bend upwards, so the list is marked not an AP.

Example 1NCERT Example 1 · p. 54

Reading a and d from fractions

Here is an AP made of fractions: $- \frac{5}{6}, - \frac{1}{3}, \frac{1}{6}, \frac{2}{3}, \dots$ What are $a$ and $d$ ?

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 54

AP or not?

Test each list. If it is an AP, extend it by two terms. (i) 9, 16, 23, 30, … (ii) 2, −3, −8, −13, … (iii) $\sqrt{5}, \sqrt{20}, \sqrt{45}, \sqrt{80}, \dots$ (iv) 12, 9, 6, 4, …

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

What is the common difference of the AP 3.6, 2.9, 2.2, 1.5, …?

Show answer

Answer -0.7

2.9 - 3.6 = - 0.7,

and every other pair gives

- 0.7

too. The terms fall, so

d

is negative.

Practice · 5 questions

Practice 5.1: questions with answers

1
Short answerEasyNCERT Exercise 5.1, Q1
Decide whether each situation gives an AP, and explain.
1. An auto-rickshaw charges ₹30 for the first km and ₹15 for every km after that. Look at the fare after 1, 2, 3, … km.
  Show solution
  Fares: 30, 45, 60, 75, … Each km adds ₹15, so it is an AP with $d = 15.$
2. Every hour a pump drains one-third of whatever water is in a tank. Look at the water left after each hour.
  Show solution
  If the tank starts with $V$ litres, the amounts are $V, \frac{2}{3} V, \frac{4}{9} V, \dots$ The drops $\frac{1}{3} V,$ $\frac{2}{9} V, \dots$ are not equal (each amount is multiplied by $\frac{2}{3}$ ), so it is not an AP.
3. A stadium stand has 40 seats in its front row and 4 more in each row behind. Look at the seats in each row.
  Show solution
  Seats: 40, 44, 48, … Each row adds 4, so it is an AP with $d = 4.$
4. ₹20,000 is left in a bank account that earns 6% compound interest a year. Look at the balance at the end of each year.
  Show solution
  Amounts: 20000, 21200, 22472, … The increases 1200 and 1272 are not equal, because interest is earned on the interest too. It is not an AP.
2
Short answerEasyNCERT Exercise 5.1, Q2
List four terms of each AP, starting from the given $a$ and adding the given $d$ each time.
1. $a = - 6,$ $d = 4$
  Show solution
  $- 6, - 2, 2, 6$
2. $a = 15,$ $d = - 2.5$
  Show solution
  $15, 12.5, 10, 7.5$
3. $a = \frac{1}{2},$ $d = - \frac{1}{3}$
  Show solution
  In sixths: $\frac{3}{6}, \frac{1}{6}, - \frac{1}{6}, - \frac{3}{6},$ that is $\frac{1}{2}, \frac{1}{6}, - \frac{1}{6}, - \frac{1}{2} .$
4. $a = 0,$ $d = - 1.75$
  Show solution
  $0, - 1.75, - 3.5, - 5.25$
3
Short answerEasyNCERT Exercise 5.1, Q3
Read off $a$ and $d$ for each AP.
1. 12, 5, −2, −9, … : $a =$ , $d =$
  Show answer
  Answer (1) 12, (2) -7
  $d = 5 - 12 = - 7.$
2. −8, −2, 4, 10, … : $a =$ , $d =$
  Show answer
  Answer (1) -8, (2) 6
  $d = - 2 - (- 8) = 6.$
3. $\frac{2}{5}, \frac{7}{5}, \frac{12}{5}, \frac{17}{5}, \dots$ : $d =$
  Show answer
  Answer 1
  $d = \frac{7}{5} - \frac{2}{5} = \frac{5}{5} = 1,$ and $a = \frac{2}{5} .$
4. 1.3, 2.1, 2.9, 3.7, … : $d =$
  Show answer
  Answer 0.8
  $d = 2.1 - 1.3 = 0.8,$ and $a = 1.3.$
4
Short answerMediumNCERT Exercise 5.1, Q4
Sort these lists into APs and non-APs. For each AP, state $d$ and continue it for three more terms.
1. 3, 6, 12, 24, …
  Show solution
  Differences 3, 6, 12 are not equal (each term doubles). Not an AP.
2. $\frac{9}{4}, 2, \frac{7}{4}, \frac{3}{2}, \dots$
  Show solution
  Each difference is $- \frac{1}{4} .$ AP with $d = - \frac{1}{4};$ next terms $\frac{5}{4}, 1, \frac{3}{4} .$
3. −0.5, −3, −5.5, −8, …
  Show solution
  Each difference is $- 2.5.$ AP with $d = - 2.5;$ next terms $- 10.5, - 13, - 15.5.$
4. 1, 1.1, 1.21, 1.331, …
  Show solution
  Differences 0.1, 0.11, 0.121 are not equal (each term is 1.1 times the one before). Not an AP.
5
Short answerMediumNCERT Exercise 5.1, Q4
Sort these lists into APs and non-APs. For each AP, state $d$ and continue it for three more terms.
1. $2 - \sqrt{3}, 2, 2 + \sqrt{3}, 2 + 2 \sqrt{3}, \dots$
  Show solution
  Each difference is $\sqrt{3} .$ AP with $d = \sqrt{3};$ next terms $2 + 3 \sqrt{3},$ $2 + 4 \sqrt{3},$ $2 + 5 \sqrt{3} .$
2. $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots$
  Show solution
  The differences are $- \frac{1}{6},$ $- \frac{1}{12},$ $- \frac{1}{20}$ : not equal. Not an AP.
3. $7^{2}, 13^{2}, 17^{2}, 409, \dots$
  Show solution
  The terms are 49, 169, 289, 409. Each difference is 120, so it is an AP with $d = 120;$ next terms 529, 649, 769.
4. $p, 3 p, 5 p, 7 p, \dots$
  Show solution
  Each difference is $2 p .$ AP with $d = 2 p;$ next terms $9 p, 11 p, 13 p .$

5.3 $n$ th Term of an AP

Your school is building an auditorium. Row 1 will have 18 seats, and every row will have 4 seats more than the row in front of it. How many seats will row 15 have? You could add 4 again and again, row after row, but there is a faster way.

Write each row as 18 plus some fours. Row 2 has $18 + 1 \times 4$ seats, row 3 has $18 + 2 \times 4$ and row 4 has $18 + 3 \times 4.$ The number of fours is always one less than the row number, because a 4 is added only when you move back one row. So row 15 has $18 + 14 \times 4 = 74$ seats, and row 30 would have $18 + 29 \times 4 = 134.$

3/4Row 5 is four jumps on: $18 + 4 \times 4 = 34.$

Fig. 5.3To reach the

n

th term you make

n - 1

jumps of

d

from the first term.

Six bars stand for rows 1 to 6 with 18, 22, 26, 30, 34 and 38 seats. Step 1 highlights the first bar, 18, the first term a. Step 2 marks one jump of plus 4 from row 1 to row 2, giving 18 plus 4. Step 3 marks the jumps from row 1 to row 5 and numbers them 1, 2, 3, 4: row 5 is four jumps after row 1, so it has 18 plus 4 times 4, which is 34. Step 4 states the rule: to reach row n you make n minus 1 jumps, so the nth term is a plus (n minus 1) times d.

DerivationThe

n

th term of an AP

Let an AP have first term $a$ and common difference $d .$ Every step forward adds one $d$ :

\begin{aligned} a_{1} & = a \\ a_{2} & = a + d \\ a_{3} & = a + 2 d \\ a_{4} & = a + 3 d \end{aligned}

In each line the number of $d$ s is one less than the position. To reach the $n$ th term from the first you take $n - 1$ steps, so

$n$th term of an AP

a_{n} = a + (n - 1) d

(5.4)

where

$a$: the first term
$d$: the common difference
$n$: the position of the term (1, 2, 3, …)

Because it produces every term, $a_{n}$ is known as the general term. For a finite AP the last term gets its own letter, $l$ : with $m$ terms, $l = a_{m} .$ Any one of $a_{n},$ $a,$ $d$ and $n$ can be the unknown, and the examples below solve for each in turn.

InteractiveAP explorer

An interactive explorer for arithmetic progressions. You set the first term a, the common difference d and the number of terms n with minus and plus buttons or by typing, or pick a preset such as auditorium seats 18, plus 4. In the Terms view a bar chart shows the first n terms, bars above the line for positive terms and below it for negative terms, and the last bar is outlined; a Play button adds the bars one at a time. A panel works out the nth term as a plus (n minus 1) times d with your numbers. In the Sum view a button adds a second, dashed orange piece to each bar, taken from the list written backwards, so every column reaches the same height, the first term plus the last term; the panel works out S n as n halves times (a plus l). A table lists every term with the running total, and a note points out when an earlier running total equals the last one.

Fig. 5.4Choose a first term, a common difference and a number of terms; read off the

n

th term and the sum.

Example 3NCERT Example 3 · p. 58

Finding a far-off term

What is the 16th term of 8, 14, 20, …?

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 58

Which term is it?

In the AP 64, 59, 54, …, where does −11 appear? Does 0 appear at all?

Try it yourself first. Then reveal the solution one step at a time.

Two terms of an AP are enough to find the whole AP. The gap between them is a whole number of jumps, as Fig. 5.5 shows.

3/3Three jumps back from $a_{4}$ : $a = 11 - 9 = 2.$

Fig. 5.5From the 4th term to the 9th term there are

9 - 4 = 5

jumps.

Positions 1 to 9 of an AP are drawn as boxes in a row. Step 1 fills in only box 4 with 11 and box 9 with 26; the rest are question marks. Step 2 draws five equal jump arrows from box 4 to box 9 and notes that 5 jumps cover 26 minus 11, which is 15, so each jump d is 3. Step 3 walks back three jumps from box 4 to box 1, so the first term is 11 minus 9, which is 2, and fills every box: 2, 5, 8, 11, 14, 17, 20, 23, 26.

Example 5NCERT Example 5 · p. 58

An AP from two of its terms

The 4th term of an AP is 11 and its 9th term is 26. Find the AP.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Find the 20th term of the AP 4, 9, 14, …

Show answer

Answer 99

a_{20} = 4 + 19 \times 5 = 99.

Using 20 jumps instead of 19 gives the wrong answer 104.

Example 6NCERT Example 6 · p. 59

Is it a term?

Does 254 belong to the list 8, 15, 22, 29, …?

Try it yourself first. Then reveal the solution one step at a time.

Example 7NCERT Example 7 · p. 59

Counting with the formula

How many two-digit numbers leave a remainder of 1 when divided by 7?

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 59

Counting from the end

In the AP 3, 7, 11, …, 119, which number is 8th when you count back from 119?

Try it yourself first. Then reveal the solution one step at a time.

2/3The 8th from the end is position $30 - 8 + 1 = 23.$

Fig. 5.6The

k

th term from the end of an AP with

n

terms is its

(n - k + 1)

th term.NCERT Example 8 · p. 59

Step 1 shows the AP as a row of boxes with positions 1, 2, 3 and so on up to 30 underneath, the first boxes holding 3, 7, 11 and the last holding 119. Step 2 adds a second count above the boxes, running from the right: 1st, 2nd, 3rd from the end, and the 8th from the end lands on position 23, marked with a ring. Step 3 adds an arrow from right to left: read backwards the AP starts at 119 and goes down by 4, and the ringed box holds 91.

Example 9NCERT Example 9 · p. 60

Simple interest makes an AP

Neha deposits ₹6000 at 7% a year simple interest. Show that the amounts in her account at the end of years 1, 2, 3, … form an AP, and find the amount after 15 years.

Try it yourself first. Then reveal the solution one step at a time.

Example 10NCERT Example 10 · p. 61

Rows of diyas

For Diwali, diyas are set out in rows: 41 in the first row, 37 in the second, 33 in the third, and so on, and the last row has 9. Find the number of rows.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Where does 28 appear in the AP 100, 94, 88, …?

the 11ththe 12ththe 13ththe 14th

Show answer

Answer (C) the 13th

6 (n - 1) = 72,

n - 1 = 12

and

n = 13.

The answer 12th comes from forgetting to add 1 back.

5.4 Sum of First $n$ Terms of an AP

Back to the auditorium. Suppose it has 25 rows. How many seats are there in all? The rows hold 18, 22, 26, … seats, so we need $18 + 22 + 26 + \dots$ to 25 terms. Adding them one by one is slow and easy to get wrong. We want a formula.

The pairing trick works for every AP, not just for 1, 2, 3, …. Here is why.

DerivationSum of the first

n

terms

Take any AP. Let its first term be $a,$ its common difference $d,$ and let $S_{n}$ stand for $a_{1} + a_{2} + \dots + a_{n} .$ Call the $n$ th term $l,$ so $l = a + (n - 1) d .$ Write the total forwards, then backwards (the dots stand for the terms in between):

\begin{aligned} S_{n} & = a + \dots + l \\ S_{n} & = l + \dots + a \end{aligned}

Add the two lines column by column. In each column a $+ d$ meets a $- d$ : the first column is $a + l,$ the second is $(a + d) + (l - d) = a + l,$ the third is $(a + 2 d) + (l - 2 d) = a + l,$ and so on. There are $n$ columns, all equal to $a + l$ :

2 S_{n} = n (a + l)

Divide by 2 to get (5.5); then put $l = a + (n - 1) d$ to get (5.6).

Sum when the last term is known

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

(5.5)

Sum of the first $n$ terms

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

(5.6)

where

$n$: the number of terms added
$a$: the first term
$d$: the common difference
$l$: the last ( $n$ th) term

4/4 $2 S_{5} = 5 \times 16 = 80,$ so $S_{5} = 40.$

Fig. 5.7Two copies of the sum, one reversed, fill a rectangle

n

columns wide and

a + l

high.NCERT p. 64

Step 1 shows five blue bars of heights 2, 5, 8, 11 and 14, the AP to be added. Step 2 shows the same five values in reverse order, 14, 11, 8, 5, 2, as dashed orange bars hanging from a line above. Step 3 slides the orange bars down onto the blue ones, so the columns read 2 plus 14, 5 plus 11, 8 plus 8, 11 plus 5 and 14 plus 2, every one 16 high. Step 4 outlines the rectangle, 5 columns by 16, holding 80 units: twice the sum, so the sum is 40, which is 5 over 2 times (2 plus 14).

Use (5.5) when you know the first and last terms, and (5.6) when you know $a$ and $d .$ For the auditorium $a = 18,$ $d = 4$ and $n = 25,$ so the last row has $18 + 24 \times 4 = 114$ seats and

\begin{aligned} S_{25} & = \frac{25}{2} (18 + 114) \\ = 25 \times 66 = 1650 \end{aligned}

The hall seats 1650 people. As with $a_{n},$ any one of $S_{n},$ $a,$ $d$ and $n$ can be the unknown, as the examples below show.

One more useful fact. The sum of the first $n$ terms is the sum of the first $n - 1$ terms plus one more term, the $n$ th. So you can get any term from a formula for the sums (and $a_{1} = S_{1}$ ):

$n$th term from the sums

a_{n} = S_{n} - S_{n - 1}

(5.7)

Quick check

1 of 2

Put the steps of the reverse-and-add argument in order.

Each column is $a + l$
Add the two lines column by column
Divide $n (a + l)$ by 2
Write $S_{n}$ forwards
Write $S_{n}$ backwards

Show answer

Answer

Write $S_{n}$ forwards
Write $S_{n}$ backwards
Add the two lines column by column
Each column is $a + l$
Divide $n (a + l)$ by 2

Forwards and backwards give

2 S_{n} = n (a + l),

because each of the

n

columns adds up to

a + l .

Halving gives

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

Example 11NCERT Example 11 · p. 65

A sum with a negative common difference

Add the first 18 terms of 15, 11, 7, …

Try it yourself first. Then reveal the solution one step at a time.

Example 12NCERT Example 12 · p. 65

From a sum to a term

An AP starts with 5, and its first 12 terms add up to 456. What is its 15th term?

Try it yourself first. Then reveal the solution one step at a time.

Example 13NCERT Example 13 · p. 66

Two answers for the number of terms

Starting from the first term, how many terms of 32, 28, 24, … add up to 132?

Try it yourself first. Then reveal the solution one step at a time.

2/3 $8 + 4 + 0 + (- 4) + (- 8) = 0$ : these terms cancel.

Fig. 5.8Terms 7 to 11 cancel out, so the first 6 and the first 11 terms have the same sum.NCERT Example 13 · p. 66

Eleven bars stand on a horizontal line for the terms 32, 28, 24, 20, 16, 12, 8, 4, 0, −4, −8; the last two hang below the line. Step 1 shades the first six bars and labels their total 132. Step 2 highlights bars 7 to 11, the terms 8, 4, 0, −4 and −8, and links 8 with −8 and 4 with −4, showing they cancel, with 0 in the middle. Step 3 shades all eleven bars and labels the total 132 again: S6 equals S11.

Example 14NCERT Example 14 · p. 66

Sums of counting numbers

Add (i) all the whole numbers from 1 to 250 and (ii) all the whole numbers from 1 to $n .$

Try it yourself first. Then reveal the solution one step at a time.

Example 15NCERT Example 15 · p. 67

A sum from a formula for the terms

A list is given by the rule $a_{n} = 7 - 3 n .$ Show that it is an AP and add its first 20 terms.

Try it yourself first. Then reveal the solution one step at a time.

Example 16NCERT Example 16 · p. 67

A growing school library

The number of new books a school library buys goes up by the same amount every year. It bought 340 new books in its 2nd year and 460 in its 6th year. Find (i) the books bought in the 1st year, (ii) the books bought in the 12th year and (iii) the total bought in the first 8 years.

Try it yourself first. Then reveal the solution one step at a time.

Example

Repaying a loan in rising instalments

Kiran borrows ₹12,600 interest-free from a self-help group. She repays ₹500 in the first month and pays ₹100 more each month than the month before. How many months does she take to repay the loan, and what is her last instalment?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

An AP has 19 terms. Its first term is 6 and its last term is 60. Its sum is

570

627

660

1254

Show answer

Answer (B) $627$

S_{19} = \frac{19}{2} \times 66 = 19 \times 33 = 627.

1254 forgets to halve; 570 and 660 use 19 times one end only.

1/3Practice 5.3 Q19: the radii grow 1, 2, 3, … cm.

Fig. 5.9Pictures for Practice 5.3 Q19 and Q21 and Practice 5.4 Q5.NCERT Fig. 5.4–5.8

Step 1 shows a spiral built from quarter circles. The first has radius 1 cm, and each next quarter circle turns the same way with a radius 1 cm larger, labelled 1, 2, 3 and 4. Step 2 shows a fielding drill: a set of stumps at the left and cricket balls in a straight line to the right, the first 6 m from the stumps and the rest 4 m apart, with dashed arrows for one run out to a ball and back. Step 3 shows the side view of ghat steps by a river, seven of the twelve drawn: each step is 0.5 m deep and 0.2 m higher than the one below, and each step is a solid block down to the ground.

Practice · 21 questions

Practice 5.3: questions with answers

1
Short answerEasyNCERT Exercise 5.3, Q1
Add the given number of terms of each AP.
1. 3, 9, 15, … to 12 terms
  Show solution
  $S_{12} = 6 (6 + 11 \times 6) = 6 \times 72 = 432.$
2. −40, −35, −30, … to 15 terms
  Show solution
  $S_{15} = \frac{15}{2} (- 80 + 14 \times 5) = \frac{15}{2} \times (- 10) = - 75.$
3. 0.4, 1.3, 2.2, … to 50 terms
  Show solution
  $d = 0.9.$ $S_{50} = 25 (0.8 + 49 \times 0.9) = 25 \times 44.9 = 1122.5.$
4. $\frac{3}{4}, \frac{1}{2}, \frac{1}{4}, \dots$ to 9 terms
  Show solution
  $d = - \frac{1}{4} .$ $S_{9} = \frac{9}{2} (\frac{3}{2} - 2) = \frac{9}{2} \times (- \frac{1}{2}) = - \frac{9}{4} .$
2
Short answerMediumNCERT Exercise 5.3, Q2
Find each sum.
1. $4 + 6 \frac{1}{2} + 9 + \dots + 54$
  Show solution
  $d = 2 \frac{1}{2} .$ $54 = 4 + 2.5 (n - 1)$ gives $n = 21.$ $S = \frac{21}{2} \times 58 = 609.$
2. $50 + 47 + 44 + \dots + 2$
  Show solution
  $2 = 50 - 3 (n - 1)$ gives $n = 17.$ $S = \frac{17}{2} \times 52 = 442.$
3. $- 2 + (- 6) + (- 10) + \dots + (- 118)$
  Show solution
  $- 118 = - 2 - 4 (n - 1)$ gives $n = 30.$ $S = 15 \times (- 120) = - 1800.$
3
Short answerMediumNCERT Exercise 5.3, Q3
Each part lists facts about an AP. Work out the two quantities it asks for.
1. $a = 4,$ $d = 5,$ $a_{n} = 89$ : find $n$ and $S_{n}$
  Show solution
  $89 = 4 + 5 (n - 1)$ gives $n = 18.$ $S_{18} = 9 (4 + 89) = 837.$
2. $a = 9,$ $a_{15} = 51$ : find $d$ and $S_{15}$
  Show solution
  $14 d = 42,$ so $d = 3.$ $S_{15} = \frac{15}{2} (9 + 51) = 450.$
3. $a_{10} = 47,$ $d = 4$ : find $a$ and $S_{10}$
  Show solution
  $a = 47 - 9 \times 4 = 11.$ $S_{10} = 5 (11 + 47) = 290.$
4. $a_{4} = 17,$ $S_{12} = 264$ : find $d$ and $a_{12}$
  Show solution
  $a + 3 d = 17$ and $6 (2 a + 11 d) = 264,$ so $2 a + 11 d = 44.$ Then $2 (17 - 3 d) + 11 d = 44$ gives $5 d = 10,$ $d = 2,$ $a = 11$ and $a_{12} = 11 + 22 = 33.$
4
Short answerHardNCERT Exercise 5.3, Q3
Each part lists facts about an AP. Work out the two quantities it asks for.
1. $d = 4,$ $S_{8} = 136$ : find $a$ and $a_{8}$
  Show solution
  $4 (2 a + 28) = 136$ gives $a = 3.$ $a_{8} = 3 + 28 = 31.$
2. $a = 3,$ $d = 6,$ $S_{n} = 192$ : find $n$ and $a_{n}$
  Show solution
  $\frac{n}{2} [6 + 6 (n - 1)] = 3 n^{2} = 192,$ so $n^{2} = 64$ and $n = 8.$ $a_{8} = 3 + 42 = 45.$
3. $a = 6,$ $a_{n} = 72,$ $S_{n} = 468$ : find $n$ and $d$
  Show solution
  $\frac{n}{2} (6 + 72) = 468$ gives $n = 12.$ $72 = 6 + 11 d$ gives $d = 6.$
4. $a_{n} = 3,$ $d = 2,$ $S_{n} = - 12$ : find $n$ and $a$
  Show solution
  $a = 3 - 2 (n - 1) = 5 - 2 n .$ Then $S_{n} = \frac{n}{2} (5 - 2 n + 3) = n (4 - n) = - 12,$ so $n^{2} - 4 n - 12 = 0,$ $(n - 6) (n + 2) = 0$ and $n = 6.$ So $a = - 7.$
Short answerMediumNCERT Exercise 5.3, Q4
Adding the terms of 7, 11, 15, … in order from the start, after how many terms does the total reach exactly 900?
Show solution
$900 = \frac{n}{2} [14 + 4 (n - 1)] = n (2 n + 5),$ so $2 n^{2} + 5 n - 900 = 0.$
The discriminant is $25 + 7200 = 7225 = 85^{2},$ so $n = \frac{- 5 + 85}{4} = 20$ (the other root is negative).
Short answerMediumNCERT Exercise 5.3, Q5
An AP starts at 8, ends at 68 and its terms add up to 608. How many terms does it have, and what is $d$ ?
Show solution
$608 = \frac{n}{2} (8 + 68) = 38 n,$ so $n = 16.$
$68 = 8 + 15 d$ gives $d = 4.$
Short answerMediumNCERT Exercise 5.3, Q6
An AP runs from 11 to 263 in steps of $d = 6.$ Find how many terms it has and their total.
Show solution
$263 = 11 + 6 (n - 1)$ gives $n - 1 = 42,$ so $n = 43.$
$S_{43} = \frac{43}{2} (11 + 263) = 43 \times 137 = 5891.$
Short answerEasyNCERT Exercise 5.3, Q7
An AP has $d = 5$ and its 25th term is 130. What do its first 25 terms add up to?
Show solution
$a = 130 - 24 \times 5 = 10.$ $S_{25} = \frac{25}{2} (10 + 130) = 1750.$
Short answerMediumNCERT Exercise 5.3, Q8
The 2nd and 3rd terms of an AP are 11 and 15. Find $S_{40} .$
Show solution
$d = 4$ and $a = 11 - 4 = 7.$ $S_{40} = 20 (14 + 39 \times 4) = 20 \times 170 = 3400.$
Short answerHardNCERT Exercise 5.3, Q9
An AP has $S_{6} = 48$ and $S_{12} = 168.$ Find a formula for $S_{n} .$
Show solution
$3 (2 a + 5 d) = 48$ gives $2 a + 5 d = 16;$ $6 (2 a + 11 d) = 168$ gives $2 a + 11 d = 28.$
Subtracting, $6 d = 12,$ so $d = 2$ and $a = 3.$
$S_{n} = \frac{n}{2} [6 + 2 (n - 1)] = n (n + 2) = n^{2} + 2 n .$ Check: $S_{6} = 48$ and $S_{12} = 168.$
11
Short answerMediumNCERT Exercise 5.3, Q10
Each rule below defines a list $a_{1}, a_{2}, a_{3}, \dots$ Prove that the list is an AP and add its first 12 terms.
1. $a_{n} = 5 + 3 n$
  Show solution
  $a_{n + 1} - a_{n} = 3$ for every $n,$ so it is an AP with $d = 3.$ $a_{1} = 8,$ $a_{12} = 41,$ so $S_{12} = 6 (8 + 41) = 294.$
2. $a_{n} = 11 - 4 n$
  Show solution
  $a_{n + 1} - a_{n} = - 4$ for every $n,$ so it is an AP with $d = - 4.$ $a_{1} = 7,$ $a_{12} = - 37,$ so $S_{12} = 6 (7 - 37) = - 180.$
12
Short answerHardNCERT Exercise 5.3, Q11
For an AP, $S_{n} = 3 n^{2} + 2 n$ for every $n .$ Find:
1. the first term
  Show solution
  $a_{1} = S_{1} = 3 + 2 = 5.$
2. the second term
  Show solution
  $S_{2} = 12 + 4 = 16,$ so $a_{2} = 16 - 5 = 11.$
3. the $n$ th term
  Show solution
  $a_{n} = (3 n^{2} + 2 n) - [3 (n - 1)^{2} + 2 (n - 1)] = 6 n - 1.$ (Check: $a_{1} = 5,$ $a_{2} = 11.$ )
4. the 10th term
  Show solution
  $a_{10} = 6 \times 10 - 1 = 59.$
Short answerEasyNCERT Exercise 5.3, Q12
Add up the first 30 positive multiples of 7.
Show solution
$S_{30} = \frac{30}{2} (7 + 210) = 15 \times 217 = 3255.$
Short answerEasyNCERT Exercise 5.3, Q13
What do the first 20 multiples of 9 (that is, 9, 18, 27, …) add up to?
Show solution
$S_{20} = 10 (9 + 180) = 1890.$
Short answerMediumNCERT Exercise 5.3, Q14
Add all the odd numbers from 101 to 199.
Show solution
They are 101, 103, …, 199. $199 = 101 + 2 (n - 1)$ gives $n = 50.$
$S_{50} = 25 (101 + 199) = 7500.$
Short answerEasyNCERT Exercise 5.3, Q15
Meena starts a savings plan: ₹100 in the first week, ₹120 in the second, ₹140 in the third, and so on. How much will she have saved after 26 weeks?
₹
Show solution
$a = 100,$ $d = 20,$ $n = 26.$ $S_{26} = 13 (200 + 25 \times 20) = 13 \times 700 = 9100.$ She saves ₹9100.
Short answerMediumNCERT Exercise 5.3, Q16
A quiz club has ₹1600 for 8 prizes. Each prize is ₹30 smaller than the one ranked just above it. What is each prize worth?
Show solution
$1600 = 4 [2 a + 7 \times (- 30)]$ gives $2 a - 210 = 400,$ so $a = 305.$
The prizes (in ₹) are 305, 275, 245, 215, 185, 155, 125 and 95.
Short answerMediumNCERT Exercise 5.3, Q17
In an eco-club drive, every section of each class from Class 6 to Class 12 plants as many saplings as its class number (a Class 6 section plants 6). Each class has 4 sections. How many saplings are planted?
Show solution
One section of each class plants $6 + 7 + \dots + 12 = \frac{7}{2} (6 + 12) = 63.$
With 4 sections per class: $4 \times 63 = 252$ saplings.
NumericalMediumNCERT Exercise 5.3, Q18
A spiral is drawn from 12 quarter circles (Fig. 5.9, step 1). The first has radius 1 cm and each next one has a radius 1 cm larger. Find the total length of the spiral. (Take $π = \frac{22}{7} .$ )
cm
Show solution
The lengths are $\frac{π}{2} \times 1,$ $\frac{π}{2} \times 2,$ …, $\frac{π}{2} \times 12,$ an AP.
Total $= \frac{π}{2} (1 + 2 + \dots + 12) = \frac{π}{2} \times 78 = 39 π .$
With $π = \frac{22}{7}$ : $39 \times \frac{22}{7} = \frac{858}{7} \approx 122.57 c m .$
Short answerHardNCERT Exercise 5.3, Q19
A builder stacks 150 bricks with 18 in the bottom row, 17 in the row above it, 16 in the next, and so on. How tall is the stack in rows, and what does the top row hold?
Show solution
$150 = \frac{n}{2} [36 - (n - 1)] = \frac{n (37 - n)}{2},$ so $n^{2} - 37 n + 300 = 0$ and $(n - 12) (n - 25) = 0.$
With 25 rows the top row would hold $18 - 24 = - 6$ bricks, which is impossible. So there are 12 rows and the top row has $18 - 11 = 7$ bricks.
NumericalMediumNCERT Exercise 5.3, Q20
In a fielding drill (Fig. 5.9, step 2), ten balls lie in a straight line from the stumps: the first is 6 m away and the others are 4 m apart. A fielder starts at the stumps, runs to the nearest ball, brings it back to the stumps, then fetches the next, and so on until all ten are back. How far does she run?
m
Show solution
The balls are 6, 10, 14, …, 42 m away, an AP with $a = 6,$ $d = 4,$ 10 terms. Their sum is $5 (6 + 42) = 240.$
Each trip is there and back, so she runs $2 \times 240 = 480 m .$

NCERT marks its Exercise 5.4 as optional: it is not set for the board examination. The questions below practise the same skills, for when you want a challenge.

Practice · 5 questions

Practice 5.4 (optional): questions with answers

Short answerMediumNCERT Exercise 5.4 (optional), Q1
The AP 150, 143, 136, … eventually goes below zero. Which term is the first negative one?
Show solution
$150 - 7 (n - 1) < 0$ gives $n - 1 > 21.4,$ so the smallest such $n$ is 23.
Check: $a_{22} = 150 - 147 = 3$ and $a_{23} = - 4.$
Short answerHardNCERT Exercise 5.4 (optional), Q2
In an AP, $a_{4} + a_{8} = 20$ and $a_{4} \times a_{8} = 64.$ Find $S_{12} .$
Show solution
$a_{4} + a_{8} = 2 a_{6} = 20,$ so $a_{6} = 10.$ Write $a_{4} = 10 - 2 d$ and $a_{8} = 10 + 2 d .$
$(10 - 2 d) (10 + 2 d) = 100 - 4 d^{2} = 64,$ so $d^{2} = 9$ and $d = 3$ or $d = - 3.$
Then $a = 10 - 5 d$ and $S_{12} = 6 (2 a + 11 d) = 6 (20 + d) .$ For $d = 3,$ $S_{12} = 138;$ for $d = - 3,$ $S_{12} = 102.$ Both APs fit the facts.
NumericalMediumNCERT Exercise 5.4 (optional), Q3
A wooden wall rack has shelves 20 cm apart. The shelves get shorter by the same amount each time, from 90 cm at the bottom to 30 cm at the top, and the top and bottom shelves are 1.4 m apart. What total length of wood do the shelves need?
cm
Show solution
There are $\frac{140}{20} + 1 = 8$ shelves.
Total $= \frac{8}{2} (90 + 30) = 480 c m = 4.8 m .$
ProofHardNCERT Exercise 5.4 (optional), Q4
A long street has houses numbered 1 to 288. Find the house $x$ for which all the numbers below $x$ add up to the same total as all the numbers above $x .$
Show solution
We need $\frac{(x - 1) x}{2} = \frac{288 \times 289}{2} - \frac{x (x + 1)}{2} .$
Multiply by 2 and collect: $x^{2} - x + x^{2} + x = 288 \times 289,$ so $x^{2} = 144 \times 289 = 41616$ and $x = 12 \times 17 = 204.$
Check: $1 + \dots + 203 = 20706$ and $205 + \dots + 288 = 20706.$ A puzzle like this appeared in a magazine in 1914. When P. C. Mahalanobis read it out to Srinivasa Ramanujan, he answered it for every street length at once.
NumericalHardNCERT Exercise 5.4 (optional), Q5
River ghat steps (Fig. 5.9, step 3): 12 steps, each 40 m long, rise 0.2 m and are 0.5 m deep. Each step is a solid concrete block from the ground up. Find the volume of concrete.
m³
Show solution
Step $k$ is $0.2 k \times 0.5 \times 40 = 4 k m^{3} .$
Total $= 4 (1 + 2 + \dots + 12) = 4 \times 78 = 312 m^{3} .$

5.5 Summary

3/3 $S_{6} = \frac{6}{2} (4 + 19) = 69.$

Fig. 5.10The whole chapter on one AP: equal jumps, any term, the sum.

One AP, 4, 7, 10, 13, 16, 19, is drawn as six bars. Step 1 marks the equal jumps of plus 3 between the bar tops and labels the first term a = 4 and the common difference d = 3. Step 2 highlights the sixth bar and shows that it is five jumps above the first: a6 = 4 + 5 times 3 = 19. Step 3 adds a reversed copy on top of the bars so every column reaches 4 + 19 = 23, and the sum of the six terms is 6 times 23 divided by 2, which is 69.

Recap

Arithmetic Progressions: key points and summary

In an AP you move from each term to the next by adding the same number $d$ (the common difference), which may be positive, negative or zero (Section 5.2).
To test a list, compute every gap "later minus earlier", $a_{k + 1} - a_{k}$ ; one unequal gap rules the AP out ((5.1)).
General form: $a, a + d, a + 2 d, \dots$ A finite AP has a last term $l$ ; an infinite AP does not ((5.2)).
If $a, b, c$ are in AP, $b = \frac{a + c}{2},$ the arithmetic mean of $a$ and $c$ ((5.3)).
The $n$ th term is $a_{n} = a + (n - 1) d$ : $n - 1$ jumps after the first term ((5.4)).
A number is a term only if solving $a_{n} = x$ gives a counting number $n$ (Example 4).
Two terms fix an AP: the gap between them is a whole number of jumps (Example 5).
The $k$ th term from the end of an AP with $n$ terms is its $(n - k + 1)$ th term, or $l - (k - 1) d$ (Example 8).
Writing the sum forwards and backwards gives $S_{n} = \frac{n}{2} (a + l) = \frac{n}{2} [2 a + (n - 1) d]$ (the sum derivation).
$a_{n} = S_{n} - S_{n - 1},$ and $1 + 2 + \dots + n = \frac{n (n + 1)}{2}$ ((5.7), Example 14).
A "how many terms" question leads to a quadratic in $n$ ; keep every root that is a counting number and fits the story (Example 13).

Board-exam practice

These questions follow the formats of the CBSE board paper: multiple choice, assertion–reason, short and long answers, and a case study.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 5.1, Q1	Practice 5.1 · Q1
NCERT Exercise 5.1, Q2	Practice 5.1 · Q2
NCERT Exercise 5.1, Q3	Practice 5.1 · Q3
NCERT Exercise 5.1, Q4	Practice 5.1 · Q4
NCERT Exercise 5.1, Q4	Practice 5.1 · Q5
NCERT Exercise 5.2, Q1	Practice 5.2 · Q1
NCERT Exercise 5.2, Q2	Practice 5.2 · Q2
NCERT Exercise 5.2, Q3	Practice 5.2 · Q3
NCERT Exercise 5.2, Q4	Practice 5.2 · Q4
NCERT Exercise 5.2, Q5	Practice 5.2 · Q5
NCERT Exercise 5.2, Q6	Practice 5.2 · Q6
NCERT Exercise 5.2, Q7	Practice 5.2 · Q7
NCERT Exercise 5.2, Q8	Practice 5.2 · Q8
NCERT Exercise 5.2, Q9	Practice 5.2 · Q9
NCERT Exercise 5.2, Q10	Practice 5.2 · Q10
NCERT Exercise 5.2, Q11	Practice 5.2 · Q11
NCERT Exercise 5.2, Q12	Practice 5.2 · Q12
NCERT Exercise 5.2, Q13	Practice 5.2 · Q13
NCERT Exercise 5.2, Q14	Practice 5.2 · Q14
NCERT Exercise 5.2, Q15	Practice 5.2 · Q15
NCERT Exercise 5.2, Q16	Practice 5.2 · Q16
NCERT Exercise 5.2, Q17	Practice 5.2 · Q17
NCERT Exercise 5.2, Q18	Practice 5.2 · Q18
NCERT Exercise 5.2, Q19	Practice 5.2 · Q19
NCERT Exercise 5.2, Q20	Practice 5.2 · Q20
NCERT Exercise 5.3, Q1	Practice 5.3 · Q1
NCERT Exercise 5.3, Q2	Practice 5.3 · Q2
NCERT Exercise 5.3, Q3	Practice 5.3 · Q3
NCERT Exercise 5.3, Q3	Practice 5.3 · Q4
NCERT Exercise 5.3, Q4	Practice 5.3 · Q5
NCERT Exercise 5.3, Q5	Practice 5.3 · Q6
NCERT Exercise 5.3, Q6	Practice 5.3 · Q7
NCERT Exercise 5.3, Q7	Practice 5.3 · Q8
NCERT Exercise 5.3, Q8	Practice 5.3 · Q9
NCERT Exercise 5.3, Q9	Practice 5.3 · Q10
NCERT Exercise 5.3, Q10	Practice 5.3 · Q11
NCERT Exercise 5.3, Q11	Practice 5.3 · Q12
NCERT Exercise 5.3, Q12	Practice 5.3 · Q13
NCERT Exercise 5.3, Q13	Practice 5.3 · Q14
NCERT Exercise 5.3, Q14	Practice 5.3 · Q15
NCERT Exercise 5.3, Q15	Practice 5.3 · Q16
NCERT Exercise 5.3, Q16	Practice 5.3 · Q17
NCERT Exercise 5.3, Q17	Practice 5.3 · Q18
NCERT Exercise 5.3, Q18	Practice 5.3 · Q19
NCERT Exercise 5.3, Q19	Practice 5.3 · Q20
NCERT Exercise 5.3, Q20	Practice 5.3 · Q21
NCERT Exercise 5.4 (optional), Q1	Practice 5.4 (optional) · Q1
NCERT Exercise 5.4 (optional), Q2	Practice 5.4 (optional) · Q2
NCERT Exercise 5.4 (optional), Q3	Practice 5.4 (optional) · Q3
NCERT Exercise 5.4 (optional), Q4	Practice 5.4 (optional) · Q4
NCERT Exercise 5.4 (optional), Q5	Practice 5.4 (optional) · Q5

Quick check: MCQs on Arithmetic Progressions

5.1 Introduction2 questions
5.2 Arithmetic Progressions4 questions
5.3 $n$th Term of an AP2 questions
5.3 $n$th Term of an AP3 questions
5.4 Sum of First $n$ Terms of an AP2 questions
5.4 Sum of First $n$ Terms of an AP3 questions

Glossary · 7

Key terms (English–Hindi)

arithmetic meanसमांतर माध्यsamaantar maadhya: The number halfway between two numbers $a$ and $c,$ namely $\frac{a + c}{2} .$ First seen in the chapter →
arithmetic progressionसमांतर श्रेढ़ीsamaantar shredhi: A list where every step from one term to the next adds the same number.

common differenceसार्व अंतरsaarv antar: The fixed number $d$ added at each step of an AP: any term minus the term before it.

finite APपरिमित समांतर श्रेढ़ीparimit samaantar shredhi: An AP with a last term, so it has a definite number of terms.First seen in the chapter →

general termव्यापक पदvyaapak pad: The $n$ th term $a_{n} = a + (n - 1) d,$ a formula that gives every term of the AP.First seen in the chapter →

infinite APअपरिमित समांतर श्रेढ़ीaparimit samaantar shredhi: An AP that goes on for ever, with no last term.First seen in the chapter →

termपदpad: One number in a list; $a_{n}$ is the term in position $n .$ First seen in the chapter →

Arithmetic Progressions: Class 10 Maths · Chapter 5

What you will learn in Arithmetic Progressions

5.1 Introduction

5.2 Arithmetic Progressions

Practice 5.1: questions with answers

5.3 $n$ th Term of an AP

Practice 5.2: questions with answers

5.4 Sum of First $n$ Terms of an AP

Practice 5.3: questions with answers

Practice 5.4 (optional): questions with answers

5.5 Summary

Arithmetic Progressions: key points and summary

Board-exam practice

Practice questions with answers (NCERT “Board-exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Arithmetic Progressions

Key terms (English–Hindi)

You reached the end

5.1 Introduction

5.2 Arithmetic Progressions

5.3 nth Term of an AP

5.4 Sum of First n Terms of an AP

5.5 Summary

Board-exam practice

Find your NCERT exercise question

Quick check: MCQs on Arithmetic Progressions

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5.3 $n$ th Term of an AP

5.4 Sum of First $n$ Terms of an AP