Class 10 Mathematics Ch 5 Short Notes: Arithmetic Progressions

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1 Definitions to learn by heart

2 Formula sheet

3 Is it an AP? Method and example

4 The nth term: four question types

5 The sum: four question types

6 Word problems at a glance

The whole of Chapter 5 on a few pages, for last-minute revision: every definition, every formula, and one worked example for each type of question. The full notes, with every explanation and all the NCERT examples, are available separately.

1 Definitions to learn by heart

Term: each number in a list.
Arithmetic progression (AP): a list of numbers in which each term is obtained by adding a fixed number to the preceding term (the term just before it), except the first term.
Common difference $d$ : that fixed number. It can be positive, negative or zero. $d = a_{k+1} – a_k$ , where $a_k$ is any term and $a_{k+1}$ the term just after it (later term minus earlier term).
General form: $a,\ a+d,\ a+2d,\ a+3d, \ldots$ where $a$ is the first term.
Finite AP: stops after a certain number of terms, so it has a last term $l$ . Infinite AP: goes on without end and has no last term.
Arithmetic mean: if $a, b, c$ are in AP, then $b = \frac{a+c}{2}$ .

2 Formula sheet

To find	Formula	Use it when
$n$ th term (general term)	$a_n = a + (n-1)d$	a term, or which term a number is
$k$ th term from the end ( $n$ terms)	$(n – k + 1)$ th term, or $l – (k-1)d$ (step back $k-1$ times from the last term $l$ )	counting back from the last term
Sum of first $n$ terms	$S_n = \frac{n}{2}[2a + (n-1)d]$	$a$ , $d$ and $n$ are known
Sum using last term	$S_n = \frac{n}{2}(a + l)$	first and last terms known, $d$ not given
$n$ th term from sums	$a_n = S_n – S_{n-1}$ (and $a_1 = S_1$ )	$S_n$ is given as an expression in $n$
Sum of first $n$ positive integers	$\frac{n(n+1)}{2}$	1 + 2 + 3 + . . . + $n$

Three rules to remember

(1) $n$ counts positions, so it must be a positive integer: a fractional or negative $n$ means “not a term”. (2) Always subtract the earlier term from the later one when finding $d$ . (3) When the sum formula gives a quadratic equation in $n$ (one with an $n^2$ term), it can have two valid answers (e.g. $n = 4$ or $13$ for 24, 21, 18, . . . with sum 78); reject an answer only if it is not a positive integer or makes no sense in the story.

3 Is it an AP? Method and example

Find $a_2 – a_1$ , $a_3 – a_2$ , $a_4 – a_3$ . All equal: AP, and that value is $d$ . Any two different: not an AP.

Example

1, −1, −3, −5, . . .: differences −2, −2, −2, so it is an AP with $d = -2$ ; next terms −7, −9.
But −2, 2, −2, 2, . . .: differences 4, −4, so it is not an AP.
Not APs from the book: 1, 1, 2, 3, 5, 8 (from the third term on, each number is the sum of the two before it); $1^2, 2^2, 3^2, \ldots$ (squares); a savings scheme where the amount is multiplied by $\frac{5}{4}$ every 3 years.

4 The nth term: four question types

Find a term

10th term of 2, 7, 12, . . .: $a_{10} = 2 + 9 \times 5 = 47$ .

Which term is a given number (or is it a term at all)?

21, 18, 15, . . .: $-81 = 21 + (n-1)(-3) \Rightarrow n = 35$ . Is 301 a term of 5, 11, 17, . . .? $301 = 5 + (n-1) \times 6 = 6n – 1 \Rightarrow n = \frac{302}{6} = \frac{151}{3}$ , not a positive integer, so no.

Two conditions give the AP

3rd term 5, 7th term 9: $a + 2d = 5$ , $a + 6d = 9 \Rightarrow d = 1$ , $a = 3$ . AP: 3, 4, 5, 6, . . .

Term from the end

11th term from the end of 10, 7, 4, . . ., −62 (25 terms): it is the $(25 – 11 + 1) = 15$ th term, $10 + 14(-3) = -32$ . Or reverse the AP ( $a = -62$ , $d = 3$ ): $-62 + 10 \times 3 = -32$ .

5 The sum: four question types

Where the formula comes from (Gauss): write the sum forwards and backwards and add; each of the $n$ pairs gives $2a + (n-1)d$ , so $2S_n = n[2a + (n-1)d]$ , where $S_n$ is the sum of the first $n$ terms. For 1 to 100: $2S = 100 \times 101$ , $S = 5050$ .

Find the sum

First 22 terms of 8, 3, −2, . . .: $S_{22} = 11[16 + 21(-5)] = 11(-89) = -979$ .

Sum given, find a missing quantity

$S_{14} = 1050$ , $a = 10$ : $1050 = \frac{14}{2}[2 \times 10 + 13d] = 7(20 + 13d) = 140 + 91d \Rightarrow d = 10$ , so $a_{20} = 10 + 19 \times 10 = 200$ .

Sum given, find the number of terms

24, 21, 18, . . . with sum 78: $78 = \frac{n}{2}(51 – 3n) \Rightarrow n^2 – 17n + 52 = 0 \Rightarrow n = 4$ or $13$ . Both are valid, because the 5th to 13th terms add up to 0.

Sum given as an expression in n, find a term

$S_n = 4n – n^2$ (Exercise 5.3, Q11): $a_1 = S_1 = 3$ , and $a_n = S_n – S_{n-1} = (4n – n^2) – [4(n-1) – (n-1)^2] = 5 – 2n$ , so $a_{10} = 5 – 20 = -15$ .

6 Word problems at a glance

Book’s situation	AP	Result
Reena’s salary, ₹8000 + ₹500 a year	$a = 8000$ , $d = 500$	15th year ₹15000; 25th year ₹20000
₹1000 at 8% simple interest	$a = d = 80$	interest after 30 years ₹2400
Rose plants 23, 21, 19, . . ., 5	$a = 23$ , $d = -2$	10 rows
Shakila’s money box, ₹100 + ₹50 a year	$a = 100$ , $d = 50$ , $n = 21$	total ₹12600
TV sets: 600 in 3rd year, 700 in 7th year	$a = 550$ , $d = 25$	10th year 775; first 7 years 4375

Summary of the chapter

An AP is a list where you get each term after the first by adding a fixed number $d$ ; general form $a, a+d, a+2d, a+3d, \ldots$
A list is an AP if $a_2 – a_1$ , $a_3 – a_2$ , $a_4 – a_3$ , . . . all give the same value
$n$ th term: $a_n = a + (n-1)d$
Sum of first $n$ terms: $S_n = \frac{n}{2}[2a + (n-1)d]$
If $l$ is the last term of a finite AP: $S_n = \frac{n}{2}(a + l)$

More for this chapter

Notes Important Questions MCQ / Objective NCERT Solutions