Chapter 5 · Introduction

Chapter 5 — Arithmetic Progressions

Seats in a stadium, an auto-rickshaw fare, the money you put aside each week — whenever a quantity changes by the same amount at every step, you have an arithmetic progression, and one formula predicts any step you like.

Walk into a cricket stadium and count the seats. The front row of one stand has $20$ seats, the next row $24$, the next $28$, and so on — every row has $4$ more seats than the row in front of it. How many seats are there in the $25$th row? You could write out all $25$ numbers, but there is a quicker way. To reach the $25$th row from the first you take $24$ steps of $4$, so the $25$th row has $20+24\times4=116$ seats. And how many seats does the whole stand have? That needs one more good idea, which you will meet at the end of this chapter.

A list of numbers in which you always add the same number to get from one term to the next is called an arithmetic progression (AP for short). The fixed step is the common difference $d$. APs are everywhere: fares that rise by a fixed amount per kilometre, savings that grow by a fixed amount each month, a water tank that loses the same number of litres every hour, the dates of all the Mondays in a month. In this chapter you learn to recognise an AP, to find any term with the formula $a_n=a+(n-1)d$ — counting from the start or from the end — and to add up many terms in one go with $S_n=\tfrac n2\left[2a+(n-1)d\right]$.

Spot a fixed step→AP: $a,\ a+d,\ a+2d,\dots$→$n$th term $a_n=a+(n-1)d$→Count from the end: $l-(n-1)d$→Sum $S_n=\tfrac n2(a+l)$

🎯 By the end of this chapter you can…

decide whether a list of numbers is an arithmetic progression by checking that every difference between consecutive terms is the same, and find the common difference $d$;
write down an AP from its first term $a$ and common difference $d$, and use the fact that $a$, $b$, $c$ are in AP exactly when $2b=a+c$;
find the $n$th term $a_n=a+(n-1)d$, decide whether a given number is a term of an AP, and find which term it is;
find an AP from two of its terms, and find the $n$th term from the end of a finite AP;
find the sum of the first $n$ terms, $S_n=\tfrac n2\left[2a+(n-1)d\right]=\tfrac n2(a+l)$, and solve word problems about fares, savings, seating, instalments and stacks.

Each situation gives a list of numbers, drawn as bars. The arrow between two bars shows the step from one number to the next. Press Play to watch the bars appear one at a time, or pick a situation and predict first: are all the steps the same? If they are, the list is an arithmetic progression and you can predict any later number.

Situation:

numbers so far—

steps (differences)—

AP?—

Sections in this chapter

5.1

Arithmetic Progressions

Lists that grow or shrink by a fixed step: the common difference, finite and infinite APs, and three numbers in AP.

$d=a_{k+1}-a_k$ for every $k$

Open section →5.2

The nth Term of an AP

One formula for any term, deciding whether a number belongs to an AP, finding an AP from two terms, and counting from the end.

$a_n=a+(n-1)d$

Open section →5.3

Sum of the First n Terms of an AP

Pair the first term with the last, the second with the second-last, … and add many terms in one line.

$S_n=\tfrac n2\left[2a+(n-1)d\right]$

Open section →

✅ Before you start

You need confident arithmetic with negative numbers, fractions, decimals and simple surds such as $\sqrt8=2\sqrt2$ (Class 9 Number Systems; see also Section 1.3). You also need to solve linear equations in one unknown, such as $5+(n-1)\times3=50$, and pairs of linear equations by elimination or substitution (Section 3.2), because finding an AP from two of its terms leads to a pair. For the last section, simple quadratic equations help (Section 4.2). Warm up: what number comes next in $7, 11, 15, 19, \dots$, and in $50, 44, 38, 32, \dots$? (Answers: $23$ and $26$.)

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