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Arithmetic ProgressionsArithmetic Progressions

Reading time: ~45 min

Riya crosses the stream behind her school every morning on a row of stepping stones, and the stones sit at perfectly even gaps. Lists of numbers can be spaced out evenly in the same way. Look at these five:

(i) 1, 2 , 3 , 4 , …

(ii) 100, 70 , 40 , 10 , …

(iii) −3, −2 , −1 , 0 , …

(iv) 3, 3 , 3 , 3 , …

(v) −1.0, −1.5 , −2.0 , −2.5 , …

Each number in a list is called a term. To continue a list, you need its rule.

In (i), every term is more than the term before it.

In (ii), every term is less than the term before it.

In (iii), every term is the term before it plus 1.

In (iv), every term is 3, so each term is the one before it plus .

In (v), every term is the one before it plus −0.5 — that is, 0.5 is taken away each time.

In all five lists, each term after the first is made by adding the same fixed number to the term before it. A list like this is called an arithmetic progression, or AP for short, and the fixed number is its common difference, written d.

An arithmetic progression is a list of numbers in which every term except the first is obtained by adding a fixed number d, the common difference, to the term before it. The common difference can be positive, negative or zero.

Stepping Stones

The stones below form an AP. The green stone is the first term, 2; each blue stone is one common difference further on, and each orange hop is one d. That colour code — first term green, other terms blue, hops of d orange, sums purple — holds for the whole chapter.

Complete all three missions by changing d.

d = ${d}

So d really can be anything. A positive d makes an AP climb, a negative d makes it fall, and d = 0 gives a list that stands still — and all three are arithmetic progressions.

An AP in the Sky

In 1682 the English astronomer Edmond Halley watched a bright comet with a glowing tail cross the night sky. Searching through older records, he found bright comets that had been seen in 1531 and in 1607. The gaps between the three sightings were 1607 − 1531 = years and 1682 − 1607 = 75 years — almost the same. Halley guessed that all three were one comet, coming back about every 76 years.

If the comet came back exactly every 76 years, its visits from 1682 on would form an AP with first term 1682 and common difference 76:

1682, 1758 +76, +76, +76, +76, …

Halley died in 1742, but the comet was spotted again on Christmas night in 1758, just as he had predicted — and it has since come back in 1835, 1910 and 1986, never more than a year away from the AP. (The pull of Jupiter and Saturn nudges its orbit a little on every trip.) The comet now carries his name. By the AP its next visit is due in ; astronomers, who allow for the planets' pull, expect it in 2061.

Naming the Terms

We write the first term of an AP as a_1, the second as a_2, and so on up to the nth term a_n; the common difference is d. The AP is then a_1, a_2, a_3, \ldots, a_n, and since every gap is the same,

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

Here are some more APs from everyday life.

(a) The heights (in cm) of some students standing in a line at the morning assembly: 147, 148, 149, …, 157.

(b) The lowest temperatures (in °C) recorded on the seven days of a January week in a hill town, arranged in increasing order: −3.1, −3.0, −2.9, −2.8, −2.7, −2.6, −2.5.

(c) A loan of ₹1000 is paid back at 5% of the loan (₹50) every month. The amounts (in ₹) still owed are 950, 900, 850, 800, …, 50.

(d) A school's cash prizes (in ₹) to the toppers of Classes I to XII are 200, 250, 300, , …, 750.

(e) Saving ₹50 every month, the total savings (in ₹) after each of 10 months are 50, 100, 150, 200, 250, 300, 350, 400, 450, .

Check for yourself why each of these lists is an AP.

Any AP can be written using only its first term and its common difference:

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

Here a is the and d is the common difference. This is called the general form of an AP.

The APs (a) to (e) stop after a certain number of terms: each one has a last term. An AP like this is a finite AP. The lists (i) to (v) at the start of this section go on for ever and have no last term; they are infinite APs.

Two Numbers Fix Everything

What is the least you need to know to write down a whole AP? The first term alone is not enough, and neither is the common difference alone. You need both: the first term a and the d.

If a = 6 and d = 3, the AP is 6, 9, , 15, …

If a = 6 and d = -3, the AP is 6, 3, , −3, …

In the same way, a = -7, d = -2 gives the AP −7, −9, , −13, … and a = 1.0, d = 0.1 gives 1.0, 1.1, , 1.3, …

Likewise a = 0, d = 1\frac{1}{2} gives 0, 1\frac{1}{2}, 3, 4\frac{1}{2}, 6, \ldots, and a = 2, d = 0 gives 2, 2, 2, 2, …

Now you set the two numbers: drag a and d to build each AP.

a = ${a}    d = ${d}

The first six terms, each plotted above its position n.

However you set them, the dots of an AP come out on a straight line: each step to the right adds the same d.

Two numbers, a and d, and the whole AP follows.

Is It an AP?

Now the other way round. Given a list of numbers, how can you tell whether it is an AP, and find its a and d? The first term a is simply the . For d, remember that every term is the term before it plus d. So subtracting any term from the term right after it gives d — and for an AP, this difference must come out the same for every pair of neighbouring (consecutive) terms.

Take this list:

6, 9 +3, 12 +3, 15 +3, …

a_2 - a_1 = 9 - 6 = 3

a_3 - a_2 = 12 - 9 = 3

a_4 - a_3 = 15 - 12 = 3

Every difference is 3, so this list is an AP with a = and d = .

Now take this one:

6, 3 −3, 0 −3, −3 −3, …

a_2 - a_1 = 3 - 6 = -3

a_3 - a_2 = 0 - 3 = -3

a_4 - a_3 = -3 - 0 = -3

Again the differences agree, so this is an AP with first term 6 and d = .

In general, for an AP a_1, a_2, \ldots, a_n,

d = a_{k+1} - a_k

where a_{k+1} and a_k are the (k+1)th and kth terms. Once you know a list is an AP, one such difference is enough to find d.

But look at 1, 1, 2, 3, 5, … : its gaps 0, 1, 1, 2 are , so this list is not an AP.

Tip

Always subtract a term from the term after it, a_{k+1} - a_k. For 6, 3, 0, −3, … that means 3 - 6 = -3, not 6 - 3 = 3 — even though the later term is the smaller one. The sign of d tells you whether the AP rises or falls.

Reading a and d

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

Solution : The first term is a = , and

d = \frac{1}{2} - \frac{3}{2} =

Once we know the numbers form an AP, any two terms will give d.

Example 2 : Which of these lists of numbers form an AP? For those that do, write down 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) a_2 - a_1 = 10 - 4 = 6, a_3 - a_2 = 16 - 10 = 6 and a_4 - a_3 = 22 - 16 = 6.

The difference is the same every time, so the list is an AP with d = . Its next two terms are 22 + 6 = and 28 + 6 = .

(ii) a_2 - a_1 = -1 - 1 = -2, a_3 - a_2 = -3 - (-1) = -2 and a_4 - a_3 = -5 - (-3) = -2.

Again the difference never changes, so this is an AP with d = . Its next two terms are -5 + (-2) = and -7 + (-2) = .

(iii) a_2 - a_1 = 2 - (-2) = 4, but a_3 - a_2 = -2 - 2 = -4.

The first two differences already disagree, so this list an AP.

(iv) a_2 - a_1 = 1 - 1 = 0 and a_3 - a_2 = 1 - 1 = 0, but a_4 - a_3 = 2 - 1 = .

The differences are not all equal, so this list an AP either.

Speed Round: AP or Not?

Find the gaps between neighbouring terms in your head. If every gap is the same, pick the common difference; if not, say so. After each answer you will see the differences.

Get five right to move on.

${e}
Sina