Arithmetic Progressions

Why This Matters

Look around and you’ll see lists of numbers that grow by the same fixed step each time. Here are three.

• Reena starts a job at ₹8000 a month. She gets a ₹500 raise every year. So her salary goes 8000, 8500, 9000, … (add 500 each year).
• The rungs of a ladder get shorter by 2 cm as you go up: 45, 43, 41, 39, … (subtract 2 each step).
• Shakila puts ₹100 in a money box on her daughter’s first birthday. Each year she adds ₹50 more than the year before: 100, 150, 200, 250, … (add 50 each year).

Of course, you could just keep adding the fixed amount again and again. But what if someone asks: what is the salary in the 25th year? Or how much money is in the box on the 21st birthday? Now you’d have to add 24 or 21 times by hand. That is slow, and it’s easy to make a mistake.

This chapter gives you two shortcuts. The first is a formula for the nth term. It lets you jump straight to any term in the list without writing out all the earlier ones. The second is a formula for the sum of the first n terms. It lets you add up a whole list in a single line. And you won’t just memorise them. You’ll see exactly why each one works — including the clever trick a 10-year-old boy named Gauss used to add 1 + 2 + ⋯ + 100 in just a few seconds.

The Big Idea

An arithmetic progression (AP) is a list of numbers where each term is the one before it plus a fixed number d. That fixed number d is called the common difference. You only need to know two things to build the whole list: the first term a and the step d. Then the list is a, a + d, a + 2d, a + 3d, … . Once you know a and d, you can find any term directly using aₙ = a + (n − 1)d. And you can add up the first n terms in one shot using Sₙ = n/2 [2a + (n − 1)d]. If you happen to know the last term l, the same sum can also be written as n/2 (a + l).

Let’s Break It Down

What an AP is, and how to spot one

Before we name the pattern, let’s nail down two words we’ll use again and again — “sequence” and “term”.

A list of numbers is an arithmetic progression if you get each term by adding the same fixed number to the one before it. The first term has nothing before it, so it just starts the list. That fixed number is the common difference. We write it as d. The value of d can be positive, negative, or zero. If d is positive, the list grows (like 2, 5, 8, …). If d is negative, the list shrinks (like 9, 7, 5, …). If d is zero, every term is the same (like 4, 4, 4, …).

We give the terms names: a₁, a₂, a₃, …, aₙ. So the AP is a₁, a₂, a₃, …, aₙ. Since each term is the one before it plus d, the gap between any two terms in a row is always d:

a₂ − a₁ = a₃ − a₂ = ⋯ = aₙ − aₙ₋₁ = d

So here is how to test a list. Take each term and subtract the term just before it. If you get the same answer every time, it’s an AP — and that answer is d. One thing to be careful about. Always do (later term) − (earlier term), in that order. We write this as aₖ₊₁ − aₖ. Do this even when the later term is smaller. In that case d simply comes out negative, which is fine. Figure 5.1 below shows this stepping idea on a number line.

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If you know a and d, the AP is fully decided. You can write out the whole list. The form a, a + d, a + 2d, a + 3d, … is called the general form of an AP. An AP that stops at a last term (like 45, 43, …, 31) is called finite. An AP that keeps going forever (like 1, 2, 3, …) is called infinite.

Let’s put the test to work on three lists — and where they pass, pin down a, d and the next two terms.

Here’s a famous pattern that fools many people. Try the subtraction test on it before you read the answer.

Deriving the nth term: aₙ = a + (n − 1)d

Say we want Reena’s salary in the 25th year, but we don’t want to write out all 25 terms. Let’s watch how the list is built. We start at the first term a, and we add d at each step:

• a₁ = a (the 1st term — we have added d zero times)
• a₂ = a + d = a + (2 − 1)d (added d once)
• a₃ = a₂ + d = a + 2d = a + (3 − 1)d (added d twice)
• a₄ = a₃ + d = a + 3d = a + (4 − 1)d (added d three times)

Now look at the pattern. To reach the 2nd term we added d once. To reach the 3rd term we added d twice. To reach the 4th term we added d three times. Each time, the number of additions is one less than the term number. So to reach the nth term, you start at a and add d a total of (n − 1) times. Why one fewer than n? Because the first term needed no addition at all — it’s just a. This gives us:

aₙ = a + (n − 1)d

This is called the nth term (or general term) of the AP. If the AP has a last term, we often call it l. So for the final term, l = a + (n − 1)d.

If the “(n − 1)” still feels strange, the climbing-the-ladder picture in Figure 5.2 below makes it click.

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Let’s take the formula for a spin: jump straight to the 10th term without writing out the first nine.

Often we flip the question around: instead of finding a term, we’re told the value and must find n — which means solving an equation. Here’s a quick refresher on that.

Now let’s use that equation-solving to answer the reverse question: given a value, which position does it sit at — and does the list ever hit exactly 0?

Sometimes you are not given a and d directly. Instead you are given two terms of the AP. In that case, write each term using the formula. This gives you two equations with two unknowns (a and d). Then solve them together as a pair, just like you did in Chapter 3.

Let’s see that play out: from just the 3rd and 7th terms, we can rebuild the entire AP.

You can also use the formula to check if a given number is in an AP at all. Just assume it is the nth term and solve for n. If n comes out as a positive whole number, the number is a term. But if n is a fraction or not a whole number, then the number is not a term of that AP.

Deriving the sum: Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)

Now comes the big reward: adding the first n terms without going through them one by one. Let’s call this sum Sₙ (or just S). Writing each term in its full form:

S = a + (a + d) + (a + 2d) + ⋯ + [a + (n − 1)d] … (1)

Here is the clever trick young Gauss used to add 1 + 2 + ⋯ + 100. Write the same sum a second time, but backwards. Start from the last term and end with a:

S = [a + (n − 1)d] + [a + (n − 2)d] + ⋯ + (a + d) + a … (2)

Now add line (1) and line (2) together, one column at a time (a column means the two terms sitting in the same spot, one from each line). Watch what happens. As you move right along line (1), each term goes up by d. But along line (2), each term goes down by d. So in every column, whatever extra d is added on top is taken away on the bottom. This means every column adds up to the same total: 2a + (n − 1)d.

Let’s check this. First column: a + [a + (n − 1)d] = 2a + (n − 1)d. Second column: (a + d) + [a + (n − 2)d] = 2a + (n − 1)d (the +d and the −d cancel). It’s the same total every time.

There are n terms, so there are n columns. Each column gives the same total. So when we add line (1) and line (2), we get:

2S = n × [2a + (n − 1)d]

(It’s 2S because we added the sum S to itself.) Now divide both sides by 2:

S = n/2 [2a + (n − 1)d]

That is the sum of the first n terms. Here’s one more nice form. Notice that 2a + (n − 1)d can be split as a + [a + (n − 1)d]. The part a + (n − 1)d is just the nth term. So 2a + (n − 1)d = a + (nth term). This lets us write S = n/2 [a + aₙ]. When the AP has a last term l (so the nth term aₙ is l), this becomes the neat form:

S = n/2 (a + l)

So which form do you use? Use n/2 (a + l) when you know the first term and the last term (but not d). Use n/2 [2a + (n − 1)d] when you know a, d and n. Figure 5.3 below lays out Gauss’s whole trick in three rows.

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Time to cash in the formula — adding 22 terms in a single line instead of adding them one by one.

Remember the boy from the start of the chapter? Let’s use the formula to crack the exact sum that made Gauss famous.

We can also run the sum formula backwards: instead of finding the sum, fix the target sum and hunt for how many terms it takes. This one has a surprise ending.

Two answers from one question feels odd, so let’s actually look at the terms. This AP is 24, 21, 18, … , and because d is negative the terms slide down past 0 into negative numbers. Figure 5.4 below draws all 13 terms as bars and shows exactly why both answers are right.

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There is also a useful link between the sum and the terms: aₙ = Sₙ − Sₙ₋₁. In words, the nth term equals the sum of the first n terms minus the sum of the first (n − 1) terms. This makes sense: if you take the first n terms and remove the first (n − 1) of them, only the nth term is left over. Figure 5.5 below makes this “remove all but the last” idea easy to see.

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This trick is handy when you are given a formula for the sum and asked to find the terms.

Putting it together: word problems

Most “real life” AP problems come down to three steps. First, spot the AP hiding in the question. Second, write down a and d (or a and l). Third, pick the right formula and use it.

Common Mistakes

Quick Check

Practice Problems

Easy

Medium

Challenge

Summary

You should now be able to explain:

• An AP is a list where each term is the one before it plus a fixed common difference d. The value of d can be positive, negative or zero. General form: a, a + d, a + 2d, … .
• To test a list, check that aₖ₊₁ − aₖ gives the same answer every time. Always subtract the earlier term from the later one.
• The nth term is aₙ = a + (n − 1)d. Reaching the nth term means adding d a total of (n − 1) times, since the first term needs no addition.
• The sum of the first n terms is Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l). We proved it by writing the sum, then writing it again backwards, then adding. Every column totals (a + l), and there are n columns, so 2S = n(a + l).
• Use n/2 (a + l) when you know the last term l. Use n/2 [2a + (n − 1)d] when you know a, d and n.
• n must be a positive whole number. If n comes out as a fraction, the value is not a term, or the situation is impossible.
• Two handy links: l = a + (n − 1)d for the last term, and aₙ = Sₙ − Sₙ₋₁.

What’s Next

So far you’ve been working with numbers in a line. Next, in Triangles, we move to shape and proportion. You’ll learn when two triangles have the same shape (we call this “similar”). You’ll see what that forces about the ratios of their sides. And you’ll meet some powerful results, like the Basic Proportionality Theorem and the rules for similarity. The same careful habit you used here — spot the pattern first, then prove it — works just as well in geometry.