Introduction
Look at these lists of numbers:
1, 3, 5, 7, 9,... and 10, 7, 4, 1, −2,...
In the first list, each number is 2 more than the one before it. In the second list, each number is 3 less than the one before it. Whenever a list follows this kind of steady step, the same fixed number added each time, it is called an arithmetic progression. This chapter teaches you how to recognise such lists, find any term without writing out the whole list, and add up the first n terms quickly.
What is an Arithmetic Progression?
An AP is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term. This fixed number is called the common difference and is denoted by d.
a, a + d, a + 2d, a + 3d, ...
where a is the first term (often written a1).
The common difference can be positive, negative or even zero:
- if d > 0, the AP increases (for example, 2, 5, 8, 11,... where d = 3);
- if d < 0, the AP decreases (for example, 10, 7, 4, 1,... where d = −3);
- if d = 0, every term is the same (for example, 4, 4, 4, 4,...).
To check whether a given list is an AP, see if the difference between any two consecutive terms is the same throughout.
Check whether the list 3, 8, 13, 18,... is an AP. If it is, write the next two terms.
a2 − a1 = 8 − 3 = 5; a3 − a2 = 13 − 8 = 5; a4 − a3 = 18 − 13 = 5.
The difference is the same each time (d = 5), so this is an AP. The next two terms are 18 + 5 = 23 and 23 + 5 = 28.
A useful test that follows from the definition: three numbers a, b and c are in AP if and only if 2b = a + c. That is, the middle term is the average of the other two.
The nth Term of an AP
Writing out a long AP term by term is slow. The nth term formula lets you jump straight to any term.
If the first term is a and the common difference is d, then the nth term is
an = a + (n − 1)d
This is also called the general term of the AP. If the last term of a finite AP is denoted by l, then l = a + (n − 1)d.
The formula makes sense if you think about it: to get from the 1st term to the 2nd you add d once, to the 3rd you add d twice, and so to the nth you add d exactly (n − 1) times.
Find the 20th term of the AP 7, 13, 19,...
Here a = 7 and d = 13 − 7 = 6.
a20 = a + (20 − 1)d = 7 + 19 × 6 = 7 + 114 = 121
So the 20th term is 121.
Which term of the AP 3, 8, 13, 18,... is 78?
Here a = 3, d = 5, and we want an = 78.
a + (n − 1)d = 78 ⟹ 3 + (n − 1) × 5 = 78
(n − 1) × 5 = 75 ⟹ n − 1 = 15 ⟹ n = 16
So 78 is the 16th term of this AP.
These two types of question, "find an given a and d" and "find n given an", are the most common in exams. Another common type is: "How many terms are there?" which is just finding n when the last term l is known.
Sum of the First n Terms
Adding many terms of an AP one by one is tedious. There is a short formula that does it in one step. It comes from a nice trick: write the sum forwards and backwards and add.
If the first term is a, the common difference is d and the last term is l = an, then
Sn = n/2 [ 2a + (n − 1)d ]
and, when the last term l is known,
Sn = n/2 (a + l)
Both forms are the same, the second is just a shorter version of the first when you already know l.
Why does it work? Pair the first term with the last, the second with the second-last, and so on. Each pair adds to the same number (a + l), and there are n/2 such pairs. That is exactly what Sn = n/2 (a + l) says.
Note that Sn here means the sum of the first n terms. The nth term itself can also be recovered from sums: an = Sn − Sn−1 (for n > 1).
Find the sum of the first 22 terms of the AP 8, 3, −2,...
Here a = 8, d = 3 − 8 = −5, n = 22.
S22 = 22/2 [ 2(8) + (22 − 1)(−5) ] = 11 [ 16 − 105 ] = 11 × (−89) = −979
So the sum of the first 22 terms is −979.
Worked Examples
Find the sum of the first 15 terms of the AP whose nth term is an = 2 + 3n.
This is an AP because an is linear in n. We can find a1 and a2 to get a and d, or use the formula directly. First find the first and last terms:
a1 = 2 + 3(1) = 5, a15 = 2 + 3(15) = 47
Now use Sn = n/2 (a + l):
S15 = 15/2 (5 + 47) = 15/2 × 52 = 15 × 26 = 390
So the sum is 390. You can also verify that d = a2 − a1 = 8 − 5 = 3, and S15 = 15/2 [2(5) + 14(3)] = 15/2 × 52 = 390.
A sum of ₹ 1000 is invested at 8% simple interest per year. What will the amount be after 4 years?
Simple interest each year is constant: 8% of ₹ 1000 = ₹ 80. So the amounts at the end of year 1, 2, 3, 4 form an AP: 1080, 1160, 1240, 1320,... where each term is the previous plus ₹ 80.
More directly, the amount after n years is 1000 + 80n, which is an AP with a = 1080 and d = 80. After 4 years: 1000 + 80 × 4 = ₹ 1320.
This shows how APs naturally model situations with a constant yearly increase, such as simple interest or fixed savings.
Summary
- AP definition: a list where each term (after the first) is obtained by adding a fixed number d to the preceding term. General form: a, a + d, a + 2d, a + 3d,...
- Common difference: d = a2 − a1 = a3 − a2 =... It can be positive, negative or zero. A list is an AP only if this difference is constant.
- nth term (general term): an = a + (n − 1)d. Use it to find any term or to find n when an is known. Three numbers a, b, c are in AP iff 2b = a + c.
- Sum of first n terms: Sn = n/2 [2a + (n − 1)d] = n/2 (a + l), where l = an is the last term. Also, an = Sn − Sn−1 for n > 1.
- Finite vs infinite AP: a finite AP has a last term l; an infinite AP goes on forever. Both use the same formulas for an and Sn.
Practice Questions
- Check whether the list 2, 4, 8, 16,... is an AP. Give a reason.
- Find the 30th term of the AP 10, 7, 4,...
- The 5th term of an AP is 16 and the 12th term is 37. Find the first term and the common difference.
- How many terms of the AP 9, 17, 25,... must be taken to give a sum of 636?
- The sum of the first n terms of an AP is Sn = 3n2 − 4n. Find the nth term and the common difference.
- A manufacturer produces 50 articles in the first year and increases production by 10 articles every year. How many articles will be produced in the 15th year, and what is the total production in 15 years?