Arithmetic Progressions
Why This Matters
Patterns that grow by a fixed step are everywhere. Reena starts a job at ₹8000 a month and gets a ₹500 raise every year — 8000, 8500, 9000, … . The rungs of a ladder shorten by 2 cm each step up — 45, 43, 41, 39, … . Shakila drops ₹100 in a money box on her daughter’s first birthday and ₹50 more each year — 100, 150, 200, 250, … .
In every one of these, you could of course just keep adding the fixed amount over and over. But what if someone asks: what’s the salary in the 25th year? Or how much money is in the box by the 21st birthday? Adding 24 or 21 times by hand is slow and error-prone.
This chapter gives you two shortcuts that turn that grind into a single calculation: a formula for the nth term (jump straight to any term without listing the earlier ones) and a formula for the sum of the first n terms (add up a whole list in one line). And you won’t just memorise them — you’ll see exactly why each one is true, including the beautiful trick a 10-year-old Gauss used to add 1 + 2 + ⋯ + 100 in seconds.
The Big Idea
An arithmetic progression (AP) is a list of numbers where each term is the previous one plus a fixed number d, called the common difference. Knowing just two things — the first term a and that step d — lets you write the whole list: a, a + d, a + 2d, a + 3d, … . From this you can pin down any term directly with aₙ = a + (n − 1)d, and add the first n of them in one shot with Sₙ = n/2 [2a + (n − 1)d], which is the same as n/2 (a + l) when you know the last term l.
Let’s Break It Down
What an AP is, and how to spot one
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, written d. It can be positive (the list grows), negative (the list shrinks), or zero (every term is the same).
We name the terms a₁, a₂, a₃, …, aₙ, so the AP is a₁, a₂, a₃, …, aₙ. Because each term is the previous one plus d:
a₂ − a₁ = a₃ − a₂ = ⋯ = aₙ − aₙ₋₁ = d
So to test a list, subtract each term from the one that follows it. If you always get the same number, it’s an AP — and that number is d. One catch: always subtract the earlier term from the later one (aₖ₊₁ − aₖ), even when the later term is smaller (then d just comes out negative).
If you know a and d, the AP is fixed — you can list it out. The form a, a + d, a + 2d, a + 3d, … is called the general form of an AP. An AP with a last term (like 45, 43, …, 31) is finite; one that keeps going (like 1, 2, 3, …) is infinite.
For each list, say whether it is an AP. If it is, give a, d and the next two terms: (i) 4, 10, 16, 22, … (ii) 1, −1, −3, −5, … (iii) −2, 2, −2, 2, …
- (i) Differences: 10 − 4 = 6, 16 − 10 = 6, 22 − 16 = 6. All equal, so it’s an AP with a = 4, d = 6. Next two: 22 + 6 = 28, then 28 + 6 = 34.
- (ii) Differences: −1 − 1 = −2, −3 − (−1) = −2, −5 − (−3) = −2. All equal, so it’s an AP with a = 1, d = −2. Next two: −5 + (−2) = −7, then −9.
- (iii) Differences: 2 − (−2) = 4, but −2 − 2 = −4. The differences are not the same, so this is not an AP. Final answers: (i) AP, a = 4, d = 6, next 28 and 34; (ii) AP, a = 1, d = −2, next −7 and −9; (iii) not an AP.
Is 1, 1, 2, 3, 5, 8, … (the Fibonacci list) an arithmetic progression?
Deriving the nth term: aₙ = a + (n − 1)d
Suppose we want Reena’s salary in the 25th year without writing out all 25. Watch the pattern build from the first term a, adding d each step:
- a₁ = a (the 1st term — we’ve 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)
The pattern is unmistakable: to reach the nth term you start at a and add d a total of (n − 1) times — one fewer than n, because the first term needed no addition. So:
aₙ = a + (n − 1)d
This is the nth term (or general term) of the AP. If the AP has a last term, it’s often called l, so l = a + (n − 1)d for the final term.
Find the 10th term of the AP 2, 7, 12, …
- Identify a and d: a = 2, and d = 7 − 2 = 5. We want n = 10.
- Apply aₙ = a + (n − 1)d: a₁₀ = 2 + (10 − 1) × 5 = 2 + 9 × 5 = 2 + 45.
- So the 10th term is 47.
Which term of the AP 21, 18, 15, … is −81? Is any term equal to 0?
- Here a = 21, d = 18 − 21 = −3. We want the n for which aₙ = −81.
- Set up aₙ = a + (n − 1)d: −81 = 21 + (n − 1)(−3) → −81 = 21 − 3(n − 1) → −81 = 24 − 3n.
- So −105 = −3n → n = 35. The 35th term is −81.
- For aₙ = 0: 21 + (n − 1)(−3) = 0 → 3(n − 1) = 21 → n − 1 = 7 → n = 8. Yes — the 8th term is 0.
When you’re given two terms instead of a and d, you get two equations — solve them as a pair (just like Chapter 3).
The 3rd term of an AP is 5 and the 7th term is 9. Find the AP.
- Write each term with the formula: a₃ = a + 2d = 5 … (1) and a₇ = a + 6d = 9 … (2).
- Subtract (1) from (2) to eliminate a: (a + 6d) − (a + 2d) = 9 − 5 → 4d = 4 → d = 1.
- Put d = 1 into (1): a + 2(1) = 5 → a = 3.
- With a = 3, d = 1 the AP is 3, 4, 5, 6, 7, …
You can also check whether a particular number even belongs to an AP: solve for n, and if n turns out not to be a positive whole number, the number is not a term.
Check whether 301 is a term of 5, 11, 17, 23, …
- It’s an AP with a = 5, d = 6. Suppose 301 is the nth term: 301 = 5 + (n − 1) × 6.
- Then 301 = 6n − 1 → 6n = 302 → n = 302/6 = 151/3.
- n must be a positive whole number, but 151/3 is not. So 301 is not a term of this AP.
Deriving the sum: Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l)
Now the big payoff: adding the first n terms without slogging through them one at a time. Let Sₙ (or just S) be that sum:
S = a + (a + d) + (a + 2d) + ⋯ + [a + (n − 1)d] … (1)
Here’s the trick young Gauss used to add 1 + 2 + ⋯ + 100. Write the same sum again, but in reverse order (the last term first, ending with a):
S = [a + (n − 1)d] + [a + (n − 2)d] + ⋯ + (a + d) + a … (2)
Now add (1) and (2) column by column. Look at any single column: the term from (1) is going up by d as you move right, while the term from (2) is going down by d — so in every column the two pieces add to the same total, 2a + (n − 1)d. (Check the first column: a + [a + (n − 1)d] = 2a + (n − 1)d. The second: (a + d) + [a + (n − 2)d] = 2a + (n − 1)d. Same every time — the extra d on one side is cancelled by the missing d on the other.)
There are n columns, so:
2S = n × [2a + (n − 1)d]
Divide both sides by 2:
S = n/2 [2a + (n − 1)d]
That’s the sum of the first n terms. Now notice that 2a + (n − 1)d = a + [a + (n − 1)d] = a + (the nth term). So we can also write S = n/2 [a + aₙ]. When the AP has a last term l (so aₙ = l), this becomes the neat form:
S = n/2 (a + l)
Use n/2 (a + l) when you know the first and last terms (and not d); use n/2 [2a + (n − 1)d] when you know a, d and n.
![The reverse-and-add proof. The sum S is written forwards as a, a+d, ..., l, then again backwards as l, l−d, ..., a. Each column adds to a+l, and with n columns this gives 2S = n(a+l), so S = n/2 (a+l) = n/2 [2a + (n−1)d].](/images/diagrams/ap-reverse-and-add.svg)
Find the sum of the first 22 terms of the AP 8, 3, −2, …
- Here a = 8, d = 3 − 8 = −5, n = 22.
- Use Sₙ = n/2 [2a + (n − 1)d]: S₂₂ = 22/2 × [2(8) + (22 − 1)(−5)] = 11 × [16 + 21(−5)].
- Inside: 16 + (−105) = −89. So S₂₂ = 11 × (−89) = −979.
Find the sum of the first n positive integers 1 + 2 + 3 + ⋯ + n.
- This is an AP with a = 1, common difference d = 1, and the last term l = n.
- The first-and-last form is cleanest: Sₙ = n/2 (a + l) = n/2 (1 + n).
- So 1 + 2 + ⋯ + n = n(n + 1)/2. (For n = 100: 100 × 101 / 2 = 5050 — exactly Gauss’s answer.)
How many terms of the AP 24, 21, 18, … must be taken so that their sum is 78?
- Here a = 24, d = 21 − 24 = −3, Sₙ = 78. Use Sₙ = n/2 [2a + (n − 1)d]: 78 = n/2 [48 + (n − 1)(−3)].
- Simplify the bracket: 48 − 3(n − 1) = 48 − 3n + 3 = 51 − 3n. So 78 = n/2 (51 − 3n) → 156 = n(51 − 3n) = 51n − 3n².
- Rearrange: 3n² − 51n + 156 = 0 → divide by 3 → n² − 17n + 52 = 0 → (n − 4)(n − 13) = 0.
- So n = 4 or n = 13 — both valid. The terms from the 5th to the 13th add to 0 (some positive, some negative cancel out), so the first 4 and the first 13 both sum to 78. Answer: 4 or 13 terms.
There’s also a handy link between the sum and the terms: the nth term equals the difference of consecutive sums, aₙ = Sₙ − Sₙ₋₁. (The first n terms minus the first (n − 1) terms leaves just the nth.) This lets you recover terms when you’re only told a formula for the sum.
Putting it together: word problems
Most “real life” AP problems boil down to: spot the AP, write down a and d (or a and l), then reach for the right formula.
A TV maker produced 600 sets in the 3rd year and 700 in the 7th year, with production rising by a fixed number each year. Find (i) the 1st-year production, (ii) the 10th-year production, (iii) the total over the first 7 years.
- Production forms an AP. Given a₃ = a + 2d = 600 … (1) and a₇ = a + 6d = 700 … (2).
- Subtract (1) from (2): 4d = 100 → d = 25. Then from (1): a + 50 = 600 → a = 550. So (i) the 1st year is 550 sets.
- (ii) 10th year: a₁₀ = a + 9d = 550 + 9(25) = 550 + 225 = 775 sets.
- (iii) First 7 years: S₇ = 7/2 [2(550) + (7 − 1)(25)] = 7/2 [1100 + 150] = 7/2 × 1250 = 4375 sets.
Common Mistakes
The nth term is a + nd — you add d once for every term.
It feels natural that 'n terms' means 'add the step n times', and a + nd reads more simply than a + (n − 1)d.
It's a + (n − 1)d. The FIRST term needs no addition (it's just a), so to reach the nth term you add d only (n − 1) times. Test it: with a = 2, d = 5, the 1st term must be 2 — and a + (1 − 1)d = 2 ✓, while a + 1·d = 7 ✗.
Common difference = (first term) − (second term), so for 6, 3, 0, … take d = 6 − 3 = 3.
Subtracting the smaller number from the bigger one to keep d positive feels tidier.
Always do d = (later term) − (earlier term) = aₖ₊₁ − aₖ. For 6, 3, 0, … that's d = 3 − 6 = −3, a NEGATIVE common difference. Flipping the subtraction would describe a different, increasing list.
If solving aₙ = (some value) gives n = 151/3 or n = 7.5, just round it to the nearest whole number.
A number that's close to a whole number looks like it 'almost' fits, so rounding seems harmless.
n counts position in the list, so it must be a POSITIVE INTEGER. A fractional n means the value is simply NOT a term of the AP — there's nothing to round. Report that it isn't a term.
To add the first n terms, you can use Sₙ = n/2 (a + l) even when you don't know the last term l.
The (a + l) form looks shorter and quicker, so it's tempting to use it everywhere.
Use n/2 (a + l) only when l (the nth/last term) is known. If you only have a, d and n, use Sₙ = n/2 [2a + (n − 1)d] — or first compute l = a + (n − 1)d, then use the (a + l) form.
Quick Check
For the AP 21, 18, 15, …, what is the common difference d?
An AP has first term a = 7 and common difference d = 3. What is its 8th term?
Which formula gives the sum of the first n terms of an AP when you know a, d and n?
Using the reverse-and-add trick, each of the n columns of (S written forwards) + (S written backwards) adds up to:
Practice Problems
Easy
Find the common difference and the next two terms of the AP −1.2, −3.2, −5.2, −7.2, …
d = −3.2 − (−1.2) = −2 (and indeed −5.2 − (−3.2) = −2). It’s an AP with d = −2.
Next term: −7.2 + (−2) = −9.2. After that: −9.2 + (−2) = −11.2.
So the next two terms are −9.2 and −11.2.
Which term of the AP 3, 8, 13, 18, … is 78?
Here a = 3, d = 8 − 3 = 5. Suppose the nth term is 78.
aₙ = a + (n − 1)d → 78 = 3 + (n − 1)×5 → 75 = 5(n − 1) → n − 1 = 15 → n = 16.
So 78 is the 16th term.
Medium
How many two-digit numbers are divisible by 3?
The two-digit multiples of 3 are 12, 15, 18, …, 99 — an AP with a = 12, d = 3, last term l = 99.
Use l = a + (n − 1)d: 99 = 12 + (n − 1)×3 → 87 = 3(n − 1) → n − 1 = 29 → n = 30.
So there are 30 two-digit numbers divisible by 3.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
a₁₇ = a + 16d and a₁₀ = a + 9d. “Exceeds by 7” means a₁₇ − a₁₀ = 7.
(a + 16d) − (a + 9d) = 7 → 7d = 7 → d = 1.
(Notice the a cancels — the gap between two terms depends only on d and how far apart they are: 16 − 9 = 7 steps of d.)
A sum of ₹1000 is invested at 8% simple interest per year. Do the yearly interest amounts form an AP? Find the interest at the end of 30 years.
Simple interest for T years = P × R × T / 100. With P = 1000, R = 8:
- Year 1: 1000 × 8 × 1 / 100 = 80
- Year 2: 1000 × 8 × 2 / 100 = 160
- Year 3: 240, and so on.
The amounts 80, 160, 240, … rise by 80 each year, so they form an AP with a = 80, d = 80.
Interest at the end of 30 years = a₃₀ = a + 29d = 80 + 29 × 80 = 80 × 30 = ₹2400.
Challenge
In a flower bed there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on, down to 5 in the last row. How many rows are there?
The counts 23, 21, 19, …, 5 form an AP with a = 23, d = 21 − 23 = −2, and last term l = 5.
Let there be n rows, so aₙ = 5: 5 = 23 + (n − 1)(−2).
→ 5 − 23 = −2(n − 1) → −18 = −2(n − 1) → n − 1 = 9 → n = 10.
So there are 10 rows in the flower bed.
If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.
Given S₁₄ = 1050, n = 14, a = 10. Use Sₙ = n/2 [2a + (n − 1)d]:
1050 = 14/2 [2(10) + (14 − 1)d] = 7 [20 + 13d] = 140 + 91d.
So 91d = 1050 − 140 = 910 → d = 10.
Then the 20th term a₂₀ = a + 19d = 10 + 19 × 10 = 10 + 190 = 200.
The sum of the first n terms of an AP is given by Sₙ = 4n − n². Find the first term, the second term, and the nth term.
First term: S₁ = 4(1) − 1² = 3, and S₁ is just the first term, so a = 3.
Sum of first two terms: S₂ = 4(2) − 2² = 8 − 4 = 4. The second term is S₂ − S₁ = 4 − 3 = 1.
So d = a₂ − a₁ = 1 − 3 = −2.
For the nth term, use aₙ = Sₙ − Sₙ₋₁: Sₙ₋₁ = 4(n − 1) − (n − 1)² = 4n − 4 − (n² − 2n + 1) = −n² + 6n − 5. aₙ = Sₙ − Sₙ₋₁ = (4n − n²) − (−n² + 6n − 5) = 4n − n² + n² − 6n + 5 = 5 − 2n.
(Check: a₁ = 5 − 2 = 3 ✓, a₂ = 5 − 4 = 1 ✓.)
Summary
You should now be able to explain:
- An AP is a list where each term is the previous one plus a fixed common difference d; d can be positive, negative or zero. General form: a, a + d, a + 2d, ….
- To test a list, check that aₖ₊₁ − aₖ is the same for every k — always subtracting the earlier term from the later one.
- The nth term is aₙ = a + (n − 1)d, because reaching the nth term means adding d a total of (n − 1) times (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), proved by writing the sum, reversing it, and 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 integer — a fractional n means the value is not a term / the situation is impossible.
- Handy links: l = a + (n − 1)d for the last term, and aₙ = Sₙ − Sₙ₋₁.
What’s Next
You’ve been working with numbers in a line. Next, in Triangles, the focus shifts to shape and proportion: when are two triangles the same shape (similar), what equal ratios of sides that forces, and the powerful results — like the Basic Proportionality Theorem and the conditions for similarity — that follow. The careful “spot the rule, then prove it” thinking you used here carries straight over into geometry.