What is an arithmetic progression?
A progression is a list of numbers that follows a rule. In an arithmetic progression (AP), the rule is: add the same number every time.
Each number in the list is a term. The first term is called a. The fixed number we keep adding is the common difference d.
d = any term − the term just before it. For 3, 7, 11, 15… d = 7 − 3 = 4. For 20, 17, 14… d = 17 − 20 = −3 (a negative d means the terms go down). For 5, 5, 5… d = 0.
To test a list: find the difference between every pair of neighbours. If all differences are equal, it is an AP. The list 1, 4, 9, 16 is not an AP (differences 3, 5, 7).
The nth term of an AP
Watch the jumps: a₁ = a, a₂ = a + d, a₃ = a + 2d, a₄ = a + 3d. The number of jumps is always one less than the position.
aₙ = a + (n − 1)d
This is also called the general term. It answers four kinds of questions: find a term, find which term a number is, find a or d from two terms, and check if a number belongs to the AP (n must come out a positive whole number).
nth term from the end
If an AP has last term l, the nth term from the end is l − (n − 1)d. Just walk backwards from the last term.
Sum of the first n terms
Write the sum forwards and backwards and add them. Each pair (first + last, second + second-last, …) gives the same total, 2a + (n − 1)d. There are n such pairs, so 2Sₙ = n[2a + (n − 1)d].
Sₙ = n/2 [2a + (n − 1)d]
If you know the last term l = a + (n − 1)d, this becomes Sₙ = n/2 (a + l).
The 3D shows this exactly: the green upside-down copy fills the staircase into a rectangle.
Useful facts
- Sum of the first n natural numbers: 1 + 2 + … + n = n(n + 1)/2.
- aₙ = Sₙ − Sₙ₋₁ (the nth term is the sum up to n minus the sum up to n − 1).
Daily-life problems on AP
In a story, look for something that goes up (or down) by the same amount each time. Then write a, d and n.
- "How much in the 10th month?" → use aₙ.
- "How much in total in 10 months?" → use Sₙ.
- "After how many months will the total reach…?" → use Sₙ and solve for n (often a quadratic equation; keep only the positive whole value).
Examples: savings plans, salaries with a fixed yearly increase, logs stacked in rows, seats in rows, fines that rise each day.
Try it: build your own staircase
Take coins or matchsticks. Make piles of 1, 3, 5, 7 and 9. Count all of them one by one. Now make the same piles again in reverse (9, 7, 5, 3, 1) and put each next to the first set. Every pair makes 10, and there are 5 pairs: 50. Half is 25. Did your one-by-one count match?
In the 3D, go to the last step. Predict first: with a = 6 and d = −2, what is the 5th term? Then move the sliders and check.
What is asked in the board exam
AP is part of the Algebra unit (about 20 marks). Expect: finding d or the nth term (1–2 marks), which term of an AP is a given number (2 marks), sum of n terms or finding n from a sum (3 marks), and a daily-life case study (4–5 marks).
Key formulas and definitions
- Common difference: d = aₙ₊₁ − aₙ
- nth term: aₙ = a + (n − 1)d
- nth term from the end: l − (n − 1)d
- Sum of n terms: Sₙ = n/2 [2a + (n − 1)d]
- Sum with last term: Sₙ = n/2 (a + l)
- aₙ = Sₙ − Sₙ₋₁; 1 + 2 + … + n = n(n + 1)/2
Worked examples
1. Is 5, 9, 13, 17, … an AP? If yes, find d.
Differences: 9 − 5 = 4, 13 − 9 = 4, 17 − 13 = 4. All equal, so it is an AP with d = 4.
2. Find the 20th term of 3, 8, 13, …
a = 3, d = 5, n = 20. a₂₀ = 3 + (20 − 1) × 5 = 3 + 95 = 98.
3. Which term of the AP 7, 11, 15, … is 87?
a = 7, d = 4. 7 + (n − 1) × 4 = 87 → (n − 1) × 4 = 80 → n − 1 = 20 → n = 21. So 87 is the 21st term.
4. The 4th term of an AP is 11 and the 9th term is 26. Find a and d.
a + 3d = 11 and a + 8d = 26. Subtract: 5d = 15, so d = 3. Then a = 11 − 9 = 2. The AP is 2, 5, 8, 11, …
5. Find the sum of the first 15 terms of 4, 7, 10, …
a = 4, d = 3, n = 15. S₁₅ = 15/2 [2 × 4 + 14 × 3] = 15/2 × [8 + 42] = 15/2 × 50 = 375.
6. Find the sum of all two-digit numbers divisible by 7.
They are 14, 21, …, 98. a = 14, d = 7, l = 98. 98 = 14 + (n − 1) × 7 → n = 13. S = 13/2 × (14 + 98) = 13/2 × 112 = 728.
7. How many terms of 24, 21, 18, … must be taken so that the sum is 78?
a = 24, d = −3. n/2 [48 + (n − 1)(−3)] = 78 → n(51 − 3n) = 156 → 3n² − 51n + 156 = 0 → n² − 17n + 52 = 0 → (n − 4)(n − 13) = 0. n = 4 or 13. Both work: the terms after the 4th add up to 0 up to the 13th (because some terms become negative).
8. Riya saves ₹100 in the first month, ₹150 in the second, ₹200 in the third, and so on. How much does she save in the 12th month, and in all 12 months?
a = 100, d = 50. a₁₂ = 100 + 11 × 50 = ₹650. S₁₂ = 12/2 × (100 + 650) = 6 × 750 = ₹4500.
Common mistakes
- Using n instead of n − 1: the 5th term is a + 4d, not a + 5d.
- Finding d the wrong way round: d = later term − earlier term. For 10, 7, 4… d = −3, not 3.
- Mixing up aₙ and Sₙ: "money in the 10th month" is a₁₀; "total money in 10 months" is S₁₀.
- Accepting n that is negative or a fraction. The number of terms must be a positive whole number.