Arithmetic Progressions (AP)

An arithmetic progression (AP) is a sequence where the difference between consecutive terms is constant. That difference is called the common difference (d).

1. Definition of Arithmetic Progression

A sequence a₁, a₂, a₃, … is an AP if

a₂ − a₁ = a₃ − a₂ = ⋯ = d

Here a₁ = first term, d = common difference, and a_n = nth term.

Example 1:

Sequence: 2, 5, 8, 11, … → a₁ = 2, d = 5 − 2 = 3

Example 2:

Sequence: 15, 12, 9, 6, … → a₁ = 15, d = 12 − 15 = −3

2. General Form of an AP

A general AP looks like:

a₁, a₁ + d, a₁ + 2d, a₁ + 3d, …, a₁ + (n − 1)d

So the nth term a_n is given by:

a_n = a₁ + (n − 1)d

Example 3:

Find the 10th term of AP: 3, 7, 11, …

Here a₁ = 3, d = 4 → a₁₀ = 3 + (10 − 1)·4 = 3 + 36 = 39

3. Sum of n Terms of an AP

Sum of the first n terms (S_n):

S_n = n/2 [2a₁ + (n − 1)d]

Or using the last term a_n:

S_n = n/2 [a₁ + a_n]

Example 4:

Find the sum of the first 20 terms of AP: 5, 8, 11, …

a₁ = 5, d = 3, n = 20 → a₂₀ = 5 + (20 − 1)·3 = 62

S₂₀ = 20/2 · (5 + 62) = 10 × 67 = 670

4. Finding Number of Terms

If the last term a_n is known, find n using:

n = (a_n − a₁)/d + 1

Example 5:

AP: 7, 10, 13, …, 52 → a₁ = 7, a_n = 52, d = 3

n = (52 − 7)/3 + 1 = 45/3 + 1 = 15 + 1 = 16

5. Finding Common Difference

If the first and nth terms are known:

d = (a_n − a₁) / (n − 1)

Example 6:

AP: 12, …, 48 (9th term = 48)

d = (48 − 12) / (9 − 1) = 36/8 = 4.5

6. Solving Word Problems Using AP

Example 7 — Sum Problem:

Sum of first n natural numbers = 5050 → Find n?

Natural numbers: 1, 2, 3, … (an AP with a₁ = 1, d = 1)

Use S_n = n/2 [2a₁ + (n − 1)d] → 5050 = n/2 [2(1) + (n − 1)·1] = n/2 (n + 1)

So n(n + 1) / 2 = 5050 → n(n + 1) = 10100 → n = 100

Example 8 — Salary Problem:

Salary of Rs 1000 with increment of Rs 50 per month → Total in 12 months?

a₁ = 1000, d = 50, n = 12

S₁₂ = 12/2 [2·1000 + (12 − 1)·50] = 6 [2000 + 550] = 6 · 2550 = 15,300

7. Summary Table

Concept	Formula	Example (brief)
nth term	an = a1 + (n − 1)d	3, 7, 11… → a10 = 39
Sum of n terms	Sn = n/2 [2a1 + (n − 1)d]	S20 = 670
Sum using last term	Sn = n/2 [a1 + an]	S20 = 670
Number of terms	n = (an − a1)/d + 1	16 terms (7…52, d=3)
Common difference	d = (an − a1)/(n − 1)	d = 4.5 (12…48, n=9)

8. Exercises for Practice

1. Find the 15th term of the AP: 2, 5, 8, …
2. Find the sum of the first 30 terms of the AP: 7, 11, 15, …
3. How many terms are there in the AP: 5, 9, 13, …, 101?
4. The sum of the first 20 terms of an AP is 610. The first term is 7. Find the common difference.
5. A person saves Rs 100 in the first month and increases the saving by Rs 20 every month. Find the total saving in 12 months.

(Answers are not shown — use the formulas above to solve them.)

Created: Arithmetic Progression lesson — contains definitions, formulas, worked examples, and practice problems for classroom or self-study use.