📐 Arithmetic Progressions — Class 10

nth term, sum of n terms, real-life applications with fully solved examples

1. Introduction to AP

📖 Definition

An Arithmetic Progression (AP) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term. This fixed number is called the common difference (d).

General form: a, a+d, a+2d, a+3d, ...

where a = first term and d = common difference

🌟 Real-Life Examples

• Saving ₹500 each month: 500, 1000, 1500, 2000... (d = 500)
• Taxi fare: ₹25 for first km, ₹8 for each additional km: 25, 33, 41, 49... (d = 8)
• Seats in a cinema: 15, 17, 19, 21... (each row has 2 more seats, d = 2)

1.1 Finding Common Difference

The common difference d = (any term) – (its preceding term) = a₂ – a₁ = a₃ – a₂ = ...

💡 Example: Identify AP and find d

a) 3, 7, 11, 15, 19...

d = 7–3 = 4 = 11–7 = 4 → ✓ It IS an AP with d = 4

b) 1, 1, 2, 3, 5, 8...

d = 1–1 = 0, then 2–1 = 1 → differences not equal → ✗ NOT an AP

c) –5, –1, 3, 7, 11...

d = –1–(–5) = 4 → ✓ AP with d = 4 (negative terms, positive d)

Type of AP	Common Difference	Example
Increasing AP	d > 0	2, 5, 8, 11...
Decreasing AP	d < 0	20, 15, 10, 5...
Constant AP	d = 0	7, 7, 7, 7...

2. nth Term of an AP

📖 Formula for nth Term (General Term)

aₙ = a + (n – 1)d

where: a = first term, d = common difference, n = term number, aₙ = nth term

⚡ How to Derive aₙ = a + (n–1)d

a₁ = a = a + (1–1)d

a₂ = a + d = a + (2–1)d

a₃ = a + 2d = a + (3–1)d

aₙ = a + (n–1)d ← pattern observed!

💡 Example 1: Find the 20th term of 3, 7, 11, 15...

a = 3, d = 4, n = 20

a₂₀ = 3 + (20–1) × 4 = 3 + 19 × 4 = 3 + 76 = 79

💡 Example 2: Find which term of the AP 5, 11, 17, 23... is 119?

a = 5, d = 6, aₙ = 119

119 = 5 + (n–1)×6 → 114 = (n–1)×6 → n–1 = 19 → n = 20

Answer: 119 is the 20th term

💡 Example 3: Find 31st term of AP where 11th term = 38 and 16th term = 73

a₁₁ = a + 10d = 38 ...(i)

a₁₆ = a + 15d = 73 ...(ii)

Subtracting (i) from (ii): 5d = 35 → d = 7

From (i): a + 70 = 38 → a = –32

a₃₁ = –32 + 30×7 = –32 + 210 = 178

3. Sum of n Terms of an AP

📖 Sum Formula

Sₙ = n/2 × [2a + (n–1)d]

Also written as: Sₙ = n/2 × (a + l)

where l = last term = a + (n–1)d

🌟 Story Behind the Formula — Gauss's Trick

Young Carl Friedrich Gauss was asked to add 1+2+3+...+100. He noticed:

1 + 100 = 101 | 2 + 99 = 101 | 3 + 98 = 101 ... (50 such pairs)

Sum = 50 × 101 = 5050 → This is exactly n/2 × (a + l) = 100/2 × (1 + 100) = 5050!

💡 Example 1: Find sum of first 24 terms of 5, 8, 11, 14...

a = 5, d = 3, n = 24

S₂₄ = 24/2 × [2×5 + (24–1)×3] = 12 × [10 + 69] = 12 × 79 = 948

💡 Example 2: How many terms of AP 9, 17, 25... must be taken to get sum 636?

a = 9, d = 8, Sₙ = 636

n/2 × [18 + (n–1)×8] = 636

n × [18 + 8n – 8] = 1272 → n(8n + 10) = 1272 → 8n² + 10n – 1272 = 0

4n² + 5n – 636 = 0 → n = (–5 + √(25 + 10176))/8 = (–5 + 101)/8 = 12

Answer: 12 terms

⚡ Useful Result

The nth term can also be found using:

aₙ = Sₙ – Sₙ₋₁ (for n ≥ 2)

a₁ = S₁

This is very useful when the sum formula is given and you need to find specific terms.

💡 Example 3: Sum of first n terms is Sₙ = 3n² + 5n. Find the AP.

a₁ = S₁ = 3(1)² + 5(1) = 8

S₂ = 3(4) + 5(2) = 22 → a₂ = S₂ – S₁ = 22 – 8 = 14

S₃ = 3(9) + 5(3) = 42 → a₃ = S₃ – S₂ = 42 – 22 = 20

d = a₂ – a₁ = 14 – 8 = 6

AP: 8, 14, 20, 26... (a = 8, d = 6)

4. Key Properties of AP

If a constant is added/subtracted to each term of an AP, the result is also an AP with same d.
If each term is multiplied/divided by a constant k, new d becomes kd.
If a, b, c are in AP then 2b = a + c (b is the arithmetic mean of a and c).
The sum of terms equidistant from the beginning and end of a finite AP is equal to the sum of first and last terms.

📋 Formula Sheet — AP

nth Term

aₙ = a + (n – 1)d

Sum of n Terms

Sₙ = n/2 × [2a + (n–1)d]

Sₙ = n/2 × (a + l)

nth Term from Sum

aₙ = Sₙ – Sₙ₋₁

Arithmetic Mean

AM = (a + b)/2

If a,b,c in AP: 2b = a+c