Ever wondered why the numbers 2, 5, 8, 11 keep popping up in math problems? That's a sneak peek at what a sequence does – it lines up numbers in a predictable pattern.
💡 In Simple Words: A sequence is just a list of numbers written one after another, each following a rule. When you add all the numbers of a sequence together, you get a series.
What is a Sequence?
A sequence (pronounced “see‑quence”) is an ordered list of numbers. Think of it like a train of carriages, each carriage (number) knows exactly where it sits because of a rule.
Arithmetic Progression (AP)
An arithmetic progression is a sequence where each term is obtained by adding the same amount, called the common difference, to the previous term.
- Common difference (d): the fixed amount you add each time.
- Formula for the nth term (the term in position n): aₙ = a₁ + (n‑1)d, where a₁ is the first term.
Geometric Progression (GP)
A geometric progression is a sequence where each term is obtained by multiplying the previous term by the same factor, called the common ratio.
- Common ratio (r): the fixed factor you multiply each time.
- Formula for the nth term: aₙ = a₁·rⁿ⁻¹.
How to Find the Sum of a Series
When you add the first n terms of a sequence, you get a series. The sum formulas look a bit scary at first, but they’re just shortcuts.
Sum of first n terms of an AP (Sₙ)
Sₙ = n/2 × (first term + last term) or Sₙ = n/2 × [2a₁ + (n‑1)d].
Sum of first n terms of a GP (Sₙ)
If r ≠ 1, Sₙ = a₁ × (1‑rⁿ) / (1‑r). If r = 1, the sum is simply n × a₁.
Quick Comparison: AP vs GP
| Feature | Arithmetic Progression (AP) | Geometric Progression (GP) |
|---|---|---|
| Rule | Add same number (d) | Multiply by same number (r) |
| nth term | a₁ + (n‑1)d | a₁·rⁿ⁻¹ |
| Sum formula | n/2[2a₁+(n‑1)d] | a₁(1‑rⁿ)/(1‑r) (r≠1) |
Step‑by‑Step Method to Solve a Series Question
Worked Example: Arithmetic Series
Find the sum of the first 15 terms of the series 3, 7, 11, …
- Identify it as an AP because the difference between terms is constant.
- Common difference d = 7‑3 = 4.
- First term a₁ = 3, n = 15.
- Use the sum formula Sₙ = n/2[2a₁+(n‑1)d].
- S₁₅ = 15/2[2·3 + (15‑1)·4] = 7.5[6 + 56] = 7.5 × 62 = 465.
Worked Example: Geometric Series
Find the sum of the first 5 terms of 2, 6, 18, …
- It’s a GP because each term is multiplied by 3.
- Common ratio r = 6/2 = 3, a₁ = 2, n = 5.
- Sum formula Sₙ = a₁(1‑rⁿ)/(1‑r) = 2(1‑3⁵)/(1‑3) = 2(1‑243)/‑2 = 2(‑242)/‑2 = 242.
📝 Likely Exam Questions
- Find the 20th term of an AP whose first term is 5 and common difference is 3.
Answer: a₂₀ = 5 + (20‑1)·3 = 5 + 57 = 62. - Sum the first 10 terms of a GP with a₁ = 4 and r = 2.
Answer: S₁₀ = 4(1‑2¹⁰)/(1‑2) = 4(1‑1024)/‑1 = 4·1023 = 4092. - If the sum of the first n terms of an AP is 210 and a₁ = 5, find n when the common difference is 2.
Answer: Use Sₙ = n/2[2·5 + (n‑1)·2] = 210 → n/2[10 + 2n‑2] = 210 → n/2[2n+8] = 210 → n(n+4)=210 → n²+4n‑210=0 → (n‑14)(n+15)=0 → n=14. - Show that the series 1 + 3 + 9 + … + 3⁴ is a GP and find its sum.
Answer: r = 3, a₁ = 1, n = 5. Sum = 1(1‑3⁵)/(1‑3) = (1‑243)/‑2 = 242/2 = 121. - Find the sum of the arithmetic series 12 + 9 + 6 + … up to the term that equals -12.
Answer: d = –3, a₁ = 12, last term aₙ = –12 → n = [(–12)‑12]/(–3) + 1 = (‑24)/(‑3)+1 = 8+1 = 9. Sum = n/2(a₁ + aₙ) = 9/2(12‑12)=0.