Ever wondered why numbers can line up like a marching band, each stepping the same distance? That pattern is the heart of sequences and series.
In simple words, a sequence is a list of numbers written in order, and a series is what you get when you add those numbers together. Think of a sequence as the steps on a staircase and a series as the total height you climb.
What are Sequences and Series?
A sequence (pronounced "see-kwens") is just an ordered list of numbers. Each number in the list is called a term. If you take the terms and add them up, you create a series. Imagine a row of dominoes: the dominoes themselves are the sequence, while the total length of the row after you line them up is the series.
Arithmetic Progression (AP)
An arithmetic progression is a sequence where each term jumps by the same amount. That constant jump is called the common difference (d). For example, 2, 5, 8, 11 … adds 3 each time, so d = 3.
First term (a) is the very first number. The nth term (the term in position n) can be found with the formula:
a_n = a + (n-1)·d
Why does this work? Think of walking up stairs: you start at step a, then take (n‑1) steps, each of height d.
To add the first n terms of an AP, use the sum formula:
S_n = n/2 · (2a + (n-1)·d)
Or, a quicker version: S_n = n/2 · (first term + last term). It’s like averaging the heights of the first and last steps, then multiplying by the number of steps.
Worked Example – AP
Find the sum of the first 10 terms of the AP: 4, 9, 14, …
- First term a = 4.
- Common difference d = 9‑4 = 5.
- Number of terms n = 10.
- Last term a_n = a + (n‑1)d = 4 + 9·5 = 49.
- Sum S_n = 10/2 · (4 + 49) = 5 · 53 = 265.
Geometric Progression (GP)
A geometric progression is a sequence where each term is multiplied by the same number. That multiplier is the common ratio (r). Example: 3, 6, 12, 24 … doubles each time, so r = 2.
First term (a) is again the starting number. The nth term formula is:
a_n = a·r^{n-1}
Here the exponent (n‑1) tells you how many times you’ve multiplied by r.
To add the first n terms of a GP (when r ≠ 1), use:
S_n = a·(1‑r^n) / (1‑r)
If r is bigger than 1, the series grows quickly, like a snowball rolling downhill.
Worked Example – GP
Find the sum of the first 5 terms of the GP: 2, 6, 18, … (r = 3).
- a = 2, r = 3, n = 5.
- a_n = 2·3^{4} = 2·81 = 162.
- S_n = 2·(1‑3^{5})/(1‑3) = 2·(1‑243)/‑2 = 2·(‑242)/‑2 = 242.
Key Formulas at a Glance
| Concept | Formula |
|---|---|
| AP nth term | a_n = a + (n-1)d |
| AP sum | S_n = n/2·(2a + (n-1)d) |
| GP nth term | a_n = a·r^{n-1} |
| GP sum (r≠1) | S_n = a·(1‑r^n)/(1‑r) |
How to Find the Sum of an AP – Step by Step
When the exam asks for the sum, follow these four moves:
Quick Comparison
- Pattern: AP adds a fixed amount; GP multiplies by a fixed amount.
- Growth: AP grows linearly (straight line); GP grows exponentially (steep curve).
- nth term: AP uses a + (n‑1)d; GP uses a·r^{n-1}.
- Sum: AP uses n/2·(first+last); GP uses a·(1‑r^n)/(1‑r) when r≠1.
📝 Likely Exam Questions
- Find the 15th term of the AP 7, 12, 17, …
Answer: a=7, d=5, a_{15}=7+(15‑1)·5=67. - Sum the first 8 terms of the GP 5, 20, 80, …
Answer: a=5, r=4, n=8, S_n=5·(1‑4^8)/(1‑4)=5·(1‑65536)/‑3=109,225. - Given that the sum of the first 6 terms of an AP is 84 and a=4, find d.
Answer: Use S_n=n/2·(2a+(n‑1)d) → 84=6/2·(8+5d) → 84=3·(8+5d) → 28=8+5d → d=4. - If the 4th term of a GP is 48 and r=2, find the first term.
Answer: a·2^{3}=48 → a=48/8=6. - Show that the sum of the first n natural numbers (1,2,3,…) is n(n+1)/2.
Answer: Pair first and last: 1+n, 2+(n‑1)… each pair sums to n+1, there are n/2 pairs → S=n/2·(n+1).