📐 Geometric Sequence
General term, finding the common ratio, and locating a term
🎯 What Is a Geometric Sequence?
A geometric sequence is a sequence in which the ratio (quotient) between any two consecutive terms is constant.
In other words: you multiply by the same number to go from one term to the next.
Examples:
| \(2, 6, 18, 54, 162, ...\) Common ratio: ×3 | \(64, 32, 16, 8, 4, ...\) Common ratio: ×0.5 (or ÷2) |
⚖️ Arithmetic vs Geometric
| Arithmetic sequence | Geometric sequence |
|---|---|
| Add a fixed number (d) | Multiply by a fixed number (q) |
| 2, 5, 8, 11, ... (+3) | 2, 6, 18, 54, ... (×3) |
🔤 Basic Notation
| Symbol | Meaning | Example |
|---|---|---|
| \(a_1\) | The first term of the sequence | In 2,6,18,... → \(a_1 = 2\) |
| \(q\) | Common ratio – the number you multiply by | In 2,6,18,... → \(q = 3\) |
| \(n\) | Position (index) of the term | Third term → \(n = 3\) |
| \(a_n\) | General term – the term at position n | Term at position 10 → \(a_{10}\) |
⭐ General Term Formula
\(a_n = a_1 \cdot q^{n-1}\)
💡 Understanding the formula:
To go from the first term to the n-th term, you multiply by q exactly \(n-1\) times
🎵 To remember: "first term, times q to the power of (position minus one)"
✏️ Example 1: Finding a Term by Position
Question: In a geometric sequence \(a_1 = 3\) and \(q = 2\). Find \(a_8\).
Solution:
Substitute into the formula: \(a_n = a_1 \cdot q^{n-1}\)
\(a_8 = 3 \cdot 2^{8-1}\)
\(a_8 = 3 \cdot 2^7\)
\(a_8 = 3 \cdot 128 = 384\)
Answer: \(a_8 = 384\)
🔍 Finding the Common Ratio (q)
Method 1: From consecutive terms
\(q = \frac{a_{n+1}}{a_n}\)
Common ratio = next term divided by current term
Method 2: From any two terms
\(q = \sqrt[m-k]{\frac{a_m}{a_k}}\)
or equivalently:
\(q^{m-k} = \frac{a_m}{a_k}\)
✏️ Example 2: Finding the Common Ratio
Question: In a geometric sequence \(a_2 = 6\) and \(a_5 = 162\). Find \(q\).
Solution:
Find the ratio between the terms:
\(\frac{a_5}{a_2} = \frac{162}{6} = 27\)
Between position 2 and position 5 there are \(5-2=3\) multiplications by q:
\(q^3 = 27\)
\(q = \sqrt[3]{27} = 3\)
Answer: \(q = 3\)
💡 Explanation: from \(a_2\) to \(a_5\) we multiply 3 times by q:
\(a_2 \xrightarrow{\times q} a_3 \xrightarrow{\times q} a_4 \xrightarrow{\times q} a_5\)
📍 Finding the Position of a Term (n)
From the formula \(a_n = a_1 \cdot q^{n-1}\), isolate \(n\):
\(q^{n-1} = \frac{a_n}{a_1}\)
then solve the exponential equation (or use logarithms)
✏️ Example 3: Finding the Position
Question: In a geometric sequence \(a_1 = 5\) and \(q = 2\). At what position is the term 320?
Solution:
Substitute \(a_n = 320\) into the formula:
\(320 = 5 \cdot 2^{n-1}\)
Divide by 5:
\(2^{n-1} = 64\)
Recognise that \(64 = 2^6\):
\(2^{n-1} = 2^6\)
\(n-1 = 6\)
\(n = 7\)
Answer: the term 320 is at position 7
🏁 Finding the First Term (a₁)
Isolate \(a_1\) from the formula:
\(a_1 = \frac{a_n}{q^{n-1}}\)
✏️ Example 4:
Question: In a geometric sequence \(a_5 = 48\) and \(q = 2\). Find \(a_1\).
\(a_1 = \frac{a_5}{q^{5-1}} = \frac{48}{2^4} = \frac{48}{16} = 3\)
Answer: \(a_1 = 3\)
⚠️ Special Cases of the Common Ratio
\(q > 1\)
Sequence is increasing
2, 6, 18, 54, ...
\(0 < q < 1\)
Sequence is decreasing (approaching 0)
64, 32, 16, 8, ...
\(q < 0\)
Sequence alternates in sign
2, −6, 18, −54, ...
\(q = 1\)
Constant sequence
5, 5, 5, 5, ...
⚠️ Note: \(q \neq 0\) (multiplying by 0 cannot produce a sequence)
❓ Checking Whether a Number Belongs to the Sequence
Method: substitute the number as \(a_n\) and check whether \(n\) comes out as a natural number
✏️ Example 5:
Question: In the sequence \(a_1 = 2\), \(q = 3\). Does 486 belong to the sequence?
\(486 = 2 \cdot 3^{n-1}\)
\(3^{n-1} = 243\)
\(3^{n-1} = 3^5\) (since \(243 = 3^5\))
\(n-1 = 5\)
\(n = 6\) ✓
Answer: Yes! 486 is the 6th term of the sequence
✏️ Example 6:
Question: In the same sequence, does 100 belong to the sequence?
\(100 = 2 \cdot 3^{n-1}\)
\(3^{n-1} = 50\)
50 is not a power of 3 (the powers are: 1, 3, 9, 27, 81, 243, ...)
\(n\) does not come out as a natural number ✗
Answer: No! 100 does not belong to the sequence
📋 Formula Summary
| What to find | Formula |
|---|---|
| General term | \(a_n = a_1 \cdot q^{n-1}\) |
| Common ratio (consecutive) | \(q = \frac{a_{n+1}}{a_n}\) |
| Common ratio (any two terms) | \(q^{m-k} = \frac{a_m}{a_k}\) |
| Position of a term | \(q^{n-1} = \frac{a_n}{a_1}\) then solve |
| First term | \(a_1 = \frac{a_n}{q^{n-1}}\) |
💡 Tips for the Exam
1️⃣ (n−1) in the exponent!
Common mistake: the exponent is \(n-1\), not \(n\)
2️⃣ Know your powers
Useful to know by heart: \(2^1\) to \(2^{10}\), \(3^1\) to \(3^5\)
3️⃣ Negative ratio
If signs alternate – the ratio is negative!
4️⃣ Powers of q
When finding position – check if you get an integer power
📝 Core Formula
\(a_n = a_1 \cdot q^{n-1}\)
From this formula you can isolate any unknown variable!
Arithmetic: \(a_n = a_1 + (n-1)d\) (add)
Geometric: \(a_n = a_1 \cdot q^{n-1}\) (multiply)