A geometric sequence multiplies each term by a constant ratio \(r\). First term: \(a_1\). General term: \(a_n = a_1 r^{\,n-1}\).
Identify \(a_1\) and find \(r = \frac{a_2}{a_1}\) to describe any geometric sequence.
Goal: practise extracting the common ratio from raw sequence data.
From the sequence \(3, 6, 12, 24, \dots\) pick pairs: \(3,6\); \(6,12\); \(12,24\).
Compute \( \frac{6}{3}, \frac{12}{6}, \frac{24}{12} \).
Each fraction simplifies to 2, giving the common ratio \( r = 2 \).
If every consecutive pair gives the same quotient, the sequence is geometric.
The nth-term formula lets you jump straight to any term in the sequence.
State it clearly and interpret each variable correctly.
Find the 50th or 100th term quickly.
Describe interest, population or radioactive decay.
For the geometric sequence with first term \(a_1 = 5\) and common ratio \(r = 3\), what is \(a_5\)?
Use \(a_5 = 5 \times 3^{4}\).
Correct! \(5 \times 81 = 405\).
Revisit the exponent: \(n - 1 = 4\).
r = 2 (blue) grows, r = 0.5 (red) decays.
Both sequences begin at 1 on the graph.
With r = 2, each term doubles; the curve shoots upward, showing exponential growth.
With r = 0.5, each term halves; the curve sinks toward the x-axis, showing exponential decay.
A finite geometric series sums the first \(n\) terms of a sequence. Know this formula by heart when \(r \neq 1\).
Determine the remaining balance after \(n\) equal payments.
Estimate cumulative population after \(n\) years of geometric growth.
Write the series in expanded form.
Multiply every term by \(r\).
Subtract to eliminate the middle terms.
Factor \(S_n\) on the left.
Divide by \(1 - r\) to isolate \(S_n\).
Telescoping quickly collapses the series.
Each step scales by \(r\); think multiplication, not addition.
Jump to term \(a_n = a_1 r^{n-1}\) instantly.
Finite sum: \(S_n = a_1 \frac{1 - r^{n}}{1 - r}\).
Infinite series converges only when \(|r| < 1\).
Governs compound interest, population change and exponential decay.