The Logic of Scaling#
An arithmetic sequence grows by adding the same amount every time. A quadratic sequence has first differences that change by a constant amount (constant second difference). A geometric sequence is different from both — it grows by multiplying by the same amount every time. We call that constant multiplier the Common Ratio ($r$).
The rule: every term divided by the one before it gives $r$.
$$r = \frac{T_2}{T_1} = \frac{T_3}{T_2} = \frac{T_4}{T_3} = \dots = \frac{T_n}{T_{n-1}}$$Let’s see this with the sequence $3;\;6;\;12;\;24;\;48;\;\dots$:
| Calculation | Result |
|---|---|
| $\frac{T_2}{T_1} = \frac{6}{3}$ | $2$ |
| $\frac{T_3}{T_2} = \frac{12}{6}$ | $2$ |
| $\frac{T_4}{T_3} = \frac{24}{12}$ | $2$ |
| $\frac{T_5}{T_4} = \frac{48}{24}$ | $2$ |
Every ratio is $2$. That’s what makes this geometric. The general pattern is:
$$r = \frac{T_n}{T_{n-1}} \quad \text{(any term divided by the previous term)}$$If any of those ratios give a different answer, the sequence is not geometric.
Arithmetic vs Geometric — Side by Side#
| Arithmetic | Geometric | |
|---|---|---|
| Operation | Add $d$ each time | Multiply by $r$ each time |
| Test | $T_2 - T_1 = T_3 - T_2$ | $\frac{T_2}{T_1} = \frac{T_3}{T_2}$ |
| General term | $T_n = a + (n-1)d$ | $T_n = ar^{n-1}$ |
| Growth type | Linear (straight line) | Exponential (curve) |
| Symbol | Meaning | Example ($3;\;6;\;12;\;24;\;\dots$) |
|---|---|---|
| $a$ | First term ($T_1$) | $3$ |
| $r$ | Common ratio | $\frac{6}{3} = 2$ |
| $n$ | Position number (positive integer) | $n = 1, 2, 3, \dots$ |
| $T_n$ | Value of the term at position $n$ | $T_4 = 24$ |
Key insight: $r$ can be positive (terms keep the same sign), negative (terms alternate positive/negative), a fraction ($0 < |r| < 1$, terms shrink), or greater than 1 (terms grow). The only value $r$ cannot be is $0$ (that would make every term after the first equal to zero) or $1$ (that would make it a constant sequence, technically arithmetic with $d = 0$).
- $3;\;6;\;12;\;24;\;\dots$ → $r = 2$ (doubling, growth)
- $100;\;50;\;25;\;12.5;\;\dots$ → $r = \frac{1}{2}$ (halving, decay)
- $2;\;-6;\;18;\;-54;\;\dots$ → $r = -3$ (alternating signs, growing magnitude)
- $8;\;-4;\;2;\;-1;\;\dots$ → $r = -\frac{1}{2}$ (alternating signs, shrinking)
1. The General Term: $T_n = ar^{n-1}$#
Don’t memorise blindly — understand the logic:
| Building block | Why? |
|---|---|
| Start at $a$ | That’s your first term |
| Multiply by $r$ a total of $(n-1)$ times | To reach term $n$ you take $(n-1)$ scaling steps from term 1 |
See it step by step:
| Term | How you get there | Using the formula |
|---|---|---|
| $T_1$ | Start | $a \cdot r^0 = a$ |
| $T_2$ | $a \times r$ | $a \cdot r^1$ |
| $T_3$ | $a \times r \times r$ | $a \cdot r^2$ |
| $T_4$ | $a \times r \times r \times r$ | $a \cdot r^3$ |
| $T_n$ | $a$ multiplied by $r$, $(n-1)$ times | $a \cdot r^{n-1}$ |
Important: The exponent is $(n-1)$, not $n$. At $T_1$, you haven’t multiplied by $r$ yet, so the exponent is $0$ (and $r^0 = 1$).
Worked Example 1 — Finding the General Term#
$$a = 5, \quad r = \frac{15}{5} = 3$$$$T_n = ar^{n-1} = 5 \cdot 3^{n-1}$$$$\boxed{T_n = 5 \cdot 3^{n-1}}$$Given $5;\;15;\;45;\;135;\;\dots$, determine $T_n$.
Check: $T_1 = 5 \cdot 3^0 = 5\;\checkmark \quad T_3 = 5 \cdot 3^2 = 45\;\checkmark \quad T_4 = 5 \cdot 3^3 = 135\;\checkmark$
Worked Example 2 — Solving for $n$#
$$T_n = 5 \cdot 3^{n-1} = 10\,935$$$$3^{n-1} = \frac{10\,935}{5} = 2\,187$$Which term of $5;\;15;\;45;\;135;\;\dots$ equals $10\,935$?
We need: $3^{n-1} = 2\,187$. Since $3^7 = 2\,187$:
$$n - 1 = 7 \quad \Rightarrow \quad \boxed{n = 8}$$Check: $T_8 = 5 \cdot 3^7 = 5 \times 2\,187 = 10\,935\;\checkmark$
How to find the power: If you can’t spot it, use logarithms: $n - 1 = \log_3(2\,187) = \frac{\log 2\,187}{\log 3} = 7$.
Worked Example 3 — Two Terms Given, Find $a$ and $r$#
In a geometric sequence, $T_3 = 20$ and $T_6 = 540$. Find $a$ and $r$.
Step 1 — Use the ratio of the two terms:
$$\frac{T_6}{T_3} = \frac{ar^5}{ar^2} = r^3$$$$r^3 = \frac{540}{20} = 27 \quad \Rightarrow \quad \boxed{r = 3}$$Why this works: When you divide $T_k$ by $T_m$ in a geometric sequence, the $a$ cancels and you’re left with $r^{k-m}$. This is the geometric equivalent of the “jump logic” from arithmetic sequences.
| From | To | Powers apart | Calculation | Result |
|---|---|---|---|---|
| $T_3$ | $T_6$ | $6 - 3 = 3$ | $r^3 = \frac{540}{20} = 27$ | $r = 3\;\checkmark$ |
| $T_1$ | $T_3$ | $3 - 1 = 2$ | $r^2 = \frac{20}{a}$ | (need $a$ first) |
Step 2 — Find $a$:
$$T_3 = ar^2 = a(9) = 20 \quad \Rightarrow \quad \boxed{a = \frac{20}{9}}$$Check: $T_6 = \frac{20}{9} \cdot 3^5 = \frac{20}{9} \cdot 243 = \frac{4860}{9} = 540\;\checkmark$
Worked Example 4 — Three Consecutive Terms with Algebra#
The first three terms of a geometric sequence are $x + 1;\;x;\;x - 3$. Find $x$ and the sequence.
For a geometric sequence the common ratio is constant, so:
$$\frac{T_2}{T_1} = \frac{T_3}{T_2}$$$$\frac{x}{x+1} = \frac{x-3}{x}$$Cross-multiply:
$$x^2 = (x+1)(x-3)$$$$x^2 = x^2 - 3x + x - 3$$$$x^2 = x^2 - 2x - 3$$$$0 = -2x - 3$$$$2x = -3$$$$\boxed{x = -\frac{3}{2}}$$Substituting: $T_1 = -\frac{1}{2},\;T_2 = -\frac{3}{2},\;T_3 = -\frac{9}{2}$
$$r = \frac{T_2}{T_1} = \frac{-3/2}{-1/2} = 3\;\checkmark$$2. Sequence vs Series (Geometric)#
Just like with arithmetic, the difference between a sequence (list) and a series (sum) applies here too. See the arithmetic page for the full explanation.
| Geometric Sequence | Geometric Series | |
|---|---|---|
| Display | $3;\;6;\;12;\;24;\;\dots$ | $3 + 6 + 12 + 24 + \dots$ |
| Symbol | $T_n$ (individual term) | $S_n$ (sum of $n$ terms) |
Using the sequence $2;\;6;\;18;\;54;\;162;\;\dots$ ($a = 2$, $r = 3$):
| $n$ | Term ($T_n$) | Partial sum ($S_n$) | What $S_n$ adds |
|---|---|---|---|
| $1$ | $2$ | $S_1 = 2$ | $2$ |
| $2$ | $6$ | $S_2 = 8$ | $2 + 6$ |
| $3$ | $18$ | $S_3 = 26$ | $2 + 6 + 18$ |
| $4$ | $54$ | $S_4 = 80$ | $2 + 6 + 18 + 54$ |
| $5$ | $162$ | $S_5 = 242$ | $2 + 6 + 18 + 54 + 162$ |
Notice how fast the sums grow — this is the power of exponential scaling.
3. Geometric Series — The Sum: $S_n$#
Two Forms of the Sum Formula#
| Formula | When to use |
|---|---|
| $S_n = \dfrac{a(r^n - 1)}{r - 1}$ | When $r > 1$ or $r < -1$ (avoids double negatives) |
| $S_n = \dfrac{a(1 - r^n)}{1 - r}$ | When $\lvert r \rvert < 1$ (avoids negatives) |
Both formulas are algebraically identical — multiply top and bottom by $-1$ to switch between them. Use whichever avoids negative signs in your working.
Conceptual meaning: Unlike the arithmetic sum (which is $n \times \text{average}$), the geometric sum has no simple “average” interpretation. It comes from a clever algebraic trick — the proof below.
Proof of the Geometric Sum Formula (CAPS Required — Full Version)#
This proof is examined regularly. Marks are awarded for every line.
Step 1 — Write the series:
$$S_n = a + ar + ar^2 + ar^3 + \dots + ar^{n-1} \quad \cdots (1)$$Step 2 — Multiply both sides by $r$:
$$rS_n = ar + ar^2 + ar^3 + \dots + ar^{n-1} + ar^n \quad \cdots (2)$$Step 3 — Subtract equation (2) from equation (1):
Line up the terms and notice that almost everything cancels:
| From (1) | $a$ | $+ ar$ | $+ ar^2$ | $+ \dots$ | $+ ar^{n-1}$ | |
|---|---|---|---|---|---|---|
| From (2) | $ar$ | $+ ar^2$ | $+ \dots$ | $+ ar^{n-1}$ | $+ ar^n$ |
Step 4 — Factorise both sides:
$$S_n(1 - r) = a(1 - r^n)$$Step 5 — Divide by $(1 - r)$, provided $r \neq 1$:
$$\boxed{S_n = \frac{a(1 - r^n)}{1 - r}, \quad r \neq 1}$$Multiplying numerator and denominator by $-1$:
$$\boxed{S_n = \frac{a(r^n - 1)}{r - 1}, \quad r \neq 1}$$Exam tip: The most common mistake in this proof is forgetting to write the $rS_n$ line, or subtracting in the wrong order. Practise until every line is automatic.
Worked Example 5 — Basic Sum#
$$a = 3, \quad r = 2, \quad n = 8$$$$S_8 = \frac{3(2^8 - 1)}{2 - 1} = \frac{3(256 - 1)}{1} = 3(255)$$$$\boxed{S_8 = 765}$$Find the sum of the first 8 terms of $3;\;6;\;12;\;24;\;\dots$
Worked Example 6 — Sum with Negative Ratio#
$$a = 4, \quad r = \frac{-12}{4} = -3, \quad n = 6$$$$S_6 = \frac{4((-3)^6 - 1)}{-3 - 1} = \frac{4(729 - 1)}{-4} = \frac{4(728)}{-4} = -728$$$$\boxed{S_6 = -728}$$Find $S_6$ for the sequence $4;\;-12;\;36;\;-108;\;\dots$
Note the brackets: $(-3)^6 = 729$ (positive, because even power). Always use brackets around negative bases on your calculator.
Worked Example 7 — Finding $n$ When Given $S_n$#
$$S_n = \frac{a(r^n - 1)}{r - 1} = \frac{1(2^n - 1)}{2 - 1} = 2^n - 1$$$$2^n - 1 = 1\,023$$$$2^n = 1\,024$$A geometric series has $a = 1$, $r = 2$, and $S_n = 1\,023$. Find $n$.
Since $2^{10} = 1\,024$:
$$\boxed{n = 10}$$Check: $S_{10} = 2^{10} - 1 = 1\,024 - 1 = 1\,023\;\checkmark$
4. Sum to Infinity ($S_\infty$)#
When Can You Add Infinitely Many Terms?#
If you keep adding terms of $3 + 6 + 12 + 24 + \dots$ forever, the total grows without limit. That’s because $r = 2 > 1$ — the terms keep getting bigger.
But what about $18 + 6 + 2 + \frac{2}{3} + \frac{2}{9} + \dots$? Here $r = \frac{1}{3}$. The terms are getting smaller and smaller, approaching zero. The running total “levels off”:
| $n$ | $T_n$ | $S_n$ |
|---|---|---|
| $1$ | $18$ | $18$ |
| $2$ | $6$ | $24$ |
| $3$ | $2$ | $26$ |
| $4$ | $\frac{2}{3} \approx 0.67$ | $26.67$ |
| $5$ | $\frac{2}{9} \approx 0.22$ | $26.89$ |
| $10$ | $\approx 0.001$ | $\approx 26.9999$ |
| $20$ | $\approx 0.000\,000\,005$ | $\approx 27.000\,000$ |
The sum is creeping toward $27$ and will never exceed it. That limit is $S_\infty$.
The Convergence Condition#
$$\boxed{S_\infty \text{ exists only when } -1 < r < 1 \quad \text{(i.e., } |r| < 1)}$$| $|r|$ value | What happens to $r^n$ as $n \to \infty$ | Series behaviour | |—|—|—| | $|r| < 1$ | $r^n \to 0$ | Converges — sum levels off | | $|r| = 1$ | $r^n = \pm 1$ forever | Diverges — terms don’t shrink | | $|r| > 1$ | $r^n \to \pm\infty$ | Diverges — terms grow without bound |
Proof of $S_\infty = \frac{a}{1-r}$ (CAPS Required)#
Start from the finite sum formula:
$$S_n = \frac{a(1 - r^n)}{1 - r}$$When $|r| < 1$, as $n \to \infty$:
$$r^n \to 0 \quad \text{(because multiplying a fraction by itself repeatedly makes it smaller)}$$Examples: $\left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \approx 0.001$, and $\left(\frac{1}{2}\right)^{20} = \frac{1}{1\,048\,576} \approx 0.000\,001$
Therefore $1 - r^n \to 1 - 0 = 1$, and:
$$S_\infty = \lim_{n \to \infty} S_n = \frac{a(1 - 0)}{1 - r}$$$$\boxed{S_\infty = \frac{a}{1 - r}, \quad |r| < 1}$$Exam tip: Always state the convergence condition ($|r| < 1$) before using $S_\infty$. Markers specifically check for this.
Worked Example 8 — Basic $S_\infty$#
$$a = 18, \quad r = \frac{6}{18} = \frac{1}{3}$$Find the sum to infinity of $18 + 6 + 2 + \frac{2}{3} + \dots$
Since $|r| = \frac{1}{3} < 1$, the series converges.
$$S_\infty = \frac{18}{1 - \frac{1}{3}} = \frac{18}{\frac{2}{3}} = 18 \times \frac{3}{2} = \boxed{27}$$Worked Example 9 — Finding $r$ from $S_\infty$#
$$S_\infty = \frac{a}{1 - r}$$$$8 = \frac{4}{1 - r}$$$$1 - r = \frac{4}{8} = \frac{1}{2}$$$$\boxed{r = \frac{1}{2}}$$A convergent geometric series has $S_\infty = 8$ and $a = 4$. Find $r$.
Check: $|r| = \frac{1}{2} < 1\;\checkmark$ (convergence confirmed) and $S_\infty = \frac{4}{1 - \frac{1}{2}} = \frac{4}{\frac{1}{2}} = 8\;\checkmark$
Worked Example 10 — Recurring Decimal as $S_\infty$#
Express $0.\overline{7} = 0.777\dots$ as a common fraction.
Break it into a geometric series:
$$0.777\dots = 0.7 + 0.07 + 0.007 + 0.0007 + \dots$$$$a = 0.7 = \frac{7}{10}, \quad r = \frac{0.07}{0.7} = 0.1 = \frac{1}{10}$$Since $|r| = \frac{1}{10} < 1$:
$$S_\infty = \frac{\frac{7}{10}}{1 - \frac{1}{10}} = \frac{\frac{7}{10}}{\frac{9}{10}} = \frac{7}{10} \times \frac{10}{9} = \boxed{\frac{7}{9}}$$Worked Example 11 — More Complex Recurring Decimal#
Express $0.2\overline{36} = 0.2363636\dots$ as a common fraction.
The non-repeating part is $0.2$. The repeating part starts:
$$0.0363636\dots = 0.036 + 0.00036 + 0.0000036 + \dots$$$$a = 0.036 = \frac{36}{1000}, \quad r = \frac{0.00036}{0.036} = 0.01 = \frac{1}{100}$$$$S_\infty = \frac{\frac{36}{1000}}{1 - \frac{1}{100}} = \frac{\frac{36}{1000}}{\frac{99}{100}} = \frac{36}{1000} \times \frac{100}{99} = \frac{36}{990} = \frac{2}{55}$$$$0.2\overline{36} = \frac{1}{5} + \frac{2}{55} = \frac{11}{55} + \frac{2}{55} = \boxed{\frac{13}{55}}$$5. Finding Values of $x$ for Convergence#
A common exam question gives a geometric series in terms of $x$ and asks: “For which values of $x$ will the series converge?”
The Strategy#
- Identify $r$ in terms of $x$
- Apply the convergence condition: $-1 < r < 1$
- Solve the inequality for $x$
Worked Example 12 — Convergence Values#
For which values of $x$ will the series $(2x-1) + (2x-1)^2 + (2x-1)^3 + \dots$ converge?
This is geometric with $a = (2x-1)$ and $r = (2x-1)$.
For convergence: $-1 < 2x - 1 < 1$
$$-1 < 2x - 1 \quad \Rightarrow \quad 0 < 2x \quad \Rightarrow \quad x > 0$$$$2x - 1 < 1 \quad \Rightarrow \quad 2x < 2 \quad \Rightarrow \quad x < 1$$$$\boxed{0 < x < 1}$$Worked Example 13 — Convergence with $S_\infty$ Calculation#
Given the series $3\left(\frac{x}{2}\right) + 3\left(\frac{x}{2}\right)^2 + 3\left(\frac{x}{2}\right)^3 + \dots$:
(a) For which values of $x$ will the series converge?
(b) If $x = 1$, calculate $S_\infty$.
(a) This is geometric with $r = \frac{x}{2}$.
$$-1 < \frac{x}{2} < 1 \quad \Rightarrow \quad \boxed{-2 < x < 2}$$(b) When $x = 1$: $a = 3\left(\frac{1}{2}\right) = \frac{3}{2}$ and $r = \frac{1}{2}$
$$S_\infty = \frac{\frac{3}{2}}{1 - \frac{1}{2}} = \frac{\frac{3}{2}}{\frac{1}{2}} = \boxed{3}$$6. The Subtraction Technique (Geometric Version)#
Just like with arithmetic series, if you’re given $S_n$ as a formula, you can extract individual terms:
$$T_n = S_n - S_{n-1} \quad \text{for } n \geq 2, \qquad T_1 = S_1$$This works for any series — arithmetic, geometric, or otherwise. The logic is identical: $S_n$ includes terms 1 through $n$, and $S_{n-1}$ includes terms 1 through $(n-1)$, so the difference is just $T_n$.
Worked Example 14 — Extracting Terms from $S_n$#
Given $S_n = 2^{n+1} - 2$, find $T_n$ and show the sequence is geometric.
Step 1 — Find $T_1$:
$$T_1 = S_1 = 2^2 - 2 = 2$$Step 2 — Find $T_n$ for $n \geq 2$:
$$T_n = S_n - S_{n-1} = (2^{n+1} - 2) - (2^n - 2) = 2^{n+1} - 2^n = 2^n(2 - 1) = 2^n$$Step 3 — Verify $T_1$:
$T_1 = 2^1 = 2 = S_1\;\checkmark$ — so $T_n = 2^n$ works for all $n \geq 1$.
Step 4 — Confirm geometric:
$$\frac{T_{n+1}}{T_n} = \frac{2^{n+1}}{2^n} = 2$$The ratio is constant at $r = 2$, so the sequence is geometric with $a = 2$ and $r = 2$.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Division order: $\frac{T_1}{T_2}$ instead of $\frac{T_2}{T_1}$ | Gives you $\frac{1}{r}$ instead of $r$ | Always: later term $\div$ earlier term |
| Forgetting brackets: $-3^4$ vs $(-3)^4$ | $-3^4 = -81$ but $(-3)^4 = 81$ | Always use brackets around negative bases |
| Using $S_\infty$ when $\lvert r \rvert \geq 1$ | The series diverges — no finite sum exists | Check $\lvert r \rvert < 1$ before using $\frac{a}{1-r}$ |
| Wrong exponent: $ar^n$ instead of $ar^{n-1}$ | Off by one factor of $r$ | $T_1 = ar^0 = a$, so exponent must be $(n-1)$ |
| Confusing $r = -\frac{1}{2}$ convergence | $\lvert -\frac{1}{2} \rvert = \frac{1}{2} < 1$ so it does converge | Use absolute value when checking convergence |
| Recurring decimals: wrong $a$ or $r$ | $0.777\dots$ has $a = 0.7$, $r = 0.1$ (not $a = 7$, $r = 0.1$) | Write out the first few terms carefully |
🎥 Video Lessons#
Geometric Sequences & Series 1#
Geometric Series and Sum to Infinity#
💡 Pro Tips for Exams#
1. The Geometric Mean Shortcut#
If three terms $T_1;\;T_2;\;T_3$ are consecutive in a geometric sequence, then:
$$T_2^2 = T_1 \times T_3$$Why? Because $T_2 = T_1 \cdot r$ and $T_3 = T_1 \cdot r^2$, so $T_1 \times T_3 = T_1^2 \cdot r^2 = (T_1 r)^2 = T_2^2$.
Example: If $4;\;x;\;16$ are geometric, then $x^2 = 4 \times 16 = 64$, so $x = \pm 8$. (Both signs are valid — they give $r = 2$ or $r = -2$.)
2. The Division Trick for Two Given Terms#
When given $T_m$ and $T_k$, divide them to eliminate $a$:
$$\frac{T_k}{T_m} = \frac{ar^{k-1}}{ar^{m-1}} = r^{k-m}$$This is a single equation in $r$ — no simultaneous equations needed. It’s the geometric version of the arithmetic “jump logic.”
3. Distinguishing Arithmetic from Geometric#
If a question gives you a sequence and you’re not sure which type:
- Compute $T_2 - T_1$ and $T_3 - T_2$. If equal → arithmetic.
- Compute $\frac{T_2}{T_1}$ and $\frac{T_3}{T_2}$. If equal → geometric.
- If neither is equal, check second differences → might be quadratic.
4. Negative $r$ and Alternating Signs#
When $r < 0$, the terms alternate between positive and negative. This means:
- $S_n$ will oscillate (sometimes bigger, sometimes smaller)
- $S_\infty$ still exists if $|r| < 1$ — the oscillations get smaller and settle on a limit
- Always use brackets when computing $r^n$ on your calculator
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