Geometric Sequences and Series

🔢 Geometric Sequences & Series

Part 1 of 5 — The Common Ratio

Topics in This Part

🔑 Key Concept: A geometric sequence multiplies by the same number each step. That number — the common ratio $r$ — is the heartbeat of everything in this lesson.

What Is a Geometric Sequence?

In a geometric sequence, you get each term by multiplying the previous term by a fixed number called the common ratio, written $r$.

$2,\; 6,\; 18,\; 54,\; 162,\; \dots \qquad (r = 3)$

Each term is $3$ times the one before it.

How to spot the ratio

To find $r$, divide any term by the term before it:

$r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \dots$

🔑 Key Idea: If the ratio between every consecutive pair is the same, the sequence is geometric. If it's not constant, it isn't.

Concept Check 🎯

Pinning Down $r$ Exactly

When the ratio isn't obvious, just divide. The result can be a whole number, a fraction, or negative:

$r = \frac{\text{any term}}{\text{the term right before it}}$

A quick sanity check: pick a different pair of consecutive terms and divide again. If you get the same $r$, you've found it. If not, the sequence isn't geometric.

Find the Common Ratio 🧮

Compute $r$ for each geometric sequence. (Fractions like 1/3 are fine.)

1) $2,\,10,\,50,\,250,\dots \quad r = \,?$ 2) $48,\,24,\,12,\,6,\dots \quad r = \,?$ 3) $1,\,-3,\,9,\,-27,\dots \quad r = \,?$

Geometric vs. Arithmetic

Don't confuse the two big sequence types:

💡 Quick test: Subtract consecutive terms — constant? Arithmetic. Divide consecutive terms — constant? Geometric.

⚠️ A sequence like $2,\,4,\,6,\,10$ is neither: differences aren't constant ($2,2,4$) and ratios aren't constant ($2,1.5,1.67$).

Classify Each Sequence 🔽

🔢 Geometric Sequences & Series

Part 2 of 5 — The nth-Term Formula

🔑 The Goal: Instead of multiplying term by term forever, we want a formula that jumps straight to any term — the 10th, the 50th, the 100th — using just $a_1$ and $r$.

The Explicit Formula

The $n$th term of a geometric sequence is:

$a_n = a_1 \cdot r^{\,n-1}$

where $a_1$ is the first term and $r$ is the common ratio.

Why $n-1$? To reach the first term you multiply by $r$ zero times, to reach the second term you multiply once, and so on — always one fewer multiplication than the term number.

Worked Example: $3,\,6,\,12,\,24,\dots$ — find $a_7$

Here $a_1 = 3$ and $r = 2$:

$a_7 = 3 \cdot 2^{\,7-1} = 3 \cdot 2^{6} = 3 \cdot 64 = 192$

✅ Check: $3, 6, 12, 24, 48, 96, \mathbf{192}$ — the 7th term is indeed $192$.

Worked Example: a shrinking sequence

Find the 5th term of $80,\,40,\,20,\dots$

Here $a_1 = 80$ and $r = \frac{1}{2}$:

$a_5 = 80 \cdot \left(\tfrac{1}{2}\right)^{5-1} = 80 \cdot \left(\tfrac{1}{2}\right)^{4} = 80 \cdot \tfrac{1}{16} = 5$

Worked Example: alternating signs

Find $a_4$ of $2,\,-6,\,18,\dots$

Here $a_1 = 2$ and $r = -3$:

$a_4 = 2 \cdot (-3)^{4-1} = 2 \cdot (-3)^{3} = 2 \cdot (-27) = -54$

⚠️ Watch the parentheses: wrap a negative ratio in parentheses. For an even power the difference is real: $(-3)^2 = 9$ (positive), but $-3^2 = -(3^2) = -9$. Always write $(-3)^{n-1}$ so the sign comes out right.

Concept Check 🎯

Recursive vs. Explicit

There are two valid ways to define a geometric sequence:

Example for $3,\,6,\,12,\dots$:

• Recursive: $a_1 = 3,\;\; a_n = 2 \cdot a_{n-1}$
• Explicit: $a_n = 3 \cdot 2^{\,n-1}$

💡 Use recursive when you only need the next term; use explicit when you need a far-off term like $a_{20}$ without listing all the others.

Recursive or Explicit? 🔽

The sequence $4,\,12,\,36,\,108,\dots$ has $a_1 = 4$ and $r = 3$.

A Reliable Routine

For any "find the $n$th term" problem, run the same three steps:

1. Identify $a_1$ — the first term.
2. Find $r$ — divide the second term by the first.
3. Plug into $a_n = a_1 \cdot r^{\,n-1}$ and simplify the power carefully.

The only place students slip is the exponent: it's $n-1$, one less than the term number. Keep that front of mind as you try the next set.

Compute the Term 🧮

Use $a_n = a_1 \cdot r^{\,n-1}$.

1) $2,\,6,\,18,\dots$ Find $a_5$. $\;=\,?$ 2) $a_1 = 7,\; r = 2$. Find $a_4$. $\;=\,?$ 3) $a_1 = 100,\; r = \tfrac{1}{5}$. Find $a_3$. $\;=\,?$

🔢 Geometric Sequences & Series

Part 3 of 5 — Finite Geometric Series

🔑 From sequence to series: A series is what you get when you add up the terms of a sequence. Adding $3+6+12+24$ by hand is fine, but adding the first $20$ terms? You need a formula.

The Finite Sum Formula

The sum of the first $n$ terms of a geometric sequence is:

$S_n = a_1 \cdot \frac{1 - r^{\,n}}{1 - r} \qquad (r \ne 1)$

You only need three things: the first term $a_1$, the ratio $r$, and how many terms $n$.

Worked Example: sum the first 5 terms of $2,\,6,\,18,\dots$

Here $a_1 = 2$, $r = 3$, $n = 5$:

$S_5 = 2 \cdot \frac{1 - 3^{5}}{1 - 3} = 2 \cdot \frac{1 - 243}{-2} = 2 \cdot \frac{-242}{-2} = 2 \cdot 121 = 242$

✅ Check by hand: $2 + 6 + 18 + 54 + 162 = 242$ ✓

Worked Example: a fractional ratio

Sum the first 4 terms of $16,\,8,\,4,\,2,\dots$

Here $a_1 = 16$, $r = \frac{1}{2}$, $n = 4$:

$S_4 = 16 \cdot \frac{1 - \left(\frac{1}{2}\right)^{4}}{1 - \frac{1}{2}} = 16 \cdot \frac{1 - \frac{1}{16}}{\frac{1}{2}} = 16 \cdot \frac{\frac{15}{16}}{\frac{1}{2}} = 16 \cdot \frac{15}{16} \cdot 2 = 30$

✅ Check by hand: $16 + 8 + 4 + 2 = 30$ ✓

⚠️ Common mistake: the exponent on $r$ is $n$ (the number of terms), not $n-1$. The $n-1$ exponent belongs to the term formula, not the sum formula.

Concept Check 🎯

Sigma (Summation) Notation

Series are often written compactly with sigma notation $\Sigma$:

$\sum_{k=1}^{n} a_1 \, r^{\,k-1}$

This means: "add up $a_1 r^{k-1}$ as $k$ runs from $1$ to $n$."

Example: unpack $\displaystyle\sum_{k=1}^{4} 3 \cdot 2^{\,k-1}$

So the sum is $3 + 6 + 12 + 24 = 45$. Reading off $a_1 = 3$, $r = 2$, $n = 4$, the formula agrees: $S_4 = 3 \cdot \frac{1-2^4}{1-2} = 3 \cdot \frac{-15}{-1} = 45$. ✓

Read the Sigma Notation 🔽

For the series $\displaystyle\sum_{k=1}^{5} 6 \cdot 4^{\,k-1}$, identify each piece.

Two Routes to the Same Answer

For a short sum (3–5 terms), you can simply list the terms and add — and it's a great way to check the formula.

For a long sum (say, 20 terms), the formula is the only practical choice:

$S_n = a_1 \cdot \frac{1 - r^{\,n}}{1 - r}$

In the drill below, use the formula on at least one problem and confirm it matches the by-hand total. Remember: the exponent is $n$ (the count of terms), not $n-1$.

Sum It Up 🧮

Use $S_n = a_1 \dfrac{1 - r^{\,n}}{1 - r}$.

1) $5,\,10,\,20,\dots$ Find $S_4$. $\;=\,?$ 2) $\displaystyle\sum_{k=1}^{3} 2 \cdot 5^{\,k-1} \;=\,?$ 3) $1,\,3,\,9,\,27,\,81,\dots$ Find $S_5$. $\;=\,?$

🔢 Geometric Sequences & Series

Part 4 of 5 — Infinite Geometric Series

🔑 The surprising idea: You can add infinitely many numbers and still get a finite total — but only when the terms shrink fast enough, meaning $|r| < 1$.

When Does an Infinite Series Have a Sum?

An infinite geometric series $a_1 + a_1 r + a_1 r^2 + \dots$ either converges (adds to a finite number) or diverges (grows without bound).

When $|r| < 1$, the sum is:

$S_{\infty} = \frac{a_1}{1 - r}$

💡 Intuition: Think of $\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \dots$ Each piece is half of what's left, so you creep ever closer to $1$ but never overshoot it. Here $a_1 = \tfrac12$, $r = \tfrac12$, and $S_\infty = \frac{1/2}{1 - 1/2} = 1$.

Concept Check 🎯

Worked Examples — Sum to Infinity

Example 1: $8 + 4 + 2 + 1 + \dots$

Here $a_1 = 8$ and $r = \frac{1}{2}$ (and $|r| < 1$ ✓):

$S_{\infty} = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16$

Example 2: $9 - 3 + 1 - \frac{1}{3} + \dots$

Here $a_1 = 9$ and $r = -\frac{1}{3}$ (and $|r| < 1$ ✓):

$S_{\infty} = \frac{9}{1 - \left(-\frac{1}{3}\right)} = \frac{9}{1 + \frac{1}{3}} = \frac{9}{\frac{4}{3}} = 9 \cdot \frac{3}{4} = \frac{27}{4} = 6.75$

⚠️ Watch the sign! With $r = -\frac13$, the denominator becomes $1 - (-\frac13) = 1 + \frac13$. Subtracting a negative adds.

Set Up the Sum 🔽

You're finding $S_\infty$ for $12 + 4 + \frac{4}{3} + \dots$

Before You Compute, Check Convergence

The infinite-sum formula $S_\infty = \dfrac{a_1}{1-r}$ is only valid when $|r| < 1$. So make convergence your first move on every problem:

1. Find $r$.
2. If $|r| \ge 1$ → the series diverges; there is no finite sum.
3. If $|r| < 1$ → plug into $\dfrac{a_1}{1-r}$.

⚠️ Skipping step 2 is the classic error — applying the formula to a divergent series gives a nonsense answer.

Sum to Infinity 🧮

Use $S_\infty = \dfrac{a_1}{1 - r}$ (only valid when $|r| < 1$).

1) $20 + 10 + 5 + \dots \;\; S_\infty = \,?$ 2) $a_1 = 6,\; r = \tfrac{1}{3}.\;\; S_\infty = \,?$ 3) $a_1 = 10,\; r = -\tfrac{1}{2}.\;\; S_\infty = \,?$ (fraction or decimal ok)

🔢 Geometric Sequences & Series

Part 5 of 5 — Applications & Mastery Check

You can now (1) find a common ratio, (2) compute any term, (3) sum a finite series, and (4) sum a convergent infinite series. Let's apply it to the real world, then prove your mastery.

Real-World Geometric Models

Geometric sequences and series model anything that grows or decays by a fixed percent.

Compound situations as a ratio

A quantity changing by $p\%$ each period multiplies by a ratio $r$:

• Growth of $p\%$: $r = 1 + \dfrac{p}{100}$ (e.g. $+5\% \Rightarrow r = 1.05$)
• Decay of $p\%$: $r = 1 - \dfrac{p}{100}$ (e.g. $-20\% \Rightarrow r = 0.80$)

Worked Example: bouncing ball

A ball dropped from $24$ ft rebounds to $\frac{1}{2}$ its height each bounce. The rebound heights $12, 6, 3, \dots$ form a geometric sequence with $a_1 = 12$, $r = \frac12$. The total rebound distance is the infinite sum:

$S_\infty = \frac{12}{1 - \frac12} = 24 \text{ ft}$

💡 Decay problems ($|r|<1$) are exactly the ones whose infinite series converge — perfect for "total distance" or "total amount forever" questions.

Application Check 🎯

Quick Reference

⚠️ Remember the traps: the term formula uses $r^{n-1}$ but the sum formula uses $r^{n}$; an infinite series only converges when $\lvert r\rvert < 1$; and subtracting a negative ratio in the denominator adds.

Pick the Right Tool 🔽

Match each question to the formula you'd use.

You're Ready

Here's the whole lesson in one breath:

🔑 A geometric sequence multiplies by $r$ each step. Reach any term with $a_n = a_1 r^{\,n-1}$. Add a finite count of terms with $S_n = a_1\frac{1-r^n}{1-r}$. Add infinitely many — but only when $|r|<1$ — with $S_\infty = \frac{a_1}{1-r}$.

The Exit Quiz below pulls one question from each major skill. Take your time and check your arithmetic.

Exit Quiz ✅

Answer all three to finish the lesson.