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Geometric Sequences and Series

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Geometric Sequences and Series - Complete Interactive Lesson

Part 1: The Common Ratio

🔢 Geometric Sequences & Series

Part 1 of 5 — The Common Ratio

Topics in This Part

Section
What Is a Geometric Sequence?
Finding the Common Ratio
Geometric vs. Arithmetic

🔑 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:

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

Sequence	Ratio test	$r$
$5,\,10,\,20,\,40$	$10/5 = 2$

🔑 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 , 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)

Geometric vs. Arithmetic

Don't confuse the two big sequence types:

	Arithmetic	Geometric
Step rule	add common difference $d$	multiply by common ratio $r$
Example	$4,\,7,\,10,\,13$ ()

Classify Each Sequence 🔽

Part 2: The nth-Term Formula

🔢 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:

Part 3: Finite Geometric Series

🔢 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:

Part 4: Infinite Geometric Series

🔢 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 (adds to a finite number) or (grows without bound).

Part 5: Applications & Mastery Check

🔢 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 $p\%$ : $r = 1 +$ (e.g. )

48, 24, 12, 6, \dots r = ?

1,\,-3,\,9,\,-27,\dots \quad r = \,?

💡 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 ( $2,1.5,1.67$ ).

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 so the sign comes out right.

Concept Check 🎯

Recursive vs. Explicit

There are two valid ways to define a geometric sequence:

Form	Rule	Says…
Recursive	$a_1 = \text{(given)},\quad a_n = r \cdot a_{n-1}$	"multiply the previous term by $r$ "
Explicit	$a_n = a_1 \cdot r^{\,n-1}$

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

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

💡 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:

Identify $a_1$ — the first term.
Find $r$ — divide the second term by the first.
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 $a_5$ . $\;=\,?$ . Find . . Find .

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

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

$k$	$3 \cdot 2^{k-1}$	value
$1$	$3 \cdot^{}$

So the sum is $3 + 6 + 12 + 24 = 45$ . Reading off $a_1 = 3$ , , , the formula agrees: . ✓

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 $S_4$ . $\;=\,?$ Find .

Condition	Behavior
$\lvert r \rvert < 1$	Converges — terms shrink to $0$ ; the sum is finite
$\lvert r \rvert \ge 1$	Diverges — terms don't shrink; no finite sum

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 ✓):

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

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

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:

Find $r$ .
If $|r| \ge 1$ → the series diverges; there is no finite sum.
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) $a_{1}$

r = 1 + \frac{p}{100}

+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

Goal	Formula
Common ratio	$r = \dfrac{a_{n}}{a_{n-1}}$
$n$ th term	$a_n = a_1 \cdot r^{\,n-1}$
Finite sum	$S_n = a_1 \dfrac{1 - r^{\,n}}{1 - r}$
Infinite sum ( $\lvert r\rvert<1$ )	$S_\infty = \dfrac{a_1}{1 - r}$

⚠️ 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 ratio in the denominator .

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.

Explicit:

a_n = 3 \cdot 2^{\,n-1}

a_1 = 100,\; r = \tfrac{1}{5}

S_4 = 3 \cdot \frac{1-2^4}{1-2} = 3 \cdot \frac{-15}{-1} = 45

\displaystyle\sum_{k=1}^{3} 2 \cdot 5^{\,k-1} \;=\,?

1,\,3,\,9,\,27,\,81,\dots

a_{1} = 6, r = \frac{1}{3} . S_{\infty} = ?

a_1 = 10,\; r = -\tfrac{1}{2}.\;\; S_\infty = \,?

(fraction or decimal ok)

Geometric Sequences and Series

Geometric Sequences and Series - Complete Interactive Lesson

Part 1: The Common Ratio

🔢 Geometric Sequences & Series

Topics in This Part

What Is a Geometric Sequence?

How to spot the ratio

Pinning Down rrr Exactly

Geometric vs. Arithmetic

Part 2: The nth-Term Formula

🔢 Geometric Sequences & Series

The Explicit Formula

Part 3: Finite Geometric Series

🔢 Geometric Sequences & Series

The Finite Sum Formula

Part 4: Infinite Geometric Series

🔢 Geometric Sequences & Series

When Does an Infinite Series Have a Sum?

Part 5: Applications & Mastery Check

🔢 Geometric Sequences & Series

Real-World Geometric Models

Compound situations as a ratio

Worked Example: 3, 6, 12, 24,…3,\,6,\,12,\,24,\dots3,6,12,24,… — find a7a_7a7​

Worked Example: a shrinking sequence

Worked Example: alternating signs

Recursive vs. Explicit

A Reliable Routine

Worked Example: sum the first 5 terms of 2, 6, 18,…2,\,6,\,18,\dots2,6,18,…

Worked Example: a fractional ratio

Sigma (Summation) Notation

Example: unpack ∑k=143⋅2 k−1\displaystyle\sum_{k=1}^{4} 3 \cdot 2^{\,k-1}k=1∑4​3⋅2k−1

Two Routes to the Same Answer

Worked Examples — Sum to Infinity

Example 1: 8+4+2+1+…8 + 4 + 2 + 1 + \dots8+4+2+1+…

Example 2: 9−3+1−13+…9 - 3 + 1 - \frac{1}{3} + \dots9−3+1−31​+…

Before You Compute, Check Convergence

Worked Example: bouncing ball

Quick Reference

You're Ready

Pinning Down $r$ Exactly

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

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

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

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

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