Pre-Calc 11 · Lesson 4

Geometric Series.

A geometric series adds up terms that keep multiplying, and one clever shift-and-subtract trick collapses that whole pile into a single tidy formula.

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By the end, you can

  • 1I can identify a geometric series and find its first term t1 and common ratio r.
  • 2I can derive and use Sn = t1(rⁿ - 1)/(r - 1) to find the sum of the first n terms.
  • 3I can use the variant Sn = (r*tn - t1)/(r - 1) when I know the last term but not how many terms there are.
  • 4I can solve for an unknown (t1, r, n, or Sn) given the others, including by setting up and solving equations.
  • 5I can model real situations like fan-out systems, tournaments, and bouncing balls with a geometric series and state the assumptions I made.
01 / CONCEPTS

The core ideas.

What a geometric series is

A geometric sequence multiplies by a fixed common ratio r each step: t1, t1*r, t1*r², and so on. A geometric series is what you get when you add those terms together instead of just listing them. For example 3 + 6 + 12 + 24 is a geometric series with r = 2.

Sn=t1+t1r+t1r2+⋯+t1rn−1S_n = t_1 + t_1 r + t_1 r^2 + \cdots + t_1 r^{n-1}

Why it works

A sequence is the list of stepping stones; a series is the total distance you have walked across them. Geometric just means each stone is a fixed multiple bigger (or smaller) than the last, so the gaps grow or shrink by a constant factor rather than a constant amount.

The shift-and-subtract derivation

To add the series fast, write Sn, then write r*Sn directly below it shifted over by one term. Almost every term lines up and cancels when you subtract, leaving only two survivors. Solving gives the sum formula.

Sn=t1(rn−1)r−1,r≠1S_n = \frac{t_1(r^n - 1)}{r - 1},\quad r \neq 1

Why it works

Multiplying by r slides the whole series one slot to the right, so r*Sn is nearly identical to Sn, just missing the first term and gaining one new term at the end. Subtract and the middle vanishes like a telescope: r*Sn - Sn = t1*rⁿ - t1. Factor out Sn and divide by (r - 1). That is the entire magic, and it is why r cannot be 1 (you would divide by zero, and a constant series should just be added the plain way).

The last-term variant of the formula

Sometimes you know the final term tn and the ratio r but not how many terms there are. Since tn = t1*rⁿ⁻¹, multiplying by r gives r*tn = t1*rⁿ. Substitute that into the sum formula to skip finding n entirely.

Sn=r tn−t1r−1,r≠1S_n = \frac{r\,t_n - t_1}{r - 1},\quad r \neq 1

Why it works

The two survivors of the subtraction were t1 and t1*rⁿ. Notice t1*rⁿ is exactly r times the last term. So if you already know the last term, you do not need to hunt for n at all. This is a huge shortcut for series written as t1 + ... + tn with a known endpoint.

Finding the common ratio r

Divide any term by the one before it: r = tₖ₊₁/tₖ. If that ratio is the same for every consecutive pair, the series is geometric. If two known terms are far apart, take a root: tₘ/tₖ = r^(m-k).

r=tk+1tkr = \frac{t_{k+1}}{t_k}

Why it works

Geometric growth is repeated multiplication, so the ratio between neighbours is the DNA of the series. If t6/t3 = r³, then the gap of 3 steps means three multiplications by r, so take the cube root to recover r. This is why a single ratio can tie distant terms together.

Solving for an unknown term count n

When n is unknown, use the general term tn = t1*rⁿ⁻¹ and the known endpoint to set up an exponential equation. Rewrite both sides as powers of the same base, then match exponents.

tn=t1r n−1t_n = t_1 r^{\,n-1}

Why it works

Equal bases force equal exponents: if 3⁹ = 3ⁿ⁻¹, the only way the two can match is n - 1 = 9. This turns a scary 'how many terms' question into simple counting once everything is expressed as powers of the base r.

Modelling and stating assumptions

Real situations like fan-out calls, tournament brackets, drug doses, and bouncing balls are geometric when each step is a fixed multiple of the last. Before trusting an answer, name the assumptions: that the ratio truly stays constant and that nothing in the real world interrupts the pattern.

Why it works

Math models are honest only when their assumptions hold. A bouncing ball loses exactly 60 percent of its height each bounce only in an idealized world; real air resistance and spin nudge it. Naming assumptions is what separates a thoughtful answer from a number that looks right but means little.

02 / WORKED EXAMPLES

From easy to tricky.

EX 1easy

Find the sum of the first 10 terms of the geometric series 4 + 12 + 36 + ...

1

Identify t1, r, n

t1 = 4, r = 12/4 = 3, n = 10.

2

Write the formula

Sₙ = t1(rⁿ - 1)/(r - 1).

3

Substitute

S10 = 4(3¹⁰ - 1)/(3 - 1) = 4(59048)/2.

4

Compute

= 4(59048)/2 = 236192/2 = 118096.

EX 2easy

Find the sum of the first 10 terms of a geometric series with t1 = 5 and r = 1/2.

1

Set up

S10 = 5((1/2)¹⁰ - 1)/((1/2) - 1).

2

Evaluate the power

(1/2)¹⁰ = 1/1024, so top inside = 1/1024 - 1 = -1023/1024.

3

Denominator

(1/2) - 1 = -1/2.

4

Divide

S10 = 5 * (-1023/1024) / (-1/2) = 5 * (-1023/1024) * (-2) = 5 * 1023/512 = 5115/512.

EX 3medium

Find the sum of the series 1/27 + 1/9 + 1/3 + ... + 729 using the last term.

1

Identify what is known

t1 = 1/27, r = 3, last term tn = 729. We do not know n yet, so use the variant formula.

2

Variant formula

Sₙ = (r*tn - t1)/(r - 1).

3

Substitute

= (3 * 729 - 1/27)/(3 - 1) = (2187 - 1/27)/2.

4

Combine

2187 = 59049/27, so top = 59049/27 - 1/27 = 59048/27. Then divide by 2.

5

Result

Sₙ = 59048/(27*2) = 59048/54 = 29524/27, about 1093.48.

EX 4medium

How many terms are in the series 4 + 12 + 36 + 108 + ... + tn if the sum is 4372?

1

Identify

t1 = 4, r = 3, Sn = 4372, find n.

2

Set up the sum equation

4372 = 4(3ⁿ - 1)/(3 - 1) = 4(3ⁿ - 1)/2 = 2(3ⁿ - 1).

3

Isolate the power

4372/2 = 3ⁿ - 1, so 2186 = 3ⁿ - 1, giving 3ⁿ = 2187.

4

Match powers of 3

3⁷ = 2187, so n = 7.

EX 5hard

A geometric series has common ratio r = 1/3 and the sum of its first 5 terms is 121. Find the first term, then write the first 5 terms.

1

Write the sum factor

S5 = t1((1/3)⁵ - 1)/((1/3) - 1). The bracket factor is ((1/243) - 1)/(-2/3) = (-242/243)/(-2/3) = (242/243)(3/2) = 121/81.

2

Solve for t1

121 = t1 * (121/81), so t1 = 121 * (81/121) = 81.

3

List terms

81, 81*(1/3) = 27, 27*(1/3) = 9, 9*(1/3) = 3, 3*(1/3) = 1.

4

Check

81 + 27 + 9 + 3 + 1 = 121. Correct.

EX 6hard

In a knockout tournament with 256 entries, every match eliminates one player and the winners advance until one champion remains. How many matches are played in total?

1

Model the rounds

Two players per match, so round 1 has 256/2 = 128 matches. Each later round has half as many, so r = 1/2. The final round is a single match, tn = 1.

2

Choose the formula

Use Sn = (r*tn - t1)/(r - 1) with t1 = 128, r = 1/2, tn = 1.

3

Substitute

= ((1/2)(1) - 128)/((1/2) - 1) = (1/2 - 128)/(-1/2) = (-255/2)/(-1/2).

4

Divide

= (-255/2)(-2/1) = 255.

5

Sanity check

In any knockout, every player except the champion loses exactly one match, so 256 - 1 = 255 matches. Matches the formula.

EX 7challenge

The third term of a geometric series is 9/4 and the sixth term is -16/81. Find the second term and the sum of the first 6 terms.

1

Relate distant terms

t6 = t3 * r³, so r³ = t6/t3 = (-16/81)/(9/4) = (-16/81)(4/9) = -64/729.

2

Take the cube root

r³ = -64/729 = (-4/9)³, so r = -4/9.

3

Find t1, then t2

t3 = t1*r², so t1 = (9/4)/(16/81) = (9/4)(81/16) = 729/64. Then t2 = t1*r = (729/64)(-4/9) = -81/16.

4

Sum of 6 terms

S6 = t1(r⁶ - 1)/(r - 1). With r = -4/9: r⁶ = 4096/531441, r⁶ - 1 = -527345/531441, r - 1 = -13/9. So S6 = (729/64)(-527345/531441)/(-13/9) = 40565/5184, about 7.8.

03 / COMMON TRAPS

Where students slip.

Trap

Using the arithmetic sum formula (adding a constant difference) on a geometric series.

Fix

Geometric series multiply by a ratio, they do not add a difference. Check by dividing consecutive terms: if t2/t1 = t3/t2 you have a geometric series and must use Sn = t1(rⁿ - 1)/(r - 1), not the arithmetic n/2(t1 + tn).

Trap

Thinking the exponent on r in the sum formula is (n - 1) instead of n.

Fix

The general TERM has exponent n - 1: tn = t1*rⁿ⁻¹. But the SUM formula has exponent n: Sn = t1(rⁿ - 1)/(r - 1). Keep them separate. The sum counts a full n factors of r in its top, the term only counts n - 1 multiplications from the start.

Trap

Sign and order errors with r - 1 when r is a fraction or negative, giving a wrong (often positive instead of negative) sum.

Fix

Keep the denominator as written, r - 1, not 1 - r. If r = 1/2 then r - 1 = -1/2, and if r = -4 then r - 1 = -5. Carry the negative through carefully; flipping it silently flips the sign of your whole answer.

Trap

Plugging the number of people, levels, or rounds straight in as n without checking what n actually counts.

Fix

Decide what each term represents first. In a 256-entry knockout tournament the terms are matches per round (128, 64, ...), so t1 = 128, not 256. Always map the story to t1, r, and either n or tn before computing.

04 / QUIZ

Test yourself.

The Test

30 problems across three tiers. Auto-graded on submit. Hints available before you submit, full solutions after.

0/30Answered
Core

Core (12)

C1

Which series is geometric?

Need a hint?
  • In a geometric series the ratio of consecutive terms is constant. Test t2/t1 against t3/t2.
  • For 4 + 24 + 144 + 864: 24/4 = 6, 144/24 = 6, 864/144 = 6, a constant ratio. For 3 + 9 + 18 + 54: 9/3 = 3 but 18/9 = 2, not constant.
C2

For the geometric series 6 + 9 + 13.5 + ..., the first term is t1 = ___ and the common ratio is r = ___.

Need a hint?
  • The first term is just the first number written.
  • Divide the second term by the first: 9/6.
C3

Find S10 for the geometric series with t1 = 12, r = 2, n = 10. Give an exact value.

Need a hint?
  • Use Sn = t1(rⁿ - 1)/(r - 1).
  • 2¹⁰ = 1024, so S10 = 12(1024 - 1)/(2 - 1).
C4

Find S8 for the geometric series with t1 = 27, r = 1/3, n = 8. Give an exact fraction.

Need a hint?
  • Sn = t1(rⁿ - 1)/(r - 1) with r = 1/3.
  • (1/3)⁸ = 1/6561 and r - 1 = -2/3.
C5

Find S10 for the geometric series with t1 = 1/256, r = -4, n = 10. Give an exact fraction.

Need a hint?
  • Sn = t1(rⁿ - 1)/(r - 1). Note (-4)¹⁰ is positive.
  • (-4)¹⁰ = 1048576 and r - 1 = -5.
C6

Find S12 for the geometric series with t1 = 72, r = 1/2, n = 12. Give an exact fraction.

Need a hint?
  • Use the sum formula with r = 1/2.
  • (1/2)¹² = 1/4096 and r - 1 = -1/2.
C7

Find the sum: 27 + 9 + 3 + ... + 1/243.

Need a hint?
  • t1 = 27, r = 1/3, last term tn = 1/243. Use the variant Sn = (r*tn - t1)/(r - 1).
  • r*tn = (1/3)(1/243) = 1/729; r - 1 = -2/3.
C8

Find Sₙ for the geometric series with t1 = 5, tn = 81 920, r = 4.

Need a hint?
  • You know the last term, so use Sn = (r*tn - t1)/(r - 1).
  • = (4*81920 - 5)/(4 - 1).
C9

Find Sₙ for the geometric series with t1 = 3, tn = 46 875, r = -5.

Need a hint?
  • Use the last-term formula Sn = (r*tn - t1)/(r - 1) with r = -5.
  • r*tn = (-5)(46875) = -234375; r - 1 = -6.
C10

Which formula should you use when you know t1, r, and the last term tn, but NOT the number of terms n?

Need a hint?
  • Which formula has no n in it at all?
  • The variant comes from substituting r*tn = t1*rⁿ into the standard sum formula.
C11

Find the sum of the first 8 terms of 5 + 15 + 45 + ...

Need a hint?
  • t1 = 5, r = 3, n = 8.
  • 3⁸ = 6561.
C12

Find the sum of the first 8 terms of a geometric series with t1 = 64 and r = 1/4. Give an exact fraction.

Need a hint?
  • Use the sum formula with r = 1/4.
  • (1/4)⁸ = 1/65536 and r - 1 = -3/4.
Apply

Apply (12)

A1

The sum of 4 + 12 + 36 + 108 + ... + tn is 4372. How many terms are in the series?

Need a hint?
  • t1 = 4, r = 3. Set 4372 = 4(3ⁿ - 1)/(3 - 1).
  • Simplify to 2186 = 3ⁿ - 1, so 3ⁿ = 2187. Recognize this as a power of 3.
A2

Find n if 3 + 3² + 3³ + ... + 3ⁿ = 9840.

Need a hint?
  • This is geometric with t1 = 3 and r = 3.
  • 9840 = 3(3ⁿ - 1)/(3 - 1) = (3/2)(3ⁿ - 1). Solve for 3ⁿ.
A3

The common ratio of a geometric series is 1/3 and the sum of the first 5 terms is 121. Find the first term and write the first 5 terms.

Need a hint?
  • S5 = t1((1/3)⁵ - 1)/((1/3) - 1). Compute the bracket factor first.
  • The factor equals 121/81, so t1(121/81) = 121.
A4

For a geometric series, Sn = 33, tn = 48, and r = -2. Find the first term t1.

Need a hint?
  • Use Sn = (r*tn - t1)/(r - 1) and solve for t1.
  • 33 = ((-2)(48) - t1)/(-2 - 1) = (-96 - t1)/(-3).
A5

A fan-out system: the person in charge contacts 4 people, each of them contacts 4 more, and so on. Write the series for the number of people contacted at each level, then find how many people are notified after 10 levels.

Need a hint?
  • Each level multiplies by 4, so t1 = 4 and r = 4.
  • Sum the first 10 terms: S10 = 4(4¹⁰ - 1)/(4 - 1).
A6

Celia runs 25 km in week 1 and increases her weekly distance by 10% each week, modelled by 25 + 25(1.1) + 25(1.1)² + .... How far will she have run in total after completing 15 weeks? Round to the nearest tenth of a kilometre.

Need a hint?
  • t1 = 25, r = 1.1, n = 15.
  • S15 = 25(1.1¹⁵ - 1)/(1.1 - 1).
A7

A line of 10 beads is laid out where each successive bead has a diameter 3/4 of the previous one. The first bead is 24 mm across. Find the total length of the line of beads, to the nearest millimetre.

Need a hint?
  • t1 = 24, r = 3/4, n = 10.
  • S10 = 24((3/4)¹⁰ - 1)/((3/4) - 1).
A8

In a Scrabble tournament with 512 participants, losers are eliminated each round until one champion remains. How many matches are played in total?

Need a hint?
  • Round 1 has 512/2 = 256 matches, then halves each round. t1 = 256, r = 1/2, tn = 1.
  • Use Sn = (r*tn - t1)/(r - 1), or note that every player but the champion loses exactly once.
A9

For a geometric series, Sn = 443, n = 6, and r = 1/3. Find the first term t1, to the nearest tenth.

Need a hint?
  • Compute the bracket factor S/t1 = ((1/3)⁶ - 1)/((1/3) - 1) = 364/243.
  • Then t1 = 443 / (364/243).
A10

A tennis ball is dropped from 20 m and bounces to 40% of its previous height each bounce. Find the total vertical distance travelled (downward drops plus upward bounces) by the time it hits the floor for the sixth time, to the nearest tenth of a metre.

Need a hint?
  • Downward drops form 20 + 8 + 3.2 + ... (six terms, r = 0.4). Upward bounces are the same series but only the five bounces between hits.
  • Down: S6 = 20(0.4⁶ - 1)/(0.4 - 1). Up: 8(0.4⁵ - 1)/(0.4 - 1). Add them.
A11

An advertising campaign reaches 1000 people at the start, and the number of newly aware people grows by 40% every 10 days. Find the total number of people aware of the product after 100 days.

Need a hint?
  • t1 = 1000, r = 1.4. Over 100 days there are 10 increases, so there are 11 terms (the start plus 10 growth periods).
  • S11 = 1000(1.4¹¹ - 1)/(1.4 - 1).
A12

What is the second term of a geometric series whose third term is 9/4 and sixth term is -16/81? Then find the sum of the first 6 terms, to the nearest tenth.

Need a hint?
  • t6 = t3*r³, so r³ = (-16/81)/(9/4). Take the cube root.
  • r = -4/9. Find t1 from t3 = t1*r², then t2 = t1*r, then S6 = t1(r⁶ - 1)/(r - 1).
Extension

Extension (6)

E1

A patient takes a 200 mg ampicillin tablet every 4 hours, and about 12% of the drug in the body at the start of each 4-hour period remains at the end of it. The amount present after the nth dose is a geometric series with t1 = 200 and r = 0.12. How much ampicillin is in the body right after the third tablet, to the nearest tenth of a milligram?

Need a hint?
  • Right after the 3rd dose the body holds the newest 200 mg plus what remains of the earlier two doses. Sum three terms: t1 = 200, r = 0.12, n = 3.
  • S3 = 200(0.12³ - 1)/(0.12 - 1). Compute 0.12³ = 0.001728.
E2

A fractal starts with a circle of radius 8 cm. Each new circle has half the radius of the previous one. Through five circles, find the exact sum of the areas (as a multiple of pi).

Need a hint?
  • Area = pi*r². With radii 8, 4, 2, 1, 1/2, the areas form a geometric series. Find r for the AREAS.
  • Halving the radius quarters the area, so the area ratio is 1/4. First area = 64pi, n = 5.
E3

Three numbers a, b, c form a geometric series with a + b + c = 35 and abc = 1000. Find a, b, and c.

Need a hint?
  • In a geometric triple, the product abc equals b³ because a = b/r and c = b*r.
  • From b³ = 1000, b = 10. Then b/r + 10 + b*r = 35 gives 10/r + 10r = 25; solve the quadratic in r.
E4

The sum of the first 7 terms of a geometric series is 89, and the sum of the first 8 terms is 104. What is the value of the eighth term?

Need a hint?
  • How are S8 and S7 related to the individual eighth term?
  • S8 = S7 + t8, so t8 = S8 - S7.
E5

Tom modelled monarch butterfly growth as one butterfly producing 400, each of those producing 400, and so on, computing S5 = 1(400⁵ - 1)/(400 - 1) is about 2.566 x 10¹⁰ for the fifth generation. State one assumption he made and explain whether his sum answers the question 'how many butterflies are in the fifth generation.'

Need a hint?
  • What does each term t1*r^(k-1) count, and what does the SUM Sn count versus a single term tn?
  • The number in just the fifth generation is the single term t5 = 400⁴, while Sn adds up every generation from the first.
E6

A student claims that for any geometric series, S8 = 2 * S4. Decide whether this is always true, sometimes true, or never true, and justify with the formula.

Need a hint?
  • Write S8 and S4 with the sum formula and form the ratio S8/S4.
  • S8/S4 = (r⁸ - 1)/(r⁴ - 1) = r⁴ + 1. Set that equal to 2.