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Mathematics AI · HL · Unit 1: Number and Algebra

Unit 1Topic 1.11Key Concepts69 min read

Infinite geometric sequences and series

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Meaning of an infinite geometric series

A geometric sequence lists terms obtained by repeatedly multiplying by a constant common ratio. A geometric series adds those terms. An infinite geometric series has no final term and takes the form a+ar+ar2+ar3+⋯a+ar+ar^2+ar^3+\cdots.

We do not add infinitely many terms one at a time. Instead, we study the partial sums S1,S2,S3,…S_1,S_2,S_3,\ldots. If these finite sums approach one fixed number, the infinite series converges and that limiting number is called S∞S_\infty.

For example, 12+14+18+⋯\frac12+\frac14+\frac18+\cdots has partial sums 12,34,78,1516,…\frac12,\frac34,\frac78,\frac{15}{16},\ldots. Each partial sum is closer to 1, so the infinite series has sum 1 even though no finite partial sum equals 1.

Key terms

Infinite geometric series
An endless sum a+ar+ar2+⋯a+ar+ar^2+\cdots whose consecutive terms have a constant ratio.
Partial sum
The finite sum SnS_n of the first nn terms.
Limit
The value approached by the partial sums as nn increases without bound.
Sum to infinity
The limiting value S∞S_\infty when the partial sums converge.

Worked examples

Sequence or series?

Classify 6,3,1.5,…6,3,1.5,\ldots and 6+3+1.5+⋯6+3+1.5+\cdots.

  1. 1Commas indicate an ordered list of terms, so the first expression is a sequence.
  2. 2Plus signs indicate addition, so the second expression is a series.
  3. 3Both are geometric because each new term is obtained by multiplying by 12\frac{1}{2}.

Answer: The first is an infinite geometric sequence; the second is an infinite geometric series.

Build partial sums

Find the first four partial sums of 12+14+18+⋯\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots.

  1. 1S1=12S_1=\frac{1}{2}.
  2. 2S2=12+14=34S_2=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}.
  3. 3S3=34+18=78S_3=\frac{3}{4}+\frac{1}{8}=\frac{7}{8}.
  4. 4S4=78+116=1516S_4=\frac{7}{8}+\frac{1}{16}=\frac{15}{16}.

Answer: S1=12,S2=34,S3=78,S4=1516S_1=\frac{1}{2},S_2=\frac{3}{4},S_3=\frac{7}{8},S_4=\frac{15}{16}; they approach 1.

Write the general term

For 10+5+2.5+⋯10+5+2.5+\cdots, write the nnth term and the first nn-term sum.

  1. 1The first term is a=10a=10 and the ratio is r=12r=\frac{1}{2}.
  2. 2The nnth term is un=arn−1u_n=ar^{n-1}.
  3. 3The partial sum is Sn=u1+u2+⋯+unS_n=u_1+u_2+\cdots+u_n.

Answer: un=10(12)n−1u_n=10(\frac{1}{2})^{n-1} and Sn=∑k=0n−110(12)kS_n=\sum\limits_{k=0}^{n-1}10(\frac{1}{2})^k.

Rules and formulas

Infinite geometric series

a+ar+ar2+ar3+⋯a+ar+ar^2+ar^3+\cdots
aa
the first term
rr
the common ratio

Partial sum notation

Sn=∑k=0n−1arkS_n=\sum\limits_{k=0}^{n-1}ar^k

First term, common ratio, and geometric structure

The first term is aa. For consecutive non-zero terms, calculate the common ratio using later term divided by earlier term: r=un+1unr=\frac{u_{n+1}}{u_n}. Check more than one pair when deciding whether a displayed series is geometric.

A positive ratio keeps the signs the same. A negative ratio makes signs alternate. The magnitude ∣r∣|r| tells us whether the term sizes shrink, stay constant, or grow.

The general term is un=arn−1u_n=ar^{n-1}. A necessary condition for any infinite series to converge is un→0u_n\to0; for a non-zero geometric series, this happens exactly when ∣r∣<1|r|<1.

Key terms

First term
The initial term a=u1a=u_1 of the geometric series.
Common ratio
The constant multiplier r=un+1unr=\frac{u_{n+1}}{u_n} between consecutive non-zero terms.
Alternating series
A series whose term signs alternate, usually because r<0r<0.
Term test
If the terms do not approach zero, the infinite series cannot converge.

Worked examples

Find a positive ratio

Find aa and rr for 24+12+6+3+⋯24+12+6+3+\cdots.

  1. 1The first term is a=24a=24.
  2. 2Divide consecutive terms: 1224=12\frac{12}{24}=\frac{1}{2} and 612=12\frac{6}{12}=\frac{1}{2}.

Answer: a=24a=24 and r=12r=\frac{1}{2}.

Find a negative ratio

Find aa and rr for 18−6+2−23+⋯18-6+2-\frac{2}{3}+\cdots.

  1. 1The first term is a=18a=18.
  2. 2(−6)18=−13\frac{(-6)}{18}=\frac{-1}{3}, and 2(−6)=−13\frac{2}{(-6)}=\frac{-1}{3}.
  3. 3The negative ratio explains the alternating signs.

Answer: a=18a=18 and r=−13r=\frac{-1}{3}.

Find an unknown term

A geometric series has a=64a=64 and r=12r=\frac{1}{2}. Find its sixth term.

  1. 1Use un=arn−1u_n=ar^{n-1}.
  2. 2u6=64(12)5=6432u_6=64(\frac{1}{2})^5=\frac{64}{32}.

Answer: u6=2u_6=2.

Rules and formulas

Common ratio

r=un+1unr=\frac{u_{n+1}}{u_n}

Geometric term

un=arn−1u_n=ar^{n-1}

Convergence and the condition |r| < 1

A non-zero infinite geometric series converges exactly when ∣r∣<1|r|<1, equivalently −1<r<1-1<r<1. Repeated multiplication then makes rn→0r^n\to0, so the terms shrink toward zero and the partial sums settle at a finite limit.

If ∣r∣>1|r|>1, term magnitudes grow. If r=1r=1, the same non-zero term is added forever. If r=−1r=-1, the partial sums oscillate between two values. In every case ∣r∣≥1|r|\ge1, a non-zero geometric series diverges.

A negative ratio does not automatically cause divergence. If −1<r<0-1<r<0, term signs alternate but their magnitudes shrink, so the partial sums oscillate ever closer to one limit.

The zero series 0+0+0+⋯0+0+0+\cdots is a trivial exception that sums to zero. IB questions normally use a non-zero first term, for which ∣r∣<1|r|<1 is the complete convergence test.

Key terms

Convergent
The sequence of partial sums approaches a finite limit.
Divergent
The partial sums do not approach one finite value.
Magnitude
Absolute value; ∣r∣|r| measures the size of the repeated multiplier without its sign.
Oscillation
Movement back and forth between values, as occurs with negative ratios.

Worked examples

Classify several ratios

Decide whether geometric series with r=0.8,−0.6,1,−1.02r=0.8,-0.6,1,-1.02 converge.

  1. 1Calculate each magnitude.
  2. 2∣0.8∣=0.8<1|0.8|=0.8<1, so it converges.
  3. 3∣−0.6∣=0.6<1|-0.6|=0.6<1, so it converges.
  4. 4∣1∣=1|1|=1 and ∣−1.02∣=1.02|-1.02|=1.02, so those series diverge.

Answer: The first two converge; the last two diverge.

Alternating convergence

Does 12−3+0.75−0.1875+⋯12-3+0.75-0.1875+\cdots converge?

  1. 1The common ratio is (−3)12=−14\frac{(-3)}{12}=\frac{-1}{4}.
  2. 2∣r∣=14<1|r|=\frac{1}{4}<1.
  3. 3Alternating signs do not prevent convergence because the magnitudes shrink.

Answer: Yes, the series converges.

Use the term test

Explain why 2+3+4.5+6.75+⋯2+3+4.5+6.75+\cdots diverges.

  1. 1The common ratio is r=32r=\frac{3}{2}.
  2. 2The terms 2(32)n−12(\frac{3}{2})^{n-1} grow rather than approach zero.
  3. 3A series whose terms do not approach zero cannot converge.

Answer: It diverges because ∣r∣=1.5≥1|r|=1.5\ge1 and un↛0u_n\not\to0.

Rules and formulas

Geometric convergence condition

∣r∣<1⟺−1<r<1|r|<1\quad\Longleftrightarrow\quad -1<r<1

Vanishing powers

∣r∣<1⟹rn→0|r|<1\quad\Longrightarrow\quad r^n\to0

Deriving and finding the sum to infinity

For r≠1r\ne1, the finite geometric sum is Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}. If ∣r∣<1|r|<1, then rn→0r^n\to0, so the limit is S∞=a1−rS_\infty=\frac{a}{1-r}.

The finite formula can be derived algebraically. Write Sn=a+ar+⋯+arn−1S_n=a+ar+\cdots+ar^{n-1}, multiply by rr, and subtract the aligned equations. All middle terms cancel, leaving (1−r)Sn=a(1−rn)(1-r)S_n=a(1-r^n).

A positive ratio gives partial sums that move toward the limit from one side when a>0a>0. A negative ratio often makes the partial sums alternate above and below the limit.

Key terms

Finite geometric sum
Sn=a(1−rn)(1−r)S_n=\frac{a(1-r^n)}{(1-r)}, the sum of the first nn terms when r≠1r\ne1.
Limiting sum
The value obtained when the finite partial sums approach a limit as n→∞n\to\infty.
Cancellation
The disappearance of matching middle terms when rSnrS_n is subtracted from SnS_n.

Worked examples

Positive sum to infinity

Find 12+6+3+1.5+⋯12+6+3+1.5+\cdots.

  1. 1a=12a=12 and r=12r=\frac{1}{2}.
  2. 2Since ∣r∣=12<1|r|=\frac{1}{2}<1, the series converges.
  3. 3S∞=12(1−12)S_\infty=\frac{12}{(1-\frac{1}{2})}.

Answer: S∞=24S_\infty=24.

Alternating sum to infinity

Find 9−3+1−13+⋯9-3+1-\frac{1}{3}+\cdots.

  1. 1a=9a=9 and r=−13r=\frac{-1}{3}.
  2. 2Since ∣r∣=13<1|r|=\frac{1}{3}<1, use the infinite-sum formula.
  3. 3S∞=9[1−(−13)]=9(43)S_\infty=\frac{9}{[1-(\frac{-1}{3})]}=\frac{9}{(\frac{4}{3})}.

Answer: S∞=274S_\infty=\frac{27}{4}.

Exact fractional sum

Find 5+1+15+125+⋯5+1+\frac{1}{5}+\frac{1}{25}+\cdots.

  1. 1The ratio is r=15r=\frac{1}{5}, which satisfies ∣r∣<1|r|<1.
  2. 2S∞=5(1−15)=5(45)S_\infty=\frac{5}{(1-\frac{1}{5})}=\frac{5}{(\frac{4}{5})}.

Answer: S∞=254S_\infty=\frac{25}{4}.

Rules and formulas

Finite geometric sum

Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

Sum to infinity

S∞=a1−r,∣r∣<1S_\infty=\frac{a}{1-r},\qquad |r|<1
S∞S_\infty
the limiting sum
aa
the first term
rr
the common ratio with magnitude less than 1

Finding unknown terms, the first term, or the common ratio

For ∣r∣<1|r|<1, the equation S∞=a(1−r)S_\infty=\frac{a}{(1-r)} gives a=S∞(1−r)a=S_\infty(1-r). If S∞≠0S_\infty\ne0, it also gives r=1−aS∞r=1-\frac{a}{S_\infty}. Check that a calculated ratio satisfies ∣r∣<1|r|<1. The zero series does not determine a unique ratio from its first term and sum.

A known term adds the equation un=arn−1u_n=ar^{n-1}. Use it together with the sum equation, then check all algebraic candidates because polynomial equations can produce more than one ratio.

When the displayed terms contain an unknown, form the ratio between consecutive terms. A true geometric series must use the same ratio throughout.

Key terms

Simultaneous conditions
Two equations, such as a term formula and an infinite-sum formula, used together to find unknown parameters.
Candidate ratio
A possible solution for rr that must be checked against the original series and ∣r∣<1|r|<1.
Rearrangement
Algebraically isolating a required variable without changing the equality.

Worked examples

Find the first term

A convergent geometric series has S∞=30S_\infty=30 and r=0.4r=0.4. Find aa.

  1. 1Use a=S∞(1−r)a=S_\infty(1-r).
  2. 2a=30(1−0.4)=30(0.6)a=30(1-0.4)=30(0.6).

Answer: a=18a=18.

Find the common ratio

A convergent geometric series has first term 14 and sum 20. Find rr.

  1. 1Use r=1−aS∞r=1-\frac{a}{S_\infty}.
  2. 2r=1−1420=1−0.7=0.3r=1-\frac{14}{20}=1-0.7=0.3.
  3. 3Check ∣0.3∣<1|0.3|<1.

Answer: r=0.3r=0.3.

Use the second term and the sum

A convergent geometric series has second term 6 and sum 24. Find aa and rr.

  1. 1The conditions are ar=6ar=6 and a(1−r)=24\frac{a}{(1-r)}=24.
  2. 2Substitute a=24(1−r)a=24(1-r) into ar=6ar=6: 24r(1−r)=624r(1-r)=6.
  3. 3This simplifies to 4r2−4r+1=0=(2r−1)24r^2-4r+1=0=(2r-1)^2, so r=12r=\frac{1}{2}.
  4. 4Then a=24(1−12)=12a=24(1-\frac{1}{2})=12.

Answer: a=12a=12 and r=12r=\frac{1}{2}.

Unknown inside the terms

The convergent series 5+x+x25+⋯5+x+\frac{x^2}{5}+\cdots has sum 254\frac{25}{4}. Find xx.

  1. 1The common ratio is r=x5r=\frac{x}{5}.
  2. 2Use 5[1−(x5)]=254\frac{5}{[1-(\frac{x}{5})]}=\frac{25}{4}.
  3. 3Cross-multiply: 20=25−5x20=25-5x, so x=1x=1.
  4. 4Then r=15r=\frac{1}{5}, which satisfies convergence.

Answer: x=1x=1.

Rules and formulas

First term from the sum

a=S∞(1−r)a=S_\infty(1-r)

Ratio from first term and sum

r=1−aS∞,S∞≠0r=1-\frac{a}{S_\infty},\qquad S_\infty\ne0

Term condition

un=arn−1u_n=ar^{n-1}

Convergent versus divergent series and remainders

A convergent series has a finite sum and can be approximated by a large partial sum. The difference between the exact limiting sum and SnS_n is the remainder RnR_n.

For a convergent geometric series, the first omitted term after SnS_n is arnar^n. The remaining tail is itself a geometric series, so Rn=arn(1−r)R_n=\frac{ar^n}{(1-r)}.

A divergent series has no finite S∞S_\infty. Writing a numerical value from a(1−r)\frac{a}{(1-r)} when ∣r∣≥1|r|\ge1 is an invalid application of the formula, not a strange sum.

Key terms

Remainder
Rn=S∞−SnR_n=S_\infty-S_n, the infinite tail after the first nn terms.
First omitted term
The term arnar^n immediately after the final term included in SnS_n.
Approximation error
The size ∣Rn∣|R_n| of the difference between a partial sum and the limiting sum.

Worked examples

Find a remainder

For 20+8+3.2+⋯20+8+3.2+\cdots, find the sum remaining after the first four terms.

  1. 1Here a=20a=20, r=0.4r=0.4, and the first omitted term is 20(0.4)420(0.4)^4.
  2. 2Use R4=20(0.4)4(1−0.4)R_4=\frac{20(0.4)^4}{(1-0.4)}.
  3. 3Simplify exactly or evaluate with full precision.

Answer: R4=6475≈0.853R_4=\frac{64}{75}\approx0.853.

Explain divergence at r=1r = 1

Explain why 4+4+4+⋯4+4+4+\cdots diverges.

  1. 1Here r=1r=1, so each partial sum adds another 4.
  2. 2Sn=4nS_n=4n, which grows without bound.
  3. 3The denominator in a(1−r)\frac{a}{(1-r)} would be zero, confirming the formula is unavailable.

Answer: The series diverges.

Explain oscillation at r=−1r = -1

Describe the partial sums of 7−7+7−7+⋯7-7+7-7+\cdots.

  1. 1The partial sums are 7,0,7,0,…7,0,7,0,\ldots.
  2. 2They do not approach one value.
  3. 3Because ∣r∣=1|r|=1, the series fails the convergence condition.

Answer: The series diverges by oscillation.

Rules and formulas

Remainder after n terms

Rn=S∞−Sn=arn1−rR_n=S_\infty-S_n=\frac{ar^n}{1-r}

Approximation relation

S∞=Sn+RnS_\infty=S_n+R_n

Applications and word problems

Infinite geometric series model repeated contributions that become smaller by a constant factor. The model is appropriate only when the repetition can continue conceptually and the ratio has magnitude less than 1.

For a bouncing ball, count the initial drop once. Each rebound height is then travelled twice—upward and downward—so the distance model is D=h+2hq+2hq2+⋯D=h+2hq+2hq^2+\cdots.

Recurring decimals are infinite geometric series because each repeated digit block shifts by a fixed power of 10. Repeated-dose models work similarly when a constant fraction remains before each new dose.

Key terms

Rebound factor
The constant fraction qq of the previous height reached after each bounce.
Recurring decimal
A decimal with a digit block that repeats forever.
Steady-state amount
The limiting total approached after many repeated additions and proportional reductions.

Worked examples

Bouncing-ball distance

A ball is dropped from 10 m and rebounds to 60% of each previous height. Find its total vertical distance.

  1. 1The initial drop contributes 10 m.
  2. 2The rebound heights form 6+3.6+2.16+⋯6+3.6+2.16+\cdots, with sum 6(1−0.6)=15\frac{6}{(1-0.6)}=15.
  3. 3Each rebound height is travelled once up and once down.
  4. 4Total distance D=10+2(15)D=10+2(15).

Answer: D=40 mD=40\text{ m}.

Recurring decimal

Write 0.36‾0.\overline{36} as a fraction.

  1. 1Write 0.36‾=36100+361002+361003+⋯0.\overline{36}=\frac{36}{100}+\frac{36}{100^2}+\frac{36}{100^3}+\cdots.
  2. 2Here a=36100a=\frac{36}{100} and r=1100r=\frac{1}{100}.
  3. 3S∞=(36100)(1−1100)=3699S_\infty=\frac{(\frac{36}{100})}{(1-\frac{1}{100})}=\frac{36}{99}.
  4. 4Simplify the fraction.

Answer: 0.36‾=4110.\overline{36}=\frac{4}{11}.

Repeated medicine dose

A patient receives 80 mg each day. Immediately before the next dose, 25% of the previous amount remains. Find the long-term amount immediately after a dose.

  1. 1The contributions present just after many doses are 80+80(0.25)+80(0.25)2+⋯80+80(0.25)+80(0.25)^2+\cdots.
  2. 2The ratio is 0.25<10.25<1.
  3. 3S∞=80(1−0.25)S_\infty=\frac{80}{(1-0.25)}.

Answer: The amount approaches 3203 mg≈106.7 mg\frac{320}{3}\text{ mg}\approx106.7\text{ mg}.

Rules and formulas

Bouncing-ball distance

D=h+2hq1−q=h(1+q1−q)D=h+\frac{2hq}{1-q}=h\left(\frac{1+q}{1-q}\right)

Repeated fixed contribution

A∞=d1−r,∣r∣<1A_\infty=\frac{d}{1-r},\qquad |r|<1
AnA_n
amount immediately after addition n in the recurrence A_(n+1)=r AnA_{n}+d
dd
fixed addition after each retention step; just before the next addition the limiting amount is rd1−r\frac{rd}{1-r}

Two-digit recurring decimal

0.ab‾=10a+b99,a,b∈{0,1,…,9}0.\overline{ab}=\frac{10a+b}{99},\qquad a,b\in\{0,1,\ldots,9\}
abab
the recurring two-digit block, not the product of a and b

Steady-state and repeated-process models

Many applications repeatedly retain a fixed fraction of an existing amount and then add a fixed contribution. If An+1=rAn+dA_{n+1}=rA_n+d with ∣r∣<1|r|<1, the long-term level LL satisfies L=rL+dL=rL+d, so L=d(1−r)L=\frac{d}{(1-r)}.

The same result can be viewed as an infinite geometric series: a new contribution is followed by retained amounts with factors of rr. Be clear about the observation time—immediately before or immediately after each new contribution—because the first term changes.

Perpetuities provide a financial application. If a payment PP is made at the end of every period and the effective interest rate per period is i>0i>0, its present value is P(1+i)+P(1+i)2+⋯=Pi\frac{P}{(1+i)}+\frac{P}{(1+i)^2}+\cdots=\frac{P}{i}.

A valid model needs a constant ratio, an unlimited conceptual repetition, and a quantity whose contributions can be added. Check whether physical limits, changing rates, or a finite time horizon make the infinite model unrealistic.

Key terms

Steady state
A limiting level unchanged by the next retain-and-add cycle.
Recurrence model
A rule such as An+1=rAn+dA_{n+1}=rA_n+d that generates each state from the previous state.
Perpetuity
An idealized sequence of equal payments continuing indefinitely.
Observation time
The point within each repeated cycle at which the modeled amount is measured.

Worked examples

Find a steady-state amount

A treatment adds 60 mg after each cycle, while 30% of the previous amount remains. Find the long-term amount immediately after a dose.

  1. 1Let An+1=0.3An+60A_{n+1}=0.3A_n+60.
  2. 2At equilibrium, L=0.3L+60L=0.3L+60.
  3. 3Thus 0.7L=600.7L=60, so L=600.7L=\frac{60}{0.7}.
  4. 4The ratio 0.3 has magnitude less than 1, so the steady state is stable.

Answer: L=6007 mg≈85.7 mgL=\frac{600}{7}\text{ mg}\approx85.7\text{ mg}.

Value a perpetuity

A scholarship pays $2400 at the end of every year. Find its present value at an effective annual interest rate of 6%.

  1. 1The discounted payments form 24001.06+2400(1.06)2+⋯\frac{2400}{1.06}+\frac{2400}{(1.06)^2}+\cdots.
  2. 2The first term is a=24001.06a=\frac{2400}{1.06} and the ratio is r=11.06r=\frac{1}{1.06}.
  3. 3Since ∣r∣<1|r|<1, sum the series or use PV=PiPV=\frac{P}{i}.

Answer: PV=24000.06=$40 000PV=\frac{2400}{0.06}=\$40\,000.

Distinguish a sequence from accumulated contributions

A tank initially contains 100 L of a marked solution. Each cycle removes 20% of what remains and replaces it with unmarked liquid. Find the total marked solution eventually removed.

  1. 1The amounts removed are 20,16,12.8,…20,16,12.8,\ldots, with a=20a=20 and r=0.8r=0.8.
  2. 2Since ∣0.8∣<1|0.8|<1, the accumulated removal is 20(1−0.8)\frac{20}{(1-0.8)}.
  3. 3This sum agrees with conservation: eventually all 100 L of the original marked solution is removed conceptually.

Answer: 100 L100\text{ L}.

Rules and formulas

Retain-and-add recurrence

An+1=rAn+dA_{n+1}=rA_n+d
rr
fraction retained per cycle
dd
fixed contribution per cycle

Stable long-term level

L=d1−r,∣r∣<1L=\frac{d}{1-r},\qquad |r|<1

Present value of an end-of-period perpetuity

PV=Pi,i>0PV=\frac{P}{i},\qquad i>0
PP
payment per period
ii
effective interest rate per period

Calculator/GDC methods and exam strategy

A GDC has no special infinity term to enter. Use the exact formula after checking ∣r∣<1|r|<1, or calculate partial sums for increasing values of nn to observe the limiting behaviour.

A finite table of Sn=a(1−rn)(1−r)S_n=\frac{a(1-r^n)}{(1-r)} illustrates numerical behaviour but does not prove convergence. Prove rn→0r^n\to0 when ∣r∣<1|r|<1. For a≠0a\ne0 and −1<r<0-1<r<0, compare consecutive odd and even indices, such as n=9,10,11,12n=9,10,11,12, to see partial sums alternate around the limit.

When an unknown ratio produces a nonlinear equation, use a numerical solver or graph intersection. Restrict the search to −1<r<1-1<r<1, record the equation entered, and verify each reported root in the original conditions.

Keep exact fractions through the algebra. If the context requires a decimal, use full GDC precision and round only the final answer.

Key terms

Numerical solver
A GDC tool that approximates roots of an equation over a specified interval.
Graph intersection
A point where two plotted expressions have equal values.
Convergence table
A table of partial sums for increasing nn used to observe approach toward a limit.
Guard digits
Extra calculator precision retained before final rounding.

Worked examples

Investigate partial sums

Use partial sums to investigate 8+2.4+0.72+⋯8+2.4+0.72+\cdots.

  1. 1Here a=8a=8 and r=0.3r=0.3.
  2. 2Enter Sn=8(1−0.3n)(1−0.3)S_n=\frac{8(1-0.3^n)}{(1-0.3)}.
  3. 3A GDC gives S5≈11.4008S_5\approx11.4008, S10≈11.4285S_{10}\approx11.4285, and S20≈11.4286S_{20}\approx11.4286.
  4. 4Compare with S∞=80.7S_\infty=\frac{8}{0.7}.

Answer: The partial sums approach 807≈11.4286\frac{80}{7}\approx11.4286.

Use a solver for an unknown ratio

A series has a=10a=10, u4=1.25u_4=1.25, and converges. Find rr.

  1. 1Use u4=ar3u_4=ar^3: 10r3=1.2510r^3=1.25.
  2. 2Solve r3=0.125r^3=0.125 with exact reasoning or a GDC.
  3. 3The real solution is r=0.5r=0.5, which satisfies ∣r∣<1|r|<1.

Answer: r=12r=\frac{1}{2}.

Check an apparently valid formula result

A student enters 6[1−(−2)]\frac{6}{[1-(-2)]} and claims 6−12+24−⋯=26-12+24-\cdots=2. Explain the error.

  1. 1The ratio is r=−2r=-2.
  2. 2Although the calculator can evaluate the algebraic fraction, ∣r∣=2≥1|r|=2\ge1.
  3. 3The terms grow and the partial sums do not approach 2.

Answer: The series diverges; the sum-to-infinity formula was used outside its domain.

Rules and formulas

Partial-sum calculator entry

Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

Solver interval

−1<r<1-1<r<1

IB AI HL-style practice with worked solutions

Attempt each problem before opening its worked solution. The set progresses from identifying convergence to solving unknowns, estimating remainders, and building applied models.

In every sum-to-infinity question, make the convergence condition visible. Exact values are preferred unless the question requests a decimal or contextual rounding.

Key terms

Exact answer
A fraction or symbolic value without premature decimal approximation.
Model validation
Checking that the common ratio and counting assumptions match the stated context.
Candidate check
Testing a calculated ratio or variable in both the original conditions and ∣r∣<1|r|<1.

Worked examples

Practice 1 — convergence

Does a non-zero geometric series with r=0.75r=0.75 converge?

  1. 1∣0.75∣=0.75<1|0.75|=0.75<1.

Answer: Yes, it converges.

Practice 2 — boundary ratio

Does a non-zero geometric series with r=−1r=-1 converge?

  1. 1∣−1∣=1|-1|=1, not less than 1.
  2. 2Its partial sums oscillate rather than approach one limit.

Answer: No, it diverges.

Practice 3 — positive infinite sum

Find 18+6+2+23+⋯18+6+2+\frac{2}{3}+\cdots.

  1. 1a=18a=18, r=13r=\frac{1}{3}, and ∣r∣<1|r|<1.
  2. 2S∞=18(1−13)=18(23)S_\infty=\frac{18}{(1-\frac{1}{3})}=\frac{18}{(\frac{2}{3})}.

Answer: 2727.

Practice 4 — alternating infinite sum

Find 10−5+2.5−1.25+⋯10-5+2.5-1.25+\cdots.

  1. 1a=10a=10, r=−12r=\frac{-1}{2}, and ∣r∣<1|r|<1.
  2. 2S∞=10[1−(−12)]S_\infty=\frac{10}{[1-(\frac{-1}{2})]}.

Answer: 203\frac{20}{3}.

Practice 5 — first term

A convergent geometric series has sum 40 and ratio 0.2. Find its first term.

  1. 1a=S∞(1−r)=40(0.8)a=S_\infty(1-r)=40(0.8).

Answer: a=32a=32.

Practice 6 — common ratio

A convergent geometric series has first term 9 and sum 12. Find rr.

  1. 1r=1−aS∞=1−912r=1-\frac{a}{S_\infty}=1-\frac{9}{12}.
  2. 2Check that the result has magnitude less than 1.

Answer: r=14r=\frac{1}{4}.

Practice 7 — unknown term

A geometric series has a=16a=16 and r=12r=\frac{1}{2}. Find u5u_5.

  1. 1u5=16(12)4u_5=16(\frac{1}{2})^4.

Answer: u5=1u_5=1.

Practice 8 — algebraic series

The convergent series 5+x+x25+⋯5+x+\frac{x^2}{5}+\cdots has sum 254\frac{25}{4}. Find xx.

  1. 1The ratio is x5\frac{x}{5}.
  2. 2Solve 5[1−(x5)]=254\frac{5}{[1-(\frac{x}{5})]}=\frac{25}{4}.
  3. 3This gives x=1x=1, hence r=15r=\frac{1}{5}.

Answer: x=1x=1.

Practice 9 — recurring decimal

Write 0.45‾0.\overline{45} as a fraction.

  1. 10.45‾=45100+451002+⋯0.\overline{45}=\frac{45}{100}+\frac{45}{100^2}+\cdots.
  2. 2Sum: (45100)(1−1100)=4599\frac{(\frac{45}{100})}{(1-\frac{1}{100})}=\frac{45}{99}.
  3. 3Simplify.

Answer: 511\frac{5}{11}.

Practice 10 — bouncing ball

A ball is dropped from 12 m and rebounds to half its previous height. Find the total distance travelled.

  1. 1Rebound heights sum to 6(1−12)=12\frac{6}{(1-\frac{1}{2})}=12 m.
  2. 2Count the initial 12 m once and each rebound height twice.
  3. 3D=12+2(12)D=12+2(12).

Answer: 36 m36\text{ m}.

Practice 11 — remainder

For 20+8+3.2+⋯20+8+3.2+\cdots, find the sum after the first four terms that has not yet been included.

  1. 1Use R4=ar4(1−r)R_4=\frac{ar^4}{(1-r)} with a=20,r=0.4a=20,r=0.4.
  2. 2R4=20(0.4)40.6R_4=\frac{20(0.4)^4}{0.6}.

Answer: 6475≈0.853\frac{64}{75}\approx0.853.

Practice 12 — divergent claim

Explain why 3+6+12+24+⋯3+6+12+24+\cdots has no sum to infinity.

  1. 1The common ratio is r=2r=2.
  2. 2∣r∣=2≥1|r|=2\ge1, and the terms do not approach zero.
  3. 3Therefore the partial sums grow without bound.

Answer: The series diverges.

Rules and formulas

Convergence and sum

∣r∣<1⟹S∞=a1−r|r|<1\quad\Longrightarrow\quad S_\infty=\frac{a}{1-r}

Term and remainder

un=arn−1,Rn=arn1−ru_n=ar^{n-1},\qquad R_n=\frac{ar^n}{1-r}

Topic summary and cheat sheet

Identify the first term and common ratio, decide whether the question concerns a term, finite partial sum, infinite sum, or remainder, and check convergence before using an infinite-sum formula.

For modelling, define what every term represents, state when it is measured, verify the constant-ratio assumption, and finish with units, accuracy, and a contextual interpretation.

Key terms

Sequence
An ordered list of terms.
Series
A sum of terms.
Convergent
Partial sums approach one finite limit.
Divergent
Partial sums do not approach a finite limit.
Remainder
The infinite tail not included in a finite partial sum.

Worked examples

Rapid convergence check

Does 7−2.8+1.12−⋯7-2.8+1.12-\cdots converge?

  1. 1The common ratio is r=−2.87=−0.4r=\frac{-2.8}{7}=-0.4.
  2. 2Since ∣r∣=0.4<1|r|=0.4<1, the series converges.

Answer: Yes, it converges.

Rapid infinite sum

Find 7−2.8+1.12−⋯7-2.8+1.12-\cdots.

  1. 1Use a=7a=7 and r=−0.4r=-0.4.
  2. 2S∞=7[1−(−0.4)]=71.4S_\infty=\frac{7}{[1-(-0.4)]}=\frac{7}{1.4}.

Answer: S∞=5S_\infty=5.

Rules and formulas

nth term

un=arn−1u_n=ar^{n-1}

Finite sum

Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}

Convergence condition

∣r∣<1|r|<1

Sum to infinity

S∞=a1−r,∣r∣<1S_\infty=\frac{a}{1-r},\qquad |r|<1

Remainder after n terms

Rn=arn1−rR_n=\frac{ar^n}{1-r}

Unknown first term

a=S∞(1−r)a=S_\infty(1-r)

Unknown common ratio

r=1−aS∞r=1-\frac{a}{S_\infty}

Topic summary

  • A geometric sequence has un=arn−1u_n=ar^{n-1}; a geometric series adds these terms.
  • The first term is aa, and for consecutive non-zero terms r=un+1unr=\frac{u_{n+1}}{u_n}.
  • An infinite sum is defined through the limit of its finite partial sums, not by adding a final infinity term.
  • A non-zero infinite geometric series converges exactly when ∣r∣<1|r|<1.
  • A negative ratio can converge; its partial sums alternate toward the limit when −1<r<0-1<r<0.
  • If ∣r∣≥1|r|\ge1, the non-zero geometric series diverges and S∞S_\infty must not be used.
  • The finite partial sum is Sn=a(1−rn)(1−r)S_n=\frac{a(1-r^n)}{(1-r)}.
  • For ∣r∣<1|r|<1, S∞=a(1−r)S_\infty=\frac{a}{(1-r)}.
  • Useful rearrangements are a=S∞(1−r)a=S_\infty(1-r) and r=1−aS∞r=1-\frac{a}{S_\infty}.
  • The remainder after nn terms is Rn=arn(1−r)R_n=\frac{ar^n}{(1-r)}.
  • Check every ratio obtained algebraically against the convergence condition.
  • In bouncing-ball models, count the initial drop once and each rebound height twice.
  • A retain-and-add model An+1=rAn+dA_{n+1}=rA_n+d approaches d(1−r)\frac{d}{(1-r)} when ∣r∣<1|r|<1.
  • An end-of-period perpetuity with payment PP and effective rate i>0i>0 has present value Pi\frac{P}{i}.
  • A GDC can inspect partial sums or solve parameter equations, but it cannot replace the convergence proof.
  • Define what each term represents, retain exact values, and state units, accuracy, and model limitations.