Skip to main content

Mathematics AI · SL · Unit 1: Number and Algebra

Unit 1Topic 1.3Key Concepts60 min read

Geometric sequences and series

Reading progress0%
Back to course
On this page

Geometric change: repeated multiplication

A geometric sequence is an ordered list in which each term is obtained by multiplying the previous term by the same constant rr, called the common ratio. For example, 5,15,45,135,…5,15,45,135,\ldots has first term a=5a=5 and ratio r=3r=3.

Geometric change is multiplicative, not additive. The differences in 5,15,45,1355,15,45,135 are not constant, but the quotients are: 155=4515=13545=3\frac{15}{5}=\frac{45}{15}=\frac{135}{45}=3. This makes geometric models suitable for repeated percentage growth or decay.

A sequence lists individual terms, while a series adds them. In an application, a term may describe the amount in one year and a series may describe the total accumulated over several years.

A two-panel diagram showing geometric growth from 10 to 80 by repeated multiplication by 2 and geometric decay from 80 to 10 by repeated multiplication by one half.Geometric change repeats the same multiplier. A ratio above 1 produces growth; a positive ratio between 0 and 1 produces decay.

Key terms

Geometric sequence
A sequence generated by repeated multiplication by a constant ratio.
First term
The starting value, usually written a=u1a=u_1.
Common ratio
The multiplier rr between consecutive terms.
Geometric series
A sum of consecutive terms from a geometric sequence.

Worked examples

Sequence or series?

Classify 12,6,3,1.5,…12,6,3,1.5,\ldots and 12+6+3+1.512+6+3+1.5.

  1. 1The comma-separated list is a sequence with a=12a=12.
  2. 2Each term is multiplied by 12\frac{1}{2}, so r=12r=\frac{1}{2}.
  3. 3The expression joined by plus signs is a finite geometric series.

Answer: The list is geometric with r=12r=\frac12; the sum is a geometric series.

Rules and formulas

Consecutive-term ratio

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

First term and common ratio

The first term is the value at position 1. To test a sequence, divide each later term by the preceding term. If all defined quotients agree, the sequence is geometric and that value is rr.

A ratio can be greater than 1, between 0 and 1, negative, 1 or 0. A negative ratio makes signs alternate. If a term is zero, quotient calculations may be undefined, so use the repeated-multiplier description or recurrence relation instead.

Do not confuse a percentage rate with its multiplier. Growth of 8%8\% gives r=1.08r=1.08; decay of 8%8\% gives r=0.92r=0.92.

Worked examples

Recognize a negative ratio

Find aa and rr for 3,−6,12,−24,…3,-6,12,-24,\ldots.

  1. 1The first term is a=3a=3.
  2. 2Compute (−6)3=−2\frac{(-6)}{3}=-2, 12(−6)=−2\frac{12}{(-6)}=-2, and (−24)12=−2\frac{(-24)}{12}=-2.
  3. 3The constant negative ratio explains the alternating signs.

Answer: a=3a=3 and r=−2r=-2.

Rules and formulas

Growth multiplier

r=1+p100r=1+\frac{p}{100}

Decay multiplier

r=1−p100r=1-\frac{p}{100}

Generating terms with recurrence notation

A geometric sequence can be defined recursively by giving a starting term and a rule for moving forward: u1=au_1=a and un+1=runu_{n+1}=ru_n.

Recurrence notation mirrors a spreadsheet: enter the first value once, then multiply the preceding cell by the fixed ratio and fill downward.

A recurrence is excellent for generating nearby terms, but the explicit nth-term formula is faster for a distant term.

Worked examples

Generate a decaying sequence

Given u1=160u_1=160 and un+1=0.75unu_{n+1}=0.75u_n, find the next three terms.

  1. 1u2=0.75(160)=120u_2=0.75(160)=120.
  2. 2u3=0.75(120)=90u_3=0.75(120)=90.
  3. 3u4=0.75(90)=67.5u_4=0.75(90)=67.5.

Answer: 160,120,90,67.5,…160,120,90,67.5,\ldots

Rules and formulas

Geometric recurrence

u1=a,un+1=runu_1=a,\qquad u_{n+1}=ru_n

Finding the nth term

Starting at u1=au_1=a, reaching unu_n requires n−1n-1 multiplications by rr. Therefore the explicit formula is un=arn−1u_n=ar^{n-1}.

The exponent is n−1n-1, not nn. Check the formula at n=1n=1: it must give u1=ar0=au_1=ar^0=a.

Keep full calculator precision during the calculation and round only the final contextual value as requested.

Worked examples

Find a distant term

For a=3a=3 and r=1.5r=1.5, find u8u_8.

  1. 1Use un=arn−1u_n=ar^{n-1}.
  2. 2Substitute: u8=3(1.5)7u_8=3(1.5)^7.
  3. 3Evaluate: u8=51.2578125u_8=51.2578125.

Answer: u8≈51.3u_8\approx51.3 to 3 significant figures.

Rules and formulas

Geometric nth term

un=arn−1u_n=ar^{n-1}

Behaviour and graphs of geometric sequences

For positive aa: if r>1r>1, terms grow; if 0<r<10<r<1, terms decay toward zero; and if r=1r=1, the sequence is constant. If r=0r=0, every term after the first is zero.

If r<0r<0, signs alternate. When −1<r<0-1<r<0, magnitudes shrink toward zero; when r<−1r<-1, magnitudes grow. Negative ratios are mathematically valid but often unsuitable for populations, prices or lengths because those quantities cannot alternate in sign.

A sequence graph consists of discrete points at integer values of nn, not a continuous curve. A smooth exponential curve can show the trend, but values between term numbers are not automatically part of the sequence.

Worked examples

Interpret without calculating every term

Describe un=80(−0.4)n−1u_n=80(-0.4)^{n-1}.

  1. 1The ratio is negative, so signs alternate.
  2. 2Since ∣r∣=0.4<1|r|=0.4<1, magnitudes decrease.
  3. 3The terms approach 00, but the sequence is not monotonic because its sign changes.

Answer: An alternating sequence whose terms approach zero.

Finding an unknown term number

To find the position of a given term, solve arn−1=Tar^{n-1}=T. Exact powers can sometimes be recognized. Otherwise use a GDC numerical solver, graph intersection or logarithms if appropriate.

For positive aa, rr and TT with r≠1r\ne1, logarithms give n=1+ln⁡(Ta)ln⁡rn=1+\frac{\ln(\frac{T}{a})}{\ln r}. A term number must be a positive integer.

For threshold questions such as “first exceeds,” solve the inequality, move to the next valid integer if necessary, then verify that the previous term fails and the chosen term succeeds.

Worked examples

Find an exact term number

In 5,10,20,40,…5,10,20,40,\ldots, which term equals 640640?

  1. 1Set 5(2)n−1=6405(2)^{n-1}=640.
  2. 2Divide by 5: 2n−1=128=272^{n-1}=128=2^7.
  3. 3Therefore n−1=7n-1=7, so n=8n=8.

Answer: 640640 is the 8th term.

Rules and formulas

Term number from logarithms

n=1+ln⁡(una)ln⁡rn=1+\frac{\ln(\frac{u_n}{a})}{\ln r}

Finding unknown terms, first terms and ratios

If two terms are known, up=arp−1u_p=ar^{p-1} and uq=arq−1u_q=ar^{q-1}. Dividing eliminates aa: uqup=rq−p\frac{u_q}{u_p}=r^{q-p}.

After finding rr, substitute into either term equation to find aa. If q−pq-p is even, a positive equation may allow both a positive and a negative ratio; use the data or context to decide whether both are valid.

Do not round a ratio early. Small rounding errors become large when the ratio is raised to a high power.

Worked examples

Use two known terms

A geometric sequence has u3=18u_3=18 and u6=486u_6=486. Find aa and rr.

  1. 1Divide: u6u3=r3=48618=27\frac{u_6}{u_3}=r^3=\frac{486}{18}=27.
  2. 2Thus r=3r=3.
  3. 3Use 18=a(3)218=a(3)^2, giving a=2a=2.

Answer: a=2a=2 and r=3r=3.

Check for two possible ratios

If u2=12u_2=12 and u4=48u_4=48, find the possible ratios.

  1. 1Divide: u4u2=r2=4\frac{u_4}{u_2}=r^2=4.
  2. 2Therefore r=2r=2 or r=−2r=-2.
  3. 3A context requiring positive terms would select the positive ratio; otherwise both need consideration.

Answer: r=±2r=\pm2.

Rules and formulas

Ratio from two terms

rq−p=uqupr^{q-p}=\frac{u_q}{u_p}

First term from a known term

a=uprp−1a=\frac{u_p}{r^{p-1}}

Geometric series and partial sums

A geometric series is formed by adding geometric-sequence terms: Sn=a+ar+ar2+⋯+arn−1S_n=a+ar+ar^2+\cdots+ar^{n-1}. The symbol SnS_n means the sum of exactly the first nn terms.

A term question and a total question are different. “Amount in year 10” requires one term; “total over the first 10 years” requires a series sum.

Always count the number of included terms carefully. From index k=mk=m through k=nk=n, inclusive, there are n−m+1n-m+1 terms.

Worked examples

Write a geometric series

Write the first five terms of the series with a=7a=7 and r=2r=2.

  1. 1The sequence terms are 7,14,28,56,1127,14,28,56,112.
  2. 2Join the five terms by addition.
  3. 3Their total is 217217.

Answer: S5=7+14+28+56+112=217S_5=7+14+28+56+112=217.

Rules and formulas

First n geometric terms

Sn=a+ar+ar2+⋯+arn−1S_n=a+ar+ar^2+\cdots+ar^{n-1}

Sum of a finite geometric series

For r≠1r\ne1, a finite geometric sum can be written in two equivalent forms: Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r} or Sn=a(rn−1)r−1S_n=\frac{a(r^n-1)}{r-1}. Choose one form and keep numerator and denominator signs consistent.

If r=1r=1, every term equals aa, so Sn=naS_n=na. The usual fraction would divide by zero and must not be used.

The finite-sum formula works for positive, fractional and negative ratios. A finite series does not require a convergence condition.

Worked examples

Calculate a finite sum

Find S8S_8 when a=6a=6 and r=1.4r=1.4.

  1. 1Use Sn=a(rn−1)r−1S_n=\frac{a(r^n-1)}{r-1}.
  2. 2Substitute: S8=6(1.48−1)0.4S_8=\frac{6(1.4^8-1)}{0.4}.
  3. 3Evaluate without intermediate rounding: S8=206.3683584S_8=206.3683584.

Answer: S8≈206S_8\approx206 to 3 significant figures.

Rules and formulas

Finite geometric sum

Sn=a(1−rn)1−r=a(rn−1)r−1S_n=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}

Constant-ratio special case

Sn=nawhen r=1S_n=na\quad\text{when }r=1

Why the finite-sum formula works

Begin with Sn=a+ar+ar2+⋯+arn−1S_n=a+ar+ar^2+\cdots+ar^{n-1}. Multiplying by rr shifts every power: rSn=ar+ar2+⋯+arnrS_n=ar+ar^2+\cdots+ar^n.

Subtracting gives Sn−rSn=a−arnS_n-rS_n=a-ar^n, so Sn(1−r)=a(1−rn)S_n(1-r)=a(1-r^n). Dividing by 1−r1-r produces the formula when r≠1r\ne1.

This derivation is also a useful memory aid: multiplying by the ratio makes the middle terms cancel.

Worked examples

Check a formula against direct addition

Check 2+6+18+542+6+18+54 using the finite-sum formula.

  1. 1Here a=2a=2, r=3r=3, and n=4n=4.
  2. 2S4=2(34−1)3−1=80S_4=\frac{2(3^4-1)}{3-1}=80.
  3. 3Direct addition also gives 2+6+18+54=802+6+18+54=80.

Answer: S4=80S_4=80.

Finding the number of terms from a sum

When SnS_n is known, substitute into the finite-sum formula and solve the resulting exponential equation for nn. Use exact powers, a GDC solver, graph intersection or logarithms.

Because nn counts terms, it must be a positive integer. A non-integer solution may mean the target total is never attained exactly; a threshold question instead requires the first integer that satisfies an inequality.

Verify by substituting the final integer into the original sum formula, not only the rearranged calculator equation.

Worked examples

Find n from a total

A geometric series has a=3a=3, r=2r=2, and Sn=189S_n=189. Find nn.

  1. 1Use 189=3(2n−1)2−1189=\frac{3(2^n-1)}{2-1}.
  2. 2Then 63=2n−163=2^n-1, so 2n=642^n=64.
  3. 3Since 64=2664=2^6, n=6n=6.

Answer: There are 66 terms.

Sigma notation for geometric sums

The capital Greek letter Σ\Sigma represents addition. A finite geometric sum may be written ∑k=0n−1ark\sum\limits_{k=0}^{n-1}ar^k. Starting at k=0k=0 makes the first term ar0=aar^0=a.

The index is a counter. Expand the first two and final terms before choosing a formula; this prevents off-by-one errors when the lower limit is not zero.

For ∑k=mncrk\sum\limits_{k=m}^{n}cr^k, the first term is crmcr^m, the ratio is rr, and there are n−m+1n-m+1 terms.

Worked examples

Evaluate a sigma sum

Evaluate ∑k=064(1.5)k\sum\limits_{k=0}^{6}4(1.5)^k.

  1. 1The first term is 44, the ratio is 1.51.5, and the indices 00 to 66 give 77 terms.
  2. 2Use S7=4(1.57−1)1.5−1S_7=\frac{4(1.5^7-1)}{1.5-1}.
  3. 3This gives S7=128.6875S_7=128.6875.

Answer: 128.6875128.6875

Rules and formulas

Finite geometric sum in sigma notation

∑k=0n−1ark=a(1−rn)1−r\sum\limits_{k=0}^{n-1}ar^k=\frac{a(1-r^n)}{1-r}

Inclusive index count

number of terms=n−m+1\text{number of terms}=n-m+1

Infinite geometric series and convergence

An infinite geometric series continues without a final term. This is a relevant extension for repeated processes such as bouncing or recurring payments. It has a finite sum only when ∣r∣<1|r|<1.

If ∣r∣<1|r|<1, then rn→0r^n\to0 as n→∞n\to\infty. Taking the limit of the finite-sum formula gives S∞=a1−rS_\infty=\frac{a}{1-r}.

If ∣r∣≥1|r|\ge1, the terms do not approach zero, so the series diverges. The condition is about ∣r∣|r|, not merely r<1r<1: for example, r=−2r=-2 does not converge.

Worked examples

Convergent alternating series

Find 24−8+83−89+⋯24-8+\frac83-\frac89+\cdots.

  1. 1The first term is a=24a=24 and r=−13r=\frac{-1}{3}.
  2. 2Since ∣r∣=13<1|r|=\frac{1}{3}<1, the infinite sum exists.
  3. 3S∞=241−(−13)=2443=18S_\infty=\frac{24}{1-(\frac{-1}{3})}=\frac{24}{\frac{4}{3}}=18.

Answer: S∞=18S_\infty=18.

Bouncing-ball distance

A ball is dropped 5 m and rebounds to 60% of each previous height. Find its total vertical distance in the idealized infinite model.

  1. 1The first 5 m downward journey occurs once.
  2. 2The rebound heights form 3+1.8+⋯3+1.8+\cdots, with sum 3(1−0.6)=7.5\frac{3}{(1-0.6)}=7.5 m.
  3. 3Each rebound height is travelled once upward and once downward, so add twice this sum.

Answer: Total distance =5+2(7.5)=20 m=5+2(7.5)=20\text{ m}.

Rules and formulas

Infinite geometric sum

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

Infinite sigma form

∑k=0∞ark=a1−r,∣r∣<1\sum\limits_{k=0}^{\infty}ar^k=\frac{a}{1-r},\qquad |r|<1

Growth and decay applications

Repeated percentage change is geometric. An increase of p%p\% uses the multiplier 1+p1001+\frac{p}{100}, while a decrease uses 1−p1001-\frac{p}{100}. The percentage rate and the multiplier are not the same number.

If A0A_0 is the amount at time t=0t=0, then after tt complete periods At=A0rtA_t=A_0r^t. This time-zero form differs from sequence notation un=arn−1u_n=ar^{n-1}; choose one convention and label it clearly.

Growth and decay models are common for populations, infections, website traffic, depreciation, medicine concentration and repeated efficiency losses. State the units and whether the model is realistic over the whole interval.

Worked examples

Repeated population growth

A culture starts with 1200 bacteria and grows by 18% each hour. Estimate the population after 6 hours.

  1. 1The multiplier is r=1+0.18=1.18r=1+0.18=1.18.
  2. 2Use the time-zero model: A6=1200(1.18)6A_6=1200(1.18)^6.
  3. 3The calculator gives 3239.4649…3239.4649\ldots.

Answer: Approximately 3239 bacteria, or 3240 to 3 significant figures.

Repeated depreciation

A car worth $32,000 loses 14% of its value each year. Find its value after 5 years.

  1. 1The retention multiplier is r=1−0.14=0.86r=1-0.14=0.86.
  2. 2V5=32000(0.86)5V_5=32000(0.86)^5.
  3. 3V5=15053.66456…V_5=15053.66456\ldots.

Answer: Approximately $15,054\$15{,}054 to the nearest dollar.

Rules and formulas

Repeated-change model

At=A0rtA_t=A_0r^t

Percentage change from ratio

percentage change=100(r−1)%\text{percentage change}=100(r-1)\%

Thresholds, integer time and model constraints

Many AI SL applications ask when a quantity first exceeds or falls below a threshold. The calculator may return a decimal time, but discrete annual, monthly or daily data require an appropriate whole-number period.

If At=A0rtA_t=A_0r^t uses t=0t=0 for the initial value, then the first sequence term corresponds to time zero. If a question labels that initial value as u1u_1, the same physical time is n=1n=1.

A model can also have a capacity, non-negativity restriction or practical stopping rule. Check these conditions before summing or extrapolating.

Worked examples

First year above a target

A population is initially 2500 and grows by 7% annually. Find the first whole year when it exceeds 4000.

  1. 1Solve 2500(1.07)t>40002500(1.07)^t>4000.
  2. 2A GDC gives the boundary t≈6.9467t\approx6.9467.
  3. 3Time is counted in complete years, so test t=7t=7; the previous value at t=6t=6 is still below the target.

Answer: The population first exceeds 4000 after 7 complete years.

When real data are only approximately geometric

Real measurements rarely have identical consecutive ratios. For positive data, an endpoint estimate over positions pp to qq is r≈(uqup)1(q−p)r\approx\left(\frac{u_q}{u_p}\right)^{\frac{1}{(q-p)}}. This spreads the total multiplicative change evenly across the intervals.

Compare predicted and observed values using residuals: en=observed−predictede_n=\text{observed}-\text{predicted}. Small, patternless residuals support the model; a curved pattern or steadily changing ratios suggests the model is missing structure.

Interpolation stays within the observed interval and is generally safer than extrapolation. A sensible conclusion states the prediction, units, rounding and at least one model limitation.

Worked examples

Estimate a common ratio from data

Weekly users are 100,149,225,333100,149,225,333. Estimate one geometric ratio using the endpoints.

  1. 1There are three intervals from the first to the fourth value.
  2. 2Use r≈(333100)13=1.493303…r\approx(\frac{333}{100})^{\frac{1}{3}}=1.493303\ldots.
  3. 3This represents approximate weekly growth of 49.3%49.3\%. Check predictions against the middle observations before relying on the model.

Answer: r≈1.493r\approx1.493, or about 49.3%49.3\% growth per week.

Rules and formulas

Approximate ratio from endpoints

r≈(uqup)1q−pr\approx\left(\frac{u_q}{u_p}\right)^{\frac{1}{q-p}}

Residual

en=observed value−predicted valuee_n=\text{observed value}-\text{predicted value}

Calculator, GDC and spreadsheet methods

Technology is expected in AI SL, but it does not replace mathematical setup. Before using a table or solver, identify the initial value, ratio, index convention and whether the question asks for a term or a sum.

For a sequence table, enter u(n)=arn−1u(n)=ar^{n-1}. For a time-zero model, enter A(t)=A0rtA(t)=A_0r^t. Use integer inputs when the context is discrete.

For an unknown term number, graph the model and target together or use a numerical solver. For imperfect data, use exponential regression, then interpret the regression multiplier and inspect whether extrapolation is reasonable.

Worked examples

Use exponential regression responsibly

A GDC fits y=82.4(1.061)xy=82.4(1.061)^x to yearly data. Interpret the model.

  1. 1The predicted value at x=0x=0 is 82.482.4.
  2. 2The multiplier 1.0611.061 represents approximate growth of 6.1%6.1\% per year.
  3. 3Before predicting, inspect the scatter plot, residuals and distance beyond the observed years.

Answer: Initial model value 82.4 with approximate annual growth of 6.1%.

An AI SL modelling workflow

A strong applied solution follows a repeatable path: define variables and time origin, decide whether change is multiplicative, identify or estimate the ratio, choose a term or sum model, calculate, validate and interpret.

Check multiplicative change by comparing ratios, not differences. If a percentage change repeats, convert it to a multiplier before writing At=A0rtA_t=A_0r^t.

Finish with a sentence in context. Include units, appropriate rounding, the required integer decision and any practical reason the model might eventually fail.

Worked examples

Choose term, sum and threshold tools

A campaign reaches 600 people in week 1 and 1.25 times as many each later week. Identify the method for week 8, the first 8 weeks combined, and the first week above 2000.

  1. 1Week 8 is one term: u8=600(1.25)7u_8=600(1.25)^7.
  2. 2The first 8 weeks combined require S8=600(1.258−1)0.25S_8=\frac{600(1.25^8-1)}{0.25}.
  3. 3The first week above 2000 requires solving 600(1.25)n−1>2000600(1.25)^{n-1}>2000 and checking the first valid integer.

Answer: Use an nth term, a finite series, and a discrete inequality respectively.

Common mistakes and IB AI SL exam strategy

Most lost marks come from indexing, using the percentage instead of the multiplier, confusing a term with a sum, applying the infinite formula without checking convergence, or accepting a non-integer term number.

A quick check is to substitute n=1n=1 into an nth-term formula. It should return aa. For a finite sum, test a small case by direct addition.

Write enough method for another student to reproduce the result: formula, substitution, unrounded technology result, required rounding and contextual statement.

Worked examples

Correct a percentage error

A quantity decreases by 12%12\% per month. A student writes At=A0(0.12)tA_t=A_0(0.12)^t. Correct the model.

  1. 1The amount retained is 100%−12%=88%100\%-12\%=88\%.
  2. 2Therefore the multiplier is 0.880.88, not 0.120.12.
  3. 3The correct model is At=A0(0.88)tA_t=A_0(0.88)^t.

Answer: Use r=0.88r=0.88.

Practice questions with worked solutions

Attempt each question before opening the steps. Questions 1–4 establish the foundations, 5–9 combine formulas, and 10–12 develop application and modelling judgment.

In an exam, give exact forms where sensible and numerical answers to the requested accuracy.

Worked examples

Practice 1 · Identify a geometric sequence

Is 96,48,24,12,…96,48,24,12,\ldots geometric? State aa and rr.

  1. 1The consecutive ratios are 4896=2448=1224=12\frac{48}{96}=\frac{24}{48}=\frac{12}{24}=\frac{1}{2}.
  2. 2The ratio is constant.

Answer: Yes; a=96a=96 and r=12r=\frac12.

Practice 2 · Recurrence relation

Write a recurrence relation for 7,−21,63,−189,…7,-21,63,-189,\ldots.

  1. 1The first term is 77.
  2. 2The common ratio is −3-3.

Answer: u1=7,un+1=−3unu_1=7,\quad u_{n+1}=-3u_n.

Practice 3 · Find a term

Find u10u_{10} when a=250a=250 and r=0.92r=0.92.

  1. 1u10=250(0.92)9u_{10}=250(0.92)^9.
  2. 2This equals 118.040…118.040\ldots.

Answer: u10≈118u_{10}\approx118 to 3 significant figures.

Practice 4 · Find the term number

Which term of 2,6,18,54,…2,6,18,54,\ldots is 43744374?

  1. 1Solve 2(3)n−1=43742(3)^{n-1}=4374.
  2. 2Then 3n−1=2187=373^{n-1}=2187=3^7.
  3. 3So n=8n=8.

Answer: It is the 8th term.

Practice 5 · Find a and r

Given u2=10u_2=10 and u5=80u_5=80, find the positive ratio and first term.

  1. 1u5u2=r3=8\frac{u_5}{u_2}=r^3=8, so r=2r=2.
  2. 2Then 10=a(2)10=a(2), so a=5a=5.

Answer: a=5a=5, r=2r=2.

Practice 6 · Finite geometric series

Find the sum of the first 9 terms when a=12a=12 and r=0.8r=0.8.

  1. 1Use S9=12(1−0.89)1−0.8S_9=\frac{12(1-0.8^9)}{1-0.8}.
  2. 2This gives S9=51.9469…S_9=51.9469\ldots.

Answer: S9≈51.9S_9\approx51.9 to 3 significant figures.

Practice 7 · Find n from a sum

A series has a=4a=4, r=3r=3, and Sn=484S_n=484. Find nn.

  1. 1484=4(3n−1)2484=\frac{4(3^n-1)}{2}.
  2. 2Thus 242=3n−1242=3^n-1, so 3n=243=353^n=243=3^5.

Answer: n=5n=5.

Practice 8 · Sigma notation

Evaluate ∑k=165(2)k−1\sum\limits_{k=1}^{6}5(2)^{k-1}.

  1. 1The first term is 55, the ratio is 22, and there are 66 terms.
  2. 2S6=5(26−1)2−1=315S_6=\frac{5(2^6-1)}{2-1}=315.

Answer: 315315

Practice 9 · Infinite geometric series

Determine whether 30+12+4.8+⋯30+12+4.8+\cdots converges and, if so, find its sum.

  1. 1The ratio is r=1230=0.4r=\frac{12}{30}=0.4.
  2. 2Since ∣r∣=0.4<1|r|=0.4<1, the series converges.
  3. 3S∞=30(1−0.4)=50S_\infty=\frac{30}{(1-0.4)}=50.

Answer: The series converges to 5050.

Practice 10 · Growth application

A social-media account has 750 followers and grows by 12% each month. Predict the number after 8 months.

  1. 1The multiplier is 1.121.12.
  2. 2A8=750(1.12)8=1856.972…A_8=750(1.12)^8=1856.972\ldots.
  3. 3Round to a whole follower.

Answer: Approximately 1857 followers.

Practice 11 · Approximate geometric model

Positive observations at years 1 and 5 are 120120 and 250250. Estimate a constant annual ratio.

  1. 1There are 5−1=45-1=4 intervals.
  2. 2r≈(250120)14=1.2014…r\approx(\frac{250}{120})^{\frac{1}{4}}=1.2014\ldots.
  3. 3This is approximately 20.1%20.1\% growth per year.

Answer: r≈1.201r\approx1.201, about 20.1%20.1\% annual growth.

Practice 12 · AI SL modelling challenge

A subscription service starts with 500 customers and grows by 15% monthly. Find the first complete month when customers exceed 1000, and estimate total monthly customer-counts from month 0 through that month inclusive.

  1. 1Use Ct=500(1.15)tC_t=500(1.15)^t and solve Ct>1000C_t>1000.
  2. 2The boundary is t≈4.959t\approx4.959, so the first complete month is t=5t=5.
  3. 3Months 00 through 55 give six terms: 500+500(1.15)+⋯+500(1.15)5500+500(1.15)+\cdots+500(1.15)^5.
  4. 4The sum is 500(1.156−1)0.15=4376.87…\frac{500(1.15^6-1)}{0.15}=4376.87\ldots.

Answer: Month 5; approximately 4377 customer-months.

Topic summary and cheat sheet

Use this final checklist to select a method quickly and present a complete AI SL solution.

A geometric model is powerful because one ratio summarizes repeated proportional change, but every applied conclusion depends on sensible indexing and model assumptions.

Rules and formulas

Common ratio

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

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}

Infinite sum

S∞=a1−rS_\infty=\frac{a}{1-r}

Repeated-change model

At=A0rtA_t=A_0r^t

Topic summary

  • A geometric sequence repeats multiplication by a constant ratio rr.
  • The first term is a=u1a=u_1.
  • When defined, r=un+1unr=\frac{u_{n+1}}{u_n}.
  • Constant ratios, not constant differences, identify geometric change.
  • Recurrence form: u1=au_1=a, un+1=runu_{n+1}=ru_n.
  • Nth term: un=arn−1u_n=ar^{n-1}.
  • The exponent is n − 1 because no multiplication is needed to reach the first term.
  • From two terms, rq−p=uqupr^{q-p}=\frac{u_q}{u_p}.
  • An even root can produce positive and negative ratio possibilities.
  • A geometric series adds terms: Sn=a+ar+⋯+arn−1S_n=a+ar+\cdots+ar^{n-1}.
  • For r≠1r\ne1, Sn=a(1−rn)(1−r)S_n=\frac{a(1-r^n)}{(1-r)}.
  • Equivalently, Sn=a(rn−1)(r−1)S_n=\frac{a(r^n-1)}{(r-1)}.
  • When r=1r=1, Sn=naS_n=na.
  • A finite sum has no convergence requirement.
  • With a nonzero first term, an infinite geometric series converges exactly when ∣r∣<1|r|<1. A zero first term gives a zero series, which converges for every finite ratio.
  • For a convergent infinite series, S∞=a(1−r)S_\infty=\frac{a}{(1-r)}.
  • Finite sigma form: ∑k=0n−1ark\sum\limits_{k=0}^{n-1}ar^k.
  • Indices mm through nn include n−m+1n-m+1 terms.
  • Growth of p%p\% uses r=1+p100r=1+\frac{p}{100}.
  • Decay of p%p\% uses r=1−p100r=1-\frac{p}{100}.
  • Time-zero model: At=A0rtA_t=A_0r^t.
  • Do not confuse a percentage rate with its multiplier.
  • For positive endpoint data, r≈(uqup)1(q−p)r\approx(\frac{u_q}{u_p})^{\frac{1}{(q-p)}}.
  • Residual equals observed value minus predicted value.
  • Term numbers and discrete time values must be integers.
  • A first-threshold answer should be checked against the previous period.
  • For positive values, logarithms can solve arn−1=Tar^{n-1}=T.
  • A GDC table should use the correct starting index.
  • Exponential regression estimates an initial coefficient and multiplier.
  • Interpolation is usually safer than extrapolation.
  • Use full calculator precision until the final answer.
  • An IB-ready answer includes setup, technology evidence, units, rounding, validation and interpretation.