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Mathematics AI · HL · Unit 1: Number and Algebra

Unit 1Topic 1.2Key Concepts57 min read

Arithmetic sequences and series

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Sequences and series: the essential distinction

A sequence is an ordered list of values called terms. We write u1,u2,u3,…u_1,u_2,u_3,\ldots, where unu_n is the value at position nn. The position is part of the information: the same values in another order form a different sequence.

A series is a sum of sequence terms. For example, 5,8,11,14,175,8,11,14,17 is a sequence, while 5+8+11+14+17=555+8+11+14+17=55 is a finite series. The notation SnS_n means the sum of the first nn terms.

In modelling, a term often represents one period or location—such as sales in week 12 or seats in row 20—whereas a series represents an accumulated total over several periods or locations.

A two-panel diagram showing the arithmetic sequence 5, 8, 11, 14, 17 with four equal plus-three steps, and the corresponding series adding to 55.A sequence lists individual ordered terms. A series adds those terms. The common difference belongs to the sequence; the sum belongs to the series.

Key terms

Sequence
An ordered list u1,u2,u3,…u_1,u_2,u_3,\ldots.
Term
One value in a sequence; unu_n is the term at position nn.
Series
The result or expression obtained by adding sequence terms.
Partial sum
The sum Sn=u1+u2+⋯+unS_n=u_1+u_2+\cdots+u_n of the first nn terms.

Worked examples

Term or total?

A theatre has 18 rows. A question asks for the seats in row 18, then the total seats in all 18 rows. Which mathematical objects are needed?

  1. 1Seats in row 18 require one term, u18u_{18}.
  2. 2Total seats in rows 1 through 18 require a series sum, S18S_{18}.

Answer: Use u18u_{18} for the last row and S18S_{18} for the accumulated total.

Rules and formulas

First n terms as a series

Sn=u1+u2+⋯+unS_n=u_1+u_2+\cdots+u_n

Arithmetic sequences and the common difference

An arithmetic sequence changes by the same additive amount at every step. This fixed amount is the common difference dd, found using later term minus earlier term: d=un+1−und=u_{n+1}-u_n.

If d>0d>0, the sequence increases. If d<0d<0, it decreases. If d=0d=0, it is constant. Arithmetic change is additive; multiplying by the same factor instead produces a geometric pattern.

Check every consecutive difference supplied. One matching pair is not sufficient evidence that an entire data set is arithmetic.

Key terms

Arithmetic sequence
A sequence with a constant difference between consecutive terms.
First term
The initial value, usually written u1u_1 or aa.
Common difference
The constant additive change d=un+1−und=u_{n+1}-u_n.

Worked examples

Test a decreasing sequence

Determine whether 31,24,17,10,3,…31,24,17,10,3,\ldots is arithmetic and state aa and dd.

  1. 1Calculate 24−31=−724-31=-7, 17−24=−717-24=-7, 10−17=−710-17=-7, and 3−10=−73-10=-7.
  2. 2Every difference is identical.
  3. 3The first listed term is a=31a=31.

Answer: It is arithmetic with a=31a=31 and d=−7d=-7.

Rules and formulas

Common difference

d=un+1−und=u_{n+1}-u_n
dd
constant difference between consecutive terms

Generating terms with recurrence notation

A recurrence relation defines each new term using the previous term. For an arithmetic sequence, state the starting value and the repeated change: u1=au_1=a, un+1=un+du_{n+1}=u_n+d.

Recurrence notation describes a step-by-step process and works naturally in spreadsheets or programs. It is convenient for generating consecutive terms but inefficient for a distant term such as the 500th.

A recurrence rule is incomplete without an initial value. The rule un+1=un+4u_{n+1}=u_n+4 describes infinitely many possible sequences until u1u_1 is specified.

Key terms

Recurrence relation
A rule that defines a term from one or more earlier terms.
Initial value
The starting term required to begin a recursive sequence.

Worked examples

Generate from a recurrence

Given u1=18u_1=18 and un+1=un−2.5u_{n+1}=u_n-2.5, write the first five terms.

  1. 1Start with u1=18u_1=18.
  2. 2Subtract 2.52.5 repeatedly: u2=15.5u_2=15.5, u3=13u_3=13, u4=10.5u_4=10.5, u5=8u_5=8.

Answer: 18,15.5,13,10.5,818,15.5,13,10.5,8.

Rules and formulas

Arithmetic recurrence

u1=a,un+1=un+du_1=a,\qquad u_{n+1}=u_n+d

Finding the nth term

To reach term nn from term 1, there are n−1n-1 equal steps. Therefore the explicit formula is un=a+(n−1)du_n=a+(n-1)d. This finds any term directly without generating all earlier terms.

Expanding gives un=dn+(a−d)u_n=dn+(a-d), which is linear in nn. The gradient is dd, but the vertical intercept at n=0n=0 is a−da-d, not usually the first term.

In an application, define what term 1 represents. If a data table starts at year 0, the indexing must match the formula rather than being silently shifted.

Key terms

Explicit formula
A rule that calculates unu_n directly from nn.
Term number
The integer position n=1,2,3,…n=1,2,3,\ldots.

Worked examples

Find a distant term

For 7,11,15,19,…7,11,15,19,\ldots, find u60u_{60}.

  1. 1Identify a=7a=7 and d=4d=4.
  2. 2Substitute into un=a+(n−1)du_n=a+(n-1)d.
  3. 3u60=7+(60−1)(4)=7+236u_{60}=7+(60-1)(4)=7+236.

Answer: u60=243u_{60}=243.

Write a formula first

Find an expression for the nth term of 42,37,32,27,…42,37,32,27,\ldots.

  1. 1a=42a=42 and d=−5d=-5.
  2. 2un=42+(n−1)(−5)u_n=42+(n-1)(-5).
  3. 3Simplify only if useful.

Answer: un=42−5(n−1)=47−5nu_n=42-5(n-1)=47-5n.

Rules and formulas

Arithmetic nth term

un=a+(n−1)du_n=a+(n-1)d
aa
first term
dd
common difference
nn
positive integer term number

Linear form

un=dn+(a−d)u_n=dn+(a-d)

Tables and graphs of arithmetic sequences

Plot term number nn horizontally and term value unu_n vertically. Arithmetic-sequence points lie on a straight line because equal increases in nn produce equal changes dd in unu_n.

Only positive integer term numbers normally belong to the sequence. A GDC may draw a continuous line between points, but values at non-integer positions may have no meaning in the model.

From a regression line un≈mn+cu_n\approx mn+c, the approximate common difference is d≈md\approx m. If indexing begins at n=1n=1, the estimated first term is a≈m+ca\approx m+c.

Key terms

Discrete graph
A graph of separate points because term numbers are integers.
Gradient
The change in term value per one-position increase; for an arithmetic sequence it equals the common difference.

Worked examples

Interpret a sequence graph

A fitted line for weekly output is un=12n+86u_n=12n+86. Identify dd and the modelled first term.

  1. 1The coefficient of nn is the gradient, so d=12d=12.
  2. 2At n=1n=1, u1=12+86=98u_1=12+86=98.

Answer: The common difference is 12 units per week and the first term is 98 units.

Finding an unknown term number

When a term value is known but its position is not, set a+(n−1)da+(n-1)d equal to the given value and solve for nn.

Then validate the solution. A term number must be a positive integer. If the algebra produces a non-integer, the target value is not an exact term of that arithmetic sequence.

For threshold language such as “first exceeds,” solve an inequality and select the first integer satisfying the condition.

Key terms

Integer constraint
The requirement that a term number nn belongs to 1,2,3,…1,2,3,\ldots.
Threshold
A boundary a model value must reach, exceed or remain below.

Worked examples

Find the position of a term

In 11,17,23,29,…11,17,23,29,\ldots, which term is 143143?

  1. 1a=11a=11, d=6d=6, so un=11+6(n−1)u_n=11+6(n-1).
  2. 2Solve 11+6(n−1)=14311+6(n-1)=143.
  3. 36(n−1)=1326(n-1)=132, so n−1=22n-1=22 and n=23n=23.

Answer: 143143 is the 23rd term.

Show that a value is not a term

Does 100100 occur in 7,15,23,31,…7,15,23,31,\ldots?

  1. 1Solve 7+8(n−1)=1007+8(n-1)=100.
  2. 2This gives n=12.625n=12.625.
  3. 3A term number cannot be fractional.

Answer: No. 100100 is not a term of the sequence.

Finding unknown terms, first term or common difference

If two non-consecutive terms upu_p and uqu_q are known, there are q−pq-p equal steps between them. Therefore d=(uq−up)(q−p)d=\frac{(u_q-u_p)}{(q-p)}. Substitute back to find aa.

Alternatively, write two equations using the nth-term formula and solve simultaneously. This method generalizes well when parameters or contextual information are involved.

Once aa and dd are found, verify both given terms. One successful substitution is not enough to detect every sign or indexing error.

Worked examples

Use two known terms

An arithmetic sequence has u6=23u_6=23 and u15=59u_{15}=59. Find aa and dd.

  1. 1There are 15−6=915-6=9 steps, and the total change is 59−23=3659-23=36.
  2. 2d=369=4d=\frac{36}{9}=4.
  3. 3Use u6=a+5du_6=a+5d: 23=a+2023=a+20.
  4. 4Hence a=3a=3.

Answer: a=3a=3 and d=4d=4.

Rules and formulas

Difference from two terms

d=uq−upq−p,q≠pd=\frac{u_q-u_p}{q-p},\quad q\ne p

Arithmetic series and finite sums

An arithmetic series adds consecutive terms of an arithmetic sequence. The sum of the first nn terms is SnS_n. Unlike an infinite geometric series, a non-constant arithmetic series does not approach a finite total as nn increases indefinitely, so AI HL uses finite sums here.

If the first and last terms are known, the average term is (a+un)2\frac{(a+u_n)}{2}. Multiplying this average by the number of terms gives Sn=n(a+un)2S_n=\frac{n(a+u_n)}{2}.

Replacing unu_n by a+(n−1)da+(n-1)d gives the equivalent formula Sn=n[2a+(n−1)d]2S_n=\frac{n[2a+(n-1)d]}{2}. Choose the form that uses the information provided most directly.

Key terms

Finite arithmetic series
The sum of a fixed number of consecutive terms from an arithmetic sequence.
Last term
The final included term unu_n in SnS_n.

Worked examples

Sum using first and last terms

Find the sum 12+17+22+⋯+10712+17+22+\cdots+107.

  1. 1Here a=12a=12, d=5d=5, and un=107u_n=107.
  2. 2Find nn: 12+5(n−1)=10712+5(n-1)=107, so n=20n=20.
  3. 3Use Sn=n(a+un)2S_n=\frac{n(a+u_n)}{2}.
  4. 4S20=20(12+107)2S_{20}=\frac{20(12+107)}{2}.

Answer: S20=1190S_{20}=1190.

Rules and formulas

Sum using first and last terms

Sn=n2(a+un)S_n=\frac{n}{2}(a+u_n)

Sum using first term and difference

Sn=n2[2a+(n−1)d]S_n=\frac{n}{2}[2a+(n-1)d]

Why the arithmetic-series formula works

Write the same sum forwards and backwards: Sn=a+(a+d)+⋯+unS_n=a+(a+d)+\cdots+u_n and Sn=un+(un−d)+⋯+aS_n=u_n+(u_n-d)+\cdots+a.

Adding corresponding positions gives a+una+u_n every time. There are nn such pairs, so 2Sn=n(a+un)2S_n=n(a+u_n), and therefore Sn=n(a+un)2S_n=\frac{n(a+u_n)}{2}.

This pairing explanation prevents formula memorization from becoming disconnected from meaning: an arithmetic-series total equals number of terms multiplied by average of first and last terms.

Key terms

Pairing argument
A derivation that pairs first with last, second with second-last, and so on.

Worked examples

Check by averaging

Explain quickly why 4+7+10+13+16=504+7+10+13+16=50.

  1. 1There are 55 terms.
  2. 2Their average is (4+16)2=10\frac{(4+16)}{2}=10.
  3. 3Total =5×10=5\times10.

Answer: The sum is 5050.

Rules and formulas

Number times average term

Sn=n(a+un2)S_n=n\left(\frac{a+u_n}{2}\right)

Finding the number of terms from a sum

If SnS_n is known, substitute into a sum formula. Because nn appears both outside and inside the bracket, the resulting equation is often quadratic.

Solve algebraically or with a GDC, then apply the domain: the number of terms must be a positive integer. Reject negative, zero or non-integer solutions unless the context uses an inequality and asks for a threshold.

Substitute the accepted integer back into the sum formula to verify the total.

Worked examples

Solve a sum condition

For an arithmetic sequence with a=8a=8 and d=3d=3, find nn if Sn=294S_n=294.

  1. 1294=n2[16+3(n−1)]294=\frac n2[16+3(n-1)].
  2. 2Simplify to 3n2+13n−588=03n^2+13n-588=0.
  3. 3Factor or use a GDC to obtain n=12n=12 or a negative root.
  4. 4Reject the negative root and verify the sum.

Answer: n=12n=12.

Sigma notation for arithmetic sums

The Greek capital letter sigma, Σ\Sigma, means “add.” In ∑r=1n[a+(r−1)d]\sum\limits_{r=1}^{n}[a+(r-1)d], the index rr takes the integer values 1,2,…,n1,2,\ldots,n, generating and adding the first nn arithmetic terms.

The lower limit gives the first index, the upper limit gives the last index, and the expression tells what to add. The number of included integers from lower limit p to upper limit q is q minus p plus 1.

The index letter is a dummy variable: ∑r=1n\sum\limits_{r=1}^{n} and ∑k=1n\sum\limits_{k=1}^{n} have the same meaning if the expression changes consistently.

Key terms

Sigma notation
Compact notation for a finite or infinite sum.
Index
The variable that takes successive integer values in a sigma expression.
Limits
The first and last index values shown below and above sigma.

Worked examples

Evaluate an arithmetic sigma sum

Evaluate ∑r=120[4+3(r−1)]\sum\limits_{r=1}^{20}[4+3(r-1)].

  1. 1The generated sequence has a=4a=4, d=3d=3, and n=20n=20.
  2. 2u20=4+19(3)=61u_{20}=4+19(3)=61.
  3. 3S20=20(4+61)2S_{20}=\frac{20(4+61)}{2}.

Answer: The sum is 650650.

Count terms with a shifted lower limit

How many terms are included in ∑r=415(2r+1)\sum\limits_{r=4}^{15}(2r+1)?

  1. 1Use 15−4+115-4+1.

Answer: There are 12 terms.

Rules and formulas

Arithmetic sum in sigma notation

∑r=1n[a+(r−1)d]=Sn\sum\limits_{r=1}^{n}[a+(r-1)d]=S_n

Number of indexed terms

count=q−p+1\text{count}=q-p+1
pp
lower integer limit
qq
upper integer limit

Application: simple interest

Under simple interest, the same interest amount PrPr is added each period, where PP is principal and rr is the interest rate per period as a decimal. The balances therefore form an arithmetic sequence.

After nn periods, An=P(1+nr)A_n=P(1+nr). If balances are listed at the end of years 1, 2, 3, and so on, the first listed balance is P(1+r)P(1+r) and the common difference is PrPr.

Do not confuse simple interest with compound interest. Compound interest multiplies by the same factor each period and is modelled geometrically.

Key terms

Principal
The initial amount invested or borrowed.
Simple interest
Interest calculated on the original principal only.
Rate per period
The percentage rate converted to a decimal and matched to the time period.

Worked examples

Model simple-interest balances

An investment of $12000\$12000 earns 3.5%3.5\% simple interest per year. Find the balance after 6 years and identify the yearly common difference.

  1. 1Convert 3.5%=0.0353.5\%=0.035.
  2. 2Yearly interest Pr=12000(0.035)=420Pr=12000(0.035)=420.
  3. 3A6=12000(1+6×0.035)A_6=12000(1+6\times0.035).

Answer: The balance is $14520\$14520, and the yearly common difference is $420\$420.

Rules and formulas

Simple-interest amount

An=P(1+nr)A_n=P(1+nr)
PP
principal
rr
rate per period as a decimal
nn
number of periods

Simple interest earned

I=PrnI=Prn

Applications with constant additive change

Arithmetic models apply when an amount changes by a fixed number each step: seats per row, regular weekly savings, straight-line depreciation, production quotas, fitness targets or a fixed annual increase.

Define the first term and what one step means. Decide whether the question asks for one period, a cumulative total, a threshold or an unknown parameter.

A negative common difference can model depletion or depreciation, but the model must stop before it predicts impossible values such as negative stock or negative length.

Key terms

Additive model
A model in which the same amount is added or subtracted each step.
Model domain
The term numbers for which the model is meaningful.

Worked examples

Seats in a theatre

A theatre has 24 seats in the first row and 3 more seats in each later row. Find the seats in row 18 and the total in the first 18 rows.

  1. 1a=24a=24, d=3d=3, n=18n=18.
  2. 2u18=24+17(3)=75u_{18}=24+17(3)=75.
  3. 3S18=18(24+75)2=891S_{18}=\frac{18(24+75)}{2}=891.

Answer: Row 18 has 75 seats, and the first 18 rows contain 891 seats.

Check a decreasing model

A tank contains 850 L and loses 45 L each hour. For how many complete hours does the arithmetic model predict a non-negative amount?

  1. 1At hour nn after the start, use Vn=850−45nV_n=850-45n.
  2. 2Solve 850−45n≥0850-45n\ge0, giving n≤18.88…n\le18.88\ldots.
  3. 3Choose complete integer hours within the domain.

Answer: The model remains non-negative for 18 complete hours after the start.

When real data are only approximately arithmetic

Real measurements rarely have perfectly equal differences. If consecutive changes are close and show no strong curvature, an arithmetic model may still be useful as an approximation.

A simple estimated difference over equally spaced data is dapprox=(uq−up)(q−p)d_{\text{approx}}=\frac{(u_q-u_p)}{(q-p)}. Using widely separated endpoints averages short-term fluctuations, although a linear regression uses all data and is usually stronger when technology is available.

For a fitted model un≈a+(n−1)du_n\approx a+(n-1)d, calculate residuals en=observed−predictede_n=\text{observed}-\text{predicted}. Small residuals with no systematic pattern support the model; a curved residual pattern suggests that constant additive change is unsuitable.

Predictions within the observed interval are generally safer than extrapolation far beyond it. Always mention relevant practical limits or changing conditions.

Key terms

Approximate common difference
A representative additive change used when observed differences are similar but not identical.
Residual
Observed value minus model-predicted value.
Extrapolation
Using a model outside the range of observed data.

Worked examples

Model imperfect data

Weekly counts are 120,129,137,146,154120,129,137,146,154. Estimate dd from the endpoints and predict week 10.

  1. 1Differences are 9,8,9,89,8,9,8, so the data are close to arithmetic.
  2. 2dapprox=(154−120)(5−1)=8.5d_{\text{approx}}=\frac{(154-120)}{(5-1)}=8.5.
  3. 3Use un≈120+8.5(n−1)u_n\approx120+8.5(n-1).
  4. 4u10≈120+8.5(9)=196.5u_{10}\approx120+8.5(9)=196.5.

Answer: The model predicts about 197 in week 10. This is an extrapolation, so the result should be treated cautiously.

Rules and formulas

Approximate difference from two positions

dapprox=uq−upq−pd_{\text{approx}}=\frac{u_q-u_p}{q-p}

Residual

en=observedn−predictedne_n=\text{observed}_n-\text{predicted}_n

Calculator, GDC and spreadsheet methods

A sequence table can generate many terms quickly. Enter the term number in one column and the explicit formula in another, or enter the first term and fill the next cells using previous term plus the fixed difference.

A graph of term number against term value should show equally spaced points on a straight line. Use a linear regression for approximately arithmetic data, then interpret its gradient and intercept using the chosen indexing.

A GDC summation command can check a finite sum, and an equation solver can find n from a term or total. However, record the defining formula and apply integer and contextual constraints yourself.

Calculator menus vary, so practise sequence, table, list, sum and regression features on the model allowed in your examinations.

Key terms

Sequence table
A technology-generated list pairing term numbers with term values.
Linear regression
A technology method fitting a straight-line model to paired data.

Worked examples

Use a GDC model responsibly

A regression gives un≈13.2n+191.6u_n\approx13.2n+191.6. Interpret the estimated common difference and first term.

  1. 1The gradient gives d≈13.2d\approx13.2 units per term.
  2. 2Because the first term is at n=1n=1, a≈13.2+191.6=204.8a\approx13.2+191.6=204.8.

Answer: The model increases by about 13.2 units per step and predicts a first term of about 204.8.

An AI HL modelling workflow

First identify the unit of one step and the meaning of term 1. Next decide whether constant additive change is exact or approximate. Then define a, d and n with units.

Choose unu_n for one term and SnS_n for an accumulated total. Use technology where efficient, but retain unrounded values until the final step.

Finally validate the result: term numbers and counts should be integers, units must match, and the model must stay within practical bounds. Conclude in a sentence that answers the application.

Key terms

Validation
Checking mathematical output against the model domain, units and real situation.

Worked examples

Choose term, sum and constraint

A workshop produces 140 units in week 1 and 12 more each week. It can produce at most 260 units in one week. What quantities would you calculate to find total production before the limit is exceeded?

  1. 1Solve 140+12(n−1)≤260140+12(n-1)\le260 for the last valid integer week.
  2. 2Then use SnS_n to total production through that week.
  3. 3Interpret the next term to confirm it exceeds the limit.

Answer: The last valid week is found from an nth-term inequality; the cumulative production is then SnS_n.

Common mistakes and IB AI HL exam strategy

Frequent errors include using ndnd instead of (n−1)d(n-1)d, reversing the sign of dd, confusing unu_n with SnS_n, and counting terms incorrectly when sigma limits do not begin at 1.

Context errors include treating simple interest as compound, rounding a non-integer exact term number, accepting a negative quadratic root, omitting units, or extrapolating an approximate model without comment.

For full communication, state the formula, substitute values, use technology transparently, retain precision, apply constraints and finish with a contextual sentence.

Worked examples

Correct an indexing error

A student models the 10th term of 6,10,14,…6,10,14,\ldots as 6+10(4)6+10(4). Correct the method.

  1. 1There are only 10−1=910-1=9 changes after the first term.
  2. 2Use u10=6+9(4)u_{10}=6+9(4).

Answer: u10=42u_{10}=42.

Practice questions with worked solutions

Attempt each question before reading the steps. The progression moves from pattern recognition to exact models, sigma notation, technology interpretation and imperfect real data.

Show formulas and working even when using a GDC. Give money to a sensible currency precision and round approximate model outputs appropriately.

Worked examples

Practice 1 · Identify the arithmetic pattern

For 48,41,34,27,…48,41,34,27,\ldots, state aa and dd.

  1. 1The first term is a=48a=48.
  2. 2Each consecutive difference is −7-7.

Answer: a=48a=48, d=−7d=-7.

Practice 2 · Recurrence relation

Write a recurrence relation for 9,14,19,24,…9,14,19,24,\ldots.

  1. 1The starting term is u1=9u_1=9.
  2. 2The common difference is 55.

Answer: u1=9u_1=9, un+1=un+5u_{n+1}=u_n+5.

Practice 3 · Find a distant term

Find u35u_{35} for a=16a=16 and d=2.5d=2.5.

  1. 1u35=16+(35−1)(2.5)u_{35}=16+(35-1)(2.5).
  2. 2u35=16+85u_{35}=16+85.

Answer: u35=101u_{35}=101.

Practice 4 · Find the term number

Which term of 11,17,23,…11,17,23,\ldots is 143143?

  1. 1Solve 11+6(n−1)=14311+6(n-1)=143.
  2. 2This gives n=23n=23.

Answer: 143143 is the 23rd term.

Practice 5 · Find a and d

Given u6=23u_6=23 and u15=59u_{15}=59, find aa and dd.

  1. 1d=(59−23)(15−6)=4d=\frac{(59-23)}{(15-6)}=4.
  2. 223=a+5(4)23=a+5(4), so a=3a=3.

Answer: a=3a=3, d=4d=4.

Practice 6 · Finite series

Find the sum of the first 30 terms when a=7a=7 and d=4d=4.

  1. 1u30=7+29(4)=123u_{30}=7+29(4)=123.
  2. 2S30=30(7+123)2S_{30}=\frac{30(7+123)}{2}.

Answer: S30=1950S_{30}=1950.

Practice 7 · Find n from a sum

For a=8a=8 and d=3d=3, find nn if Sn=294S_n=294.

  1. 1Set 294=n2[16+3(n−1)]294=\frac n2[16+3(n-1)].
  2. 2Solve 3n2+13n−588=03n^2+13n-588=0.
  3. 3The positive integer root is 12.

Answer: n=12n=12.

Practice 8 · Sigma notation

Evaluate ∑r=120[4+3(r−1)]\sum\limits_{r=1}^{20}[4+3(r-1)].

  1. 1The sequence has a=4a=4, d=3d=3, and n=20n=20.
  2. 2u20=61u_{20}=61.
  3. 3S20=20(4+61)2S_{20}=\frac{20(4+61)}{2}.

Answer: The sum is 650650.

Practice 9 · Simple interest

$8000\$8000 earns 4.2%4.2\% simple interest per year. Find the balance after 5 years.

  1. 1Use r=0.042r=0.042.
  2. 2A5=8000(1+5×0.042)A_5=8000(1+5\times0.042).

Answer: The balance is $9680\$9680.

Practice 10 · Rows of seats

A hall has 28 seats in row 1 and 4 more in each later row. Find the seats in row 16 and the total seats in the first 16 rows.

  1. 1u16=28+15(4)=88u_{16}=28+15(4)=88.
  2. 2S16=16(28+88)2S_{16}=\frac{16(28+88)}{2}.

Answer: Row 16 has 88 seats; the first 16 rows contain 928 seats.

Practice 11 · Approximate arithmetic data

Weekly subscriptions are 205,218,230,244,257205,218,230,244,257. Estimate an arithmetic model using the endpoints and predict week 9.

  1. 1dapprox=(257−205)4=13d_{\text{approx}}=\frac{(257-205)}{4}=13.
  2. 2Use un≈205+13(n−1)u_n\approx205+13(n-1).
  3. 3u9≈205+8(13)=309u_9\approx205+8(13)=309.

Answer: The model predicts about 309 subscriptions in week 9; this is an extrapolation.

Practice 12 · AI HL capacity challenge

Daily bookings start at 420420 and increase by 3535 per day. A system can process at most 10001000 bookings in one day. Find the first day the model exceeds capacity, the total bookings over all earlier valid days, and the revenue if each booking earns $12\$12.

  1. 1Solve 420+35(n−1)>1000420+35(n-1)>1000. The first integer solution is n=18n=18.
  2. 2Day 17 has u17=420+16(35)=980u_{17}=420+16(35)=980, so days 1 through 17 are valid.
  3. 3S17=17(420+980)2=11900S_{17}=\frac{17(420+980)}{2}=11900.
  4. 4Revenue =11900(12)=142800=11900(12)=142800.

Answer: Capacity is first exceeded on day 18. The valid 17 days contain 1190011900 bookings and generate $142800\$142800.

Topic summary and cheat sheet

Use this checklist to choose efficiently between a term formula, a sum formula, sigma notation and a technology-supported approximate model.

Rules and formulas

Common difference

d=un+1−und=u_{n+1}-u_n

Nth term

un=a+(n−1)du_n=a+(n-1)d

Sum from first and last terms

Sn=n2(a+un)S_n=\frac n2(a+u_n)

Sum from first term and difference

Sn=n2[2a+(n−1)d]S_n=\frac n2[2a+(n-1)d]

Sigma form

∑r=1n[a+(r−1)d]=Sn\sum\limits_{r=1}^{n}[a+(r-1)d]=S_n

Topic summary

  • A sequence is an ordered list; a series is a sum of terms.
  • unu_n is one term, while SnS_n is the sum of the first nn terms.
  • An arithmetic sequence has constant additive change.
  • The common difference is d=un+1−und=u_{n+1}-u_n.
  • Positive d increases, negative d decreases, and zero d is constant.
  • A recurrence requires both u1=au_1=a and un+1=un+du_{n+1}=u_n+d.
  • The explicit formula is un=a+(n−1)du_n=a+(n-1)d.
  • There are n−1n-1 changes between u1u_1 and unu_n.
  • A term number must normally be a positive integer.
  • From upu_p and uqu_q, use d=(uq−up)(q−p)d=\frac{(u_q-u_p)}{(q-p)}.
  • Arithmetic-sequence points form a discrete straight-line pattern.
  • In un=dn+(a−d)u_n=dn+(a-d), the gradient is dd.
  • Use Sn=n(a+un)2S_n=\frac{n(a+u_n)}{2} when the last term is known.
  • Use Sn=n[2a+(n−1)d]2S_n=\frac{n[2a+(n-1)d]}{2} when aa and dd are known.
  • An arithmetic-series sum is number of terms times average of first and last terms.
  • Finding n from a sum often produces a quadratic equation.
  • Reject roots that violate positive-integer or contextual constraints.
  • Sigma notation states an index, lower limit, upper limit and term expression.
  • From integer p through q inclusive, there are q minus p plus 1 terms.
  • Simple interest adds the same amount each period and is arithmetic.
  • For simple interest, An=P(1+nr)A_n=P(1+nr) and yearly change is PrPr.
  • Define clearly whether indexing starts at time 0 or term 1.
  • Use an nth term for one period and a series for an accumulated total.
  • Real data can be approximately arithmetic even when differences are not identical.
  • An endpoint estimate averages change over several equal intervals.
  • Linear regression uses all paired data and estimates the common difference from its gradient.
  • Residuals help evaluate whether constant additive change is appropriate.
  • Extrapolations are less reliable than predictions within observed data.
  • A GDC can generate tables, sums and solutions but does not remove the need for interpretation.
  • Always report units per step for a common difference in context.
  • Check indexing, sign, formula choice, integer constraints and units.
  • Finish an application with a contextual conclusion and model limitation where relevant.