1.2 Arithmetic sequences and series

Introduction

Many quantities change by the same amount at every step: seats added to each row of a theatre, centimetres added to each rung of a ladder, money added to a savings jar every week, interest added each year on a simple-interest loan. Lists of numbers built this way are arithmetic sequences, and adding up their terms gives an arithmetic series. Two short formulas give any term and any sum directly, so a question such as "how many seats are in row 40?" or "how much is saved in a year?" needs one substitution instead of a long list.

Real data is rarely perfectly arithmetic. The world's population grew by between 65 million and 78 million a year from 2020 to 2025, close to but not exactly a constant amount. Approximating the common difference turns such data into a simple model that can be used for prediction, and judging how far such a model can be trusted matters as much as the formulas.

Sequences and their terms

A list of numbers written in a definite order, such as the heights of the rungs on a ladder or the amount in a savings account at the end of each month, is called a sequence. The order matters: the list 3, 7, 11 is a different sequence from 11, 7, 3, even though it contains the same numbers.

Each number in the list is a term, labelled by its position. The first term is written $u_1$, the second $u_2$, and the term in position $n$ is $u_n$. The small number $n$ is the term's position (its index), not its value: in the sequence 5, 8, 11, 14, … the position $n = 3$ holds the value $u_3 = 11$.

A sequence can stop or go on forever. The number of seats in each row of a particular theatre stops at the last row, while the list of positive even numbers 2, 4, 6, 8, … has no last term. The three dots "…" at the end of a list show that the sequence continues in the same way.

The distinction matters for sums. A finite sequence always has a total, whereas the terms of an infinite sequence can only be added up to some chosen point.

The general term

Listing terms one by one is slow when a far-off term is needed. A general term is a formula that produces $u_n$ directly from the position $n$. Substituting $n = 1, 2, 3, \ldots$ gives the terms in order, and substituting any other whole number gives that term at once.

The general term $u_n = 4n + 1$, for example, gives $u_1 = 5$, $u_2 = 9$, $u_3 = 13$ and $u_{50} = 201$. The positions are always the positive integers $1, 2, 3, \ldots$, so $n$ is never a fraction or a negative number, and a value such as "$u_{2.5}$" has no meaning.

The gaps between the terms of this sequence (6, 10, …) grow, so it is not arithmetic. In an arithmetic sequence the gap stays the same.

Arithmetic sequences

In the sequence 7, 11, 15, 19, 23, … each term is 4 more than the one before. A sequence that grows or shrinks by the same fixed amount at every step is arithmetic, and that fixed amount is its common difference.

The common difference is found by subtracting any term from the term that follows it. The order of the subtraction matters: later term minus earlier term.

A positive $d$ gives an increasing sequence and a negative $d$ a decreasing one. When $d = 0$ every term is equal, which is a constant sequence. The common difference need not be a whole number: 2.5, 3.1, 3.7, 4.3, … has $d = 0.6$.

Showing that a sequence is arithmetic

Checking the first few differences suggests that a sequence is arithmetic but does not prove it, because a pattern that holds for three terms might break later. When the general term is known, a proof works with a general $n$: subtract $u_n$ from $u_{n+1}$ and show that the result does not depend on $n$.

For $u_n = 5n + 2$:

The difference is 5 for every value of $n$, so the sequence is arithmetic with $d = 5$. Repeating the same step for $u_n = 2n^2 - 1$ gives $u_{n+1} - u_n = 4n + 2$, which changes with $n$, so that sequence is not arithmetic.

The formula for the nth term

To reach the 5th term of 7, 11, 15, … from the first, the common difference is added four times: $u_5 = 7 + 4 \times 4 = 23$. In general, moving from $u_1$ to $u_n$ takes $n - 1$ steps, one fewer than the position number, because the first term is already in place before any step is taken. The number line of terms shows $n$ terms joined by only $n - 1$ jumps of size $d$.

Adding $d$ a total of $n - 1$ times to $u_1$ gives the formula for the general term of any arithmetic sequence.

Expanding the formula shows a second, useful form. For 7, 11, 15, … the general term is $u_n = 7 + (n - 1) \times 4 = 4n + 3$. Every arithmetic sequence has a general term of the form $u_n = dn + c$: the coefficient of $n$ is the common difference, and the constant is $c = u_1 - d$, the value that a "zeroth" term would have. Spotting this form identifies an arithmetic sequence at once: $u_n = 120 - 8n$ is arithmetic with $d = -8$ and $u_1 = 112$.

Graphs of arithmetic sequences

Plotting each term $u_n$ against its position $n$ shows the structure of an arithmetic sequence. Because each step to the right adds the same amount $d$, the points lie on a straight line with gradient $d$. The points are separate, not joined, because only whole-number positions exist.

The line through the points is $y = 4x + 3$, which is the general term $u_n = 4n + 3$ written as a function. Arithmetic sequences are therefore the whole-number version of linear functions, and the same idea of a constant rate of change reappears in linear models in subtopic 2.5 (linear model).

Finding a term's position and counting terms

The $n$th-term formula can be used backwards: given a value, set $u_n$ equal to it and solve for $n$. A whole-number answer gives the position of that term. A non-integer answer shows that the value is not a term.

The same idea counts the terms of a finite arithmetic sequence. The sequence 14, 21, 28, …, 350 has its last term at position $n$ where $14 + 7(n - 1) = 350$, so $n - 1 = 48$ and $n = 49$. The general rule is

The "+ 1" is there for the same reason as the $n - 1$ in the formula: 48 gaps separate 49 terms, in the same way that 4 fence panels need 5 posts.

Rounding n in context

When the answer for $n$ is not a whole number, the context decides whether to round up or down, and the usual rules for rounding to the nearest integer do not apply. Two situations recur.

• "What is the greatest number…" (rungs that fit on a ladder, complete rows that can be built, chairs in a stack under a doorway): the quantity must stay within a limit, so $n$ is rounded down. Rounding up would break the limit.

• "When does it first exceed…" (the first week a distance is longer than a marathon, the first year a value passes a target): the quantity must pass a limit, so $n$ is rounded up. Rounding down gives a term that has not yet passed it.

Checking the two whole numbers either side of the decimal answer confirms the direction, as with the first negative term of 250, 243, 236, …

Finding u₁ and d from two terms

Often only two terms are known, such as the 4th and the 11th. Each one gives an equation in the two unknowns $u_1$ and $d$, so the result is a system of linear equations in two unknowns.

There is also a quick route by hand. Between the 4th and 11th terms there are $11 - 4 = 7$ steps of size $d$, so the difference between the two terms divided by 7 gives $d$ directly. Once $d$ is known, $u_1$ follows by stepping back from either term.

The GDC's simultaneous equation solver, and how to enter a system into it, is taught in subtopic 1.8, Equations and systems using technology. Writing the two equations on the page matters in either method, because they show the method that the solver carries out.

Three consecutive terms

If $a$, $b$ and $c$ are consecutive terms of an arithmetic sequence, then $b - a = c - b$, which rearranges to

The middle term is the mean of its two neighbours. This gives a quick way to fill a gap (the term between 18 and 31 is 24.5) and an equation when the terms contain an unknown.

Generating sequences with technology

A GDC, a spreadsheet or graphing software lists the terms of a sequence quickly, which is useful for checking an answer or for finding the first term that passes a value.

• GDC table. Enter the general term as a function, for example $Y_1 = 82 + (X - 1) \times 4.5$, and view the table with a start of 1 and a step of 1. Each row is a position and its term. Many GDCs also have a sequence or recursion mode (on a TI-84 Plus CE, MODE → SEQ; on a Casio fx-CG50, MENU → Recursion) in which $u(n)$ is entered directly.

• Spreadsheet. Put the first term in cell A1 and the formula =A1+4.5 in cell A2, then fill down. Each cell adds the common difference to the cell above, which is the definition of an arithmetic sequence written as a rule.

• List command. On a TI-84 Plus CE, seq(82+(X−1)*4.5, X, 1, 30) (from 2nd STAT → OPS) produces the first 30 terms as a list.

Arithmetic series

Adding the terms of a sequence gives a series. The total number of seats in a theatre is the sum of the numbers of seats in its rows, and the total saved over a year is the sum of the weekly deposits.

Everyday English often uses the two words for the same thing, but in mathematics they differ. A sequence is a list of separate numbers, and a series is the result of adding them.

Most questions add up a fixed number of terms from the start of the sequence. The total of the first $n$ terms is written $S_n$: $S_1 = u_1$, $S_2 = u_1 + u_2$, $S_3 = u_1 + u_2 + u_3$, and so on.

Deriving the sum formula by pairing

A well-known method adds the series twice, once forwards and once backwards. For $2 + 5 + 8 + 11 + 14$, the first term plus the last is 16, the second plus the second-last is also 16, and so is every other pair: moving inwards, one term rises by $d$ while its partner falls by $d$. The staircase of squares shows this for five terms: the shaded rows hold 2, 5, 8, 11 and 14 squares, the reversed copy fills the rest, and every row ends up 16 squares long.

The same argument works for any arithmetic series. Writing $S_n$ forwards and backwards and adding the two lines column by column gives $n$ columns, each equal to $u_1 + u_n$:

Dividing by 2 gives the first form of the formula. In words, the sum is the number of terms multiplied by the mean of the first and last terms.

Replacing $u_n$ by $u_1 + (n - 1)d$ gives the second form, which needs only the first term and the common difference:

The pairing idea is old and appears in many cultures. A popular story, of uncertain reliability, tells of the young Carl Friedrich Gauss adding 1 to 100 by forming 50 pairs that each make 101, for a total of 5,050. Rules for summing arithmetic series were known long before, to Archimedes in Greece, to Zhang Qiujian in China, and to Aryabhata in India, whose Aryabhatiya (about 499 CE) treats arithmetic progressions.

Finding n when the sum is known

When the total is given and the number of terms is unknown, substituting into the sum formula gives an equation in $n$ that contains $n^2$, a quadratic equation. With a GDC there are three good ways to solve it:

1. Equation solver or polynomial solver. Rearrange to the form $an^2 + bn + c = 0$ (or enter $S_n$ − total $= 0$) and solve. Keep the positive solution that makes sense.

2. Graphs. Graph $y = S_x$ and the horizontal line $y = \text{total}$ and find their point of intersection.

3. Table. Tabulate $S_x$ and find the two consecutive values either side of the total.

Each of these finds the roots of the equation. A quadratic usually has two, and the context rules out any that are negative or impossible, as the tin display shows. When the total is "at most" or "at least" a value, the decimal answer is rounded down or up with the same reasoning as for terms.

Recovering terms from a sum formula

When a formula for $S_n$ is given, the terms can be found from it. The sum of the first $n$ terms is the sum of the first $n - 1$ terms plus the $n$th term, so

A sum formula of the form $S_n = an^2 + bn$ always belongs to an arithmetic sequence, with $d = 2a$. Expanding $\frac{n}{2}(2u_1 + (n - 1)d)$ shows why: it equals $\frac{d}{2}n^2 + \left(u_1 - \frac{d}{2}\right)n$.

Sigma notation

Writing out a long series term by term is slow, and "…" can hide where the sum starts and stops. Sigma notation packs a whole series into one expression with the capital Greek letter sigma, $\Sigma$, the Greek S, for "sum".

The expression is read as "the sum of $3k + 4$ for $k$ from 1 to 20". The letter $k$ is the index: it takes each whole-number value from the number below the sigma to the number above it, and each value of $k$ gives one term.

Mathematical notation uses letters from several alphabets. The Greek capital $\Sigma$ for sums sits beside the Latin letters of algebra, Greek $\pi$ and $\mu$, and the Hebrew $\aleph$ used for sizes of infinite sets in higher mathematics.

Reading the limits

The numbers below and above the sigma fix the first and last values of the index. They need not start at 1, and the number of terms is

So $\sum_{k=5}^{20}$ has $20 - 5 + 1 = 16$ terms, not 15 and not 20. The letter used for the index does not change the value: $\sum_{k=1}^{10} k$ and $\sum_{r=1}^{10} r$ are both 55.

The last row is a sum of 12 sevens, so it equals $7 \times 12 = 84$. A constant inside the sigma is added once for every value of the index.

Evaluating sums of arithmetic sequences

When the expression inside the sigma is of the form $ak + b$, consecutive terms differ by $a$, so the sum is an arithmetic series. Writing down the first term, last term and number of terms turns it into a partial sum that the formula can evaluate.

Writing a series in sigma notation

To write a series in sigma notation, find its general term, then find the values of the index for the first and last terms. For $4 + 9 + 14 + \cdots + 99$, the common difference is 5, so $u_k = 4 + 5(k - 1) = 5k - 1$. The last term 99 needs $5k - 1 = 99$, so $k = 20$:

The same series can be written with other limits, for example $\sum_{k=0}^{19}(5k + 4)$. Both are correct as long as they produce exactly the same terms.

Sigma notation on the GDC

GDCs evaluate a sigma sum directly. Two methods are common:

• Summation template. On a TI-84 Plus CE, MATH → summation Σ( gives a template: enter the expression, the index, and the lower and upper limits. On a Casio fx-CG50, in the Run-Matrix app, OPTN → CALC → Σ( takes the expression, the variable, the start and the end.

• sum of a list. On a TI-84 Plus CE, sum(seq(100−3X, X, 5, 20)) builds the 16 terms as a list and adds them; sum( is under 2nd STAT → MATH and seq( under 2nd STAT → OPS.

Applications of arithmetic sequences and series

Arithmetic sequences model any situation in which a quantity changes by a constant amount per step: rows of seats, stacked objects, training plans, regular savings with no interest, and the total owed on a loan charged simple interest. In each case the work starts by deciding what $n$ counts (rows, weeks, years), what $u_1$ and $d$ are, and whether the question asks for a term or a sum.

Simple interest

Money lent or borrowed earns or costs interest, a charge for the use of the money over time. The amount first invested or borrowed is the principal.

With simple interest, the interest each year is a fixed percentage of the original principal only. Interest already earned does not itself earn interest, so the same amount is added every year.

If the principal is $C$ and the interest rate is $r\%$ per year, each year's interest is $\frac{Cr}{100}$. Over $n$ years the total interest is $n$ times this:

The value of an investment (or the amount owed on a debt) after $n$ years is the principal plus the interest, $C + \frac{Crn}{100}$. Year by year these values increase by the same amount $\frac{Cr}{100}$, so they form an arithmetic sequence with common difference equal to one year's interest.

A real quoted rate shows the size of the numbers. In September 2026, UK banks quoted an average interest rate of 11.52% a year on a £5,000 personal loan. Charged as simple interest, this would cost £576 every year ($\frac{5000 \times 11.52}{100} = 576$), so the amount owed after 1, 2, 3, … years would be £5,576, £6,152, £6,728, …, an arithmetic sequence with $d = 576$. Real personal loans are repaid in monthly instalments and the interest is charged on the falling balance, which is the amortization of subtopic 1.7, so this simple-interest version is an illustration of the arithmetic pattern rather than how such a loan works.

Source: Bank of England, quoted household interest rate, £5,000 personal loan (series IUMBX67), September 2026.

The way the index is set up needs care in finance questions. In $V_n = 2400 + 84n$, the position $n = 0$ is the start, before any interest, and $V_n$ is the value after $n$ years. Writing the same values as $u_1 = 2484$ (after one year) and $d = 84$ is equally correct, but then $u_n$ is the value after $n$ years and $u_1$ is not the principal. Stating clearly what $n$ counts avoids an answer that is one year out.

Contexts with patterns, rows and plans

Diagrams of tiles, matchsticks or panes are a common way to present an arithmetic sequence: the first few terms are counted from the pictures, and the formulas take over from there. Seating plans, stacked objects and training plans work the same way. A training plan has one extra feature: the target is often given in a different unit, such as a marathon's fixed distance of 42.195 km.

Both the question and the answer of a contextual problem should be read in the context. In the training plan, week 58 is more than a year away, and a runner cannot keep adding 600 m every week indefinitely. The arithmetic model describes the plan as written, and a coach would expect to adjust it once the runs become very long.

Comparing two arithmetic plans

When two quantities each grow by a constant amount, the one with the larger common difference eventually overtakes the other, however far behind it starts. Comparing two arithmetic sequences leads to a linear inequality in $n$. Comparing their sums leads to an inequality that, after cancelling the common factor $\frac{n}{2}$, is also linear.

The contrast between parts (b) and (c) matters in context. Offer A's yearly salary is higher from year 7, but the earlier shortfall takes another five years to recover, so a person who expects to stay fewer than 12 years earns more in total with offer B.

When real data is not perfectly arithmetic

Real quantities seldom change by exactly the same amount every year, but many change by roughly the same amount. An arithmetic model replaces the varying differences with a single estimated common difference. This simplification is a mathematical model. It gives reasonable predictions a short way beyond the data, but its accuracy over a long range is doubtful.

Estimating the common difference

The differences between consecutive values give the year-to-year changes. Their mean is the natural estimate of $d$, and it has a convenient shortcut: when the differences are added, all the middle values cancel, leaving last minus first.

The mean difference is therefore the total change divided by the number of steps, which is one fewer than the number of values. The model built from it, $u_n = u_1 + (n - 1)d$, matches the first and last data values exactly.

The world's population from 2020 to 2025 is a good example.

Source: World Bank, World Development Indicators (SP.POP.TOTL), 2026. Populations rounded to the nearest million.

The differences range from 65.7 million to 78.0 million: not constant, but not far apart. The mean difference is

Taking 2020 as $n = 1$, the model is $P_n = 7.854 + 0.0723(n - 1)$ billion. The plotted data points lie close to the model's line, with the middle years slightly below it because the yearly increases were smaller at the start of the period than at the end.

The model predicts the 2030 population as the 11th term:

This prediction lies outside the years of the data, an extrapolation, so it is less reliable than an estimate for a year inside the range. It assumes that the yearly increase stays near 72 million, which is reasonable for a few years but ignores changes in birth and death rates that make the yearly increase drift over decades.

Case study: Carbon dioxide at Mauna Loa

The Mauna Loa Observatory in Hawaii has measured the concentration of carbon dioxide (CO₂) in the air since 1958, in parts per million (ppm). The annual mean rose from 401.01 ppm in 2015 to 427.35 ppm in 2025, and the yearly increases over that decade varied between 1.96 ppm and 3.53 ppm.

Source: NOAA Global Monitoring Laboratory, Mauna Loa CO₂ annual mean data, 2026.

The ten differences average

With 2015 as $n = 1$, the arithmetic model is $C_n = 401.01 + 2.634(n - 1)$. It predicts 440.52 ppm for 2030 ($n = 16$), about 441 ppm. Over the decade the model is a fair summary: although single years differ from 2.634 by up to 0.9 ppm, these variations partly cancel, so the model's values stay within about 1 ppm of the measurements.

A longer view shows where the arithmetic model fails. The mean yearly increase has itself grown decade by decade.

The rise from 1959 to 1960 was only 0.93 ppm. An arithmetic model fitted to 1959–1969, with $d \approx 0.864$, would have predicted about 373 ppm for 2025, 54 ppm below the measured 427.35 ppm. Each decade's increase is larger than the last because emissions have grown, so the data is better described by a curve than by a straight line. An arithmetic model is useful here only for short-range prediction, and its common difference must be re-estimated from recent data.

Regression fits a line to all the data points instead of only the first and last, and it also allows a straight-line model to be compared with a curved one; regression lines are taught in subtopic 4.4 (regression line of y on x).

Exam focus

Arithmetic sequences and series are examined very often, usually as a Paper 1 question of 4–7 marks in two or three parts, or as the opening parts of a longer Paper 2 question that later compares the arithmetic model with a geometric sequence (two salaries, two savings plans, two businesses whose profits grow differently). The command terms are write down ($d$ or $u_1$ read straight from given terms, usually 1 mark with no working needed), find and calculate (a substitution into a formula, with working), show that (usually an expression for $S_n$), hence (use the result just shown), determine and justify or comment (on a model's reliability or a comparison).

The markschemes follow a regular pattern. An (M1) is given for substituting into the $u_n$ or $S_n$ formula, an (A1) for a correct substitution and an A1 for the answer; a correct answer from a GDC earns the implied marks, but an incorrect one with no working earns nothing. Contexts that count objects (rungs, rows, weeks, levels) need a whole-number answer rounded in the correct direction, and a decimal such as $n = 29.7$ left unrounded loses the final mark.

The most common lost marks are counting errors: $u_1 + nd$ instead of $u_1 + (n - 1)d$, the wrong number of terms in a sigma sum, and giving a single year's value when the total over several years is asked for. Examiners' reports also mention answers given as negative numbers when the context needs a positive amount, such as "−1.5 cups of food per day" for a bag that empties by 1.5 cups a day.

When the equation to solve is quadratic, as in part (d), write the equation or inequality first and then name the GDC method ("solve on GDC", or a sketch with the intersection marked). A labelled sketch of the curve and the horizontal line, with the point of intersection marked, earns the method mark even without algebra.

Practice questions

1. For the arithmetic sequence 31, 26.5, 22, …, write down $u_1$ and $d$, and find $u_{15}$. [3]

2. Find the number of terms in the sequence 9, 15, 21, …, 357. [2]

3. The 6th term of an arithmetic sequence is 41 and the 15th term is 113. Find the first term and the common difference, and hence find $S_{30}$. [5]

4. Calculate $\displaystyle\sum_{k=1}^{30}(4k - 7)$ and $\displaystyle\sum_{r=10}^{40}(2r + 1)$

5. A principal of $6,000 is invested at 2.8% simple interest a year. (a) Calculate the value of the investment after 5 years. (b) Find the number of complete years after which the value first reaches $8,000. [5]

6. A pattern of hexagons is made from matchsticks. Pattern 1 uses 6 matchsticks, pattern 2 uses 11 and pattern 3 uses 16, and the pattern continues in the same way. (a) Find the number of matchsticks in pattern 40. (b) Find the total number of matchsticks needed to make patterns 1 to 40. (c) Find the largest pattern that can be made with 500 matchsticks. [7]

7. The sum of the first $n$ terms of a sequence is $S_n = 4n^2 - n$. (a) Write down $u_1$

8. A café's sales of a new drink in its first five weeks are 380, 402, 431, 447 and 478 cups. (a) Estimate the weekly common difference. (b) Predict the sales in week 8. (c) Comment on the reliability of your prediction. [5]

9. Offer A pays $38,000 in year 1, rising by $2,000 a year; offer B pays $42,200 in year 1, rising by $1,300 a year. Find the number of years after which the total earned under offer A first exceeds that under offer B. Justify your answer. [5]

Summary

• A sequence is an ordered list of terms $u_1, u_2, u_3, \ldots$; a series is the sum of its terms, and $S_n$ is the sum of the first $n$ terms.

• An arithmetic sequence adds the same common difference $d = u_{n+1} - u_n$ at each step; to prove a sequence is arithmetic, show that $u_{n+1} - u_n$

• The $n$th term is $u_n = u_1 + (n - 1)d$, because $n$

• The sum of the first $n$ terms is $S_n = \frac{n}{2}(u_1 + u_n) = \frac{n}{2}\left(2u_1 + (n - 1)d\right)$

• Two terms give two linear equations in $u_1$ and $d$; a given sum gives a quadratic in $n$, solved with the GDC, keeping only solutions that fit the context.

• In context, "greatest number" answers are rounded down and "first time it exceeds" answers are rounded up; check the whole numbers either side.

• $\sum_{k=a}^{b} u_k$ has $b - a + 1$

• Simple interest $I = \frac{Crn}{100}$ (not in the booklet) adds the same amount each year, so values form an arithmetic sequence.

• For real data that is approximately arithmetic, estimate $d$ as $\frac{u_n - u_1}{n - 1}$

Self-check

• To do:

  I can tell a sequence from a series, and a finite sequence from an infinite one.

• To do:

  I can use a general term to find any term and decide whether a value is a term.

• To do:

  I can write down $u_1$ and $d$ from a list, a table or a pattern diagram.

• To do:

  I can show that a sequence is arithmetic using $u_{n+1} - u_n$ for a general $n$.

• To do:

  I can explain why the $n$th-term formula contains $n - 1$ and not $n$.

• To do:

  I can find the position of a term and count the terms of a finite arithmetic sequence.

• To do:

  I can find $u_1$ and $d$ from two terms, by hand and with the GDC's equation solver.

• To do:

  I can derive the sum formula by pairing first and last terms.

• To do:

  I can choose the better form of the $S_n$ formula and use it.

• To do:

  I can find $n$ from a given sum with the GDC and reject solutions that do not fit the context.

• To do:

  I can decide whether to round $n$ up or down from the wording of a question.

• To do:

  I can read, write and evaluate a sum in sigma notation, counting the terms as top − bottom + 1.

• To do:

  I can calculate simple interest with $I = \frac{Crn}{100}$ and find a rate or a time from it.

• To do:

  I can estimate a common difference from real data, make a prediction and comment on its reliability.

• To do:

  I can write down the equations, entries and checks that show my GDC working.

$= 109$ (seats) A1

[2 marks]

The first three terms are 1, 7, 17.

(b) $u_{10} = 2(10)^2 - 1 = 199$.

(c) Set the general term equal to 161 and solve for $n$:

Because $n = 9$ is a positive whole number, 161 is a term: it is $u_9$. If the solution had been a decimal such as $n = 8.7$, the value would not be a term of the sequence.

(c) For a general $n \ge 2$:

This also gives $u_1 = 5$, so $u_n = 4n + 1$ for every $n$. Then $u_{n+1} - u_n = (4n + 5) - (4n + 1) = 4$, a constant, so the sequence is arithmetic with $d = 4$.

(b) First term $k = 5$: $100 - 15 = 85$. Last term $k = 20$: $100 - 60 = 40$. Number of terms: $20 - 5 + 1 = 16$.

With a GDC: both sums can be checked with the GDC's summation command. For (b), the GDC must be given the lower limit 5, not 1.

$= 69$ (pages) A1

[2 marks]

(c) $S_n = \frac{n}{2}\big(2 \times 12 + (n - 1) \times 3\big)$ M1

$= \frac{n}{2}(3n + 21)$ A1

$= \frac{3n^2 + 21n}{2}$ AG

Note: the line before the AG must be seen; substituting a value of $n$ to check the result earns no marks.

[2 marks]

(d) $\frac{3n^2 + 21n}{2} \ge 1250$ (accept $=$) (M1)

$n = 25.578\ldots$ (from the GDC's solver, or a graph of $y = \frac{3x^2 + 21x}{2}$ and $y = 1250$) (A1)

26 days A1

Note: Accept $S_{25} = 1200$ and $S_{26} = 1287$ seen for the (A1). An answer of 25 is incorrect.

[3 marks]

(e) $\sum_{k=11}^{20}(3k + 9)$ (or equivalent, e.g. $\sum_{k=1}^{10}(3k + 39)$) A1

$u_{11} = 42$, $u_{20} = 69$, 10 terms: $\frac{10}{2}(42 + 69)$ or $S_{20} - S_{10} = 810 - 255$ (M1)

$= 555$ (pages) A1

[3 marks]

Total [11 marks]