Arithmetic progressions

AS · P1 · 17 min

An arithmetic progression (AP) is a list of numbers that goes up or down by the same amount every step, like $5, 9, 13, 17, \ldots$ or $40, 37, 34, \ldots$ . Paper 1 needs only two formulas, one for any term and one for the sum of the first $n$ terms, but the questions hide them inside words: savings that grow by a fixed amount, lengths cut from a rod, a sum that must first exceed a target. The skill being tested is turning a sentence into an equation in $a$ and $d$ , and then solving it cleanly.

Sequences, series and the language of APs

A sequence is an ordered list of numbers, called terms. We write $u_1$ for the first term, $u_2$ for the second, and $u_n$ for the $n$ th term. A series is what you get when you add terms of a sequence: $S_n = u_1 + u_2 + \cdots + u_n$ is the sum of the first $n$ terms.

Definition

An arithmetic progression is a sequence in which each term is obtained from the previous one by adding a fixed number $d$ , called the common difference. The first term is written $a$ .

So $u_{n+1} - u_n = d$ for every $n$ . The difference can be positive (the terms increase), negative (they decrease) or a fraction.

Examples:

$3, 7, 11, 15, \ldots$ has $a = 3$ and $d = 4$ .
$20, 17.5, 15, 12.5, \ldots$ has $a = 20$ and $d = -2.5$ .
$x, x + y, x + 2y, \ldots$ has $a = x$ and $d = y$ .

To find $d$ , always calculate later term minus earlier term. For $20, 17.5, \ldots$ that is $17.5 - 20 = -2.5$ , not $2.5$ .

The nth term

Why the formula has n minus 1

Write out the terms in terms of $a$ and $d$ :

u_1 = a, \quad u_2 = a + d, \quad u_3 = a + 2d, \quad u_4 = a + 3d, \quad \ldots

To reach the $n$ th term you start at $a$ and take $n - 1$ steps of size $d$ . The $10$ th term is $a + 9d$ , not $a + 10d$ : there are only $9$ gaps between $10$ terms, just as there are $9$ gaps between $10$ fence posts.

Key result

The $n$ th term of an AP with first term $a$ and common difference $d$ is

u_n = a + (n - 1)d

Three numbers $p$ , $q$ , $r$ are consecutive terms of an AP exactly when $q - p = r - q$ , that is

2q = p + r

so the middle term is the mean of its neighbours.

Because $u_n = dn + (a - d)$ is a linear function of $n$ , the terms of an AP plotted against $n$ lie on a straight line with gradient $d$ .

y = 2x + 1 (1, 3) (2, 5) (3, 7) (4, 9) (5, 11) (6, 13) (7, 15) (8, 17)

The AP $3, 5, 7, \ldots$ plotted as points $(n, u_n)$ . They lie on $y = 2x + 1$ , a line of gradient $d = 2$ .

How many terms?

If you know the first term, the last term and $d$ , the formula tells you how many terms there are. For $7, 11, 15, \ldots, 203$ :

203 = 7 + (n - 1) \times 4 \quad\Rightarrow\quad n - 1 = 49 \quad\Rightarrow\quad n = 50

The sum of the first n terms

The pairing argument

The trick is said to go back to a young Gauss adding $1 + 2 + \cdots + 100$ . Write the sum forwards and backwards, then add the two lines:

\begin{aligned} S_n &= a + (a + d) + (a + 2d) + \cdots + (l - d) + l \\ S_n &= l + (l - d) + (l - 2d) + \cdots + (a + d) + a \end{aligned}

Here $l$ is the last term, $l = a + (n - 1)d$ . Adding the columns, every pair sums to $a + l$ , and there are $n$ pairs:

2S_n = n(a + l) \quad\Rightarrow\quad S_n = \frac{n}{2}(a + l)

In words: the sum is the number of terms times the average of the first and last terms. Substituting $l = a + (n - 1)d$ gives the second form.

Key result

The sum of the first $n$ terms of an AP is

S_n = \frac{n}{2}(a + l) = \frac{n}{2}\left[2a + (n - 1)d\right]

where $l$ is the last ( $n$ th) term. Both forms are in the list of formulae (MF19).

Use $\tfrac{n}{2}(a + l)$ when you know the last term, and $\tfrac{n}{2}\left[2a + (n - 1)d\right]$ when you know $d$ .

Because $S_n$ is a quadratic in $n$ (with no constant term), questions about "the least $n$ for which the sum exceeds a value" lead to a quadratic inequality. See Quadratic inequalities.

y = x(x + 2) (1, 3) (2, 8) (3, 15) (4, 24) (5, 35) (6, 48) (7, 63) (8, 80) (9, 99) (10, 120) (11, 143)

For the AP $3, 5, 7, \ldots$ the sums $S_n = \tfrac{n}{2}(6 + 2(n - 1)) = n(n + 2)$ lie on a parabola, not a line.

Getting terms back from a sum formula

Sometimes the question gives $S_n$ as a formula in $n$ . Since $S_n = S_{n-1} + u_n$ ,

u_n = S_n - S_{n-1} \quad (n \ge 2), \qquad u_1 = S_1

If the resulting $u_n$ is linear in $n$ , the sequence is an AP and its common difference is the coefficient of $n$ .

Solving an AP problem

Identify what is given: a term ( $u_n$ ) or a total ( $S_n$ ). Underline phrases like "the 10th payment" (a term) and "the total of the first 10 payments" (a sum).
Write each piece of information as an equation in $a$ and $d$ using $u_n = a + (n - 1)d$ or $S_n = \tfrac{n}{2}\left[2a + (n - 1)d\right]$ .
Solve the equations simultaneously, usually by subtracting to eliminate $a$ .
If $n$ is the unknown, form an equation or inequality in $n$ , solve it, and remember that $n$ must be a positive integer.
Answer the question asked, with units, and check one value by substitution.

Worked examples

Finding the first term and common difference

The fifth term of an arithmetic progression is $17$ and the twelfth term is $45$ . Find the first term, the common difference and the sum of the first $20$ terms.

Solution

Write each fact as an equation:

\begin{aligned} u_5 &= a + 4d = 17 \\ u_{12} &= a + 11d = 45 \end{aligned}

Subtracting the first from the second: $7d = 28$ , so $d = 4$ . Then $a = 17 - 16 = 1$ .

S_{20} = \frac{20}{2}\left[2(1) + 19(4)\right] = 10 \times 78 = 780

Summing between two values

Find the sum of all the multiples of $7$ between $100$ and $500$ .

Solution

The multiples of $7$ form an AP with $d = 7$ . The first one above $100$ is $105$ ( $7 \times 15$ ) and the last one below $500$ is $497$ ( $7 \times 71$ ).

Number of terms:

497 = 105 + (n - 1)(7) \quad\Rightarrow\quad n - 1 = 56 \quad\Rightarrow\quad n = 57

Using the first and last terms:

S_{57} = \frac{57}{2}(105 + 497) = \frac{57 \times 602}{2} = 17\,157

A quick check on the count: from $7 \times 15$ to $7 \times 71$ there are $71 - 15 + 1 = 57$ multiples.

When the sum is zero and the first negative term

An arithmetic progression has first term $51$ and common difference $-3$ .

(a) Find the number of terms for which the sum of the progression is zero.

(b) Find the first negative term.

Solution

(a)

S_n = \frac{n}{2}\left[2(51) + (n - 1)(-3)\right] = \frac{n}{2}(105 - 3n)

Setting $S_n = 0$ gives $n = 0$ or $n = 35$ . Since $n$ must be positive, $n = 35$ .

(b) We need $u_n < 0$ :

51 - 3(n - 1) < 0 \quad\Rightarrow\quad 3(n - 1) > 51 \quad\Rightarrow\quad n > 18

The first integer value is $n = 19$ , and $u_{19} = 51 - 3 \times 18 = -3$ .

It makes sense that the sum returns to zero at $n = 35$ : the terms run $51, 48, \ldots, 3, 0, -3, \ldots, -51$ , and the positives cancel the negatives in pairs.

Terms from a sum formula

The sum of the first $n$ terms of a sequence is given by $S_n = 3n^2 + 2n$ . Show that the sequence is an arithmetic progression and state its first term and common difference.

Solution

\begin{aligned} u_n &= S_n - S_{n-1} \\ &= 3n^2 + 2n - \left[3(n - 1)^2 + 2(n - 1)\right] \\ &= 3n^2 + 2n - \left[3n^2 - 6n + 3 + 2n - 2\right] \\ &= 6n - 1 \end{aligned}

Then $u_{n} - u_{n-1} = (6n - 1) - (6n - 7) = 6$ , a constant, so the sequence is an AP with $d = 6$ . The first term is $u_1 = S_1 = 3 + 2 = 5$ , which agrees with $6(1) - 1 = 5$ .

Least number of terms in context

Mia saves $200$ dollars in the first month. Each month after that she saves $15$ dollars more than in the previous month. Find the number of months it takes for her total savings to first exceed $10\,000$ dollars.

Solution

The monthly amounts form an AP with $a = 200$ , $d = 15$ . We need the least $n$ with $S_n > 10\,000$ :

\begin{aligned} \frac{n}{2}\left[400 + 15(n - 1)\right] &> 10\,000 \\ n(385 + 15n) &> 20\,000 \\ 15n^2 + 385n - 20\,000 &> 0 \end{aligned}

The positive root of $15n^2 + 385n - 20\,000 = 0$ is

n = \frac{-385 + \sqrt{385^2 + 4(15)(20\,000)}}{30} = \frac{-385 + \sqrt{1\,348\,225}}{30} \approx 25.87

So $n = 26$ months. Check: $S_{25} = \tfrac{25}{2}(400 + 360) = 9500$ , which is not enough, and $S_{26} = 13(400 + 375) = 10\,075$ , which is.

A relationship between two sums

The first term of an arithmetic progression is $a$ and the common difference is $d$ , where $d \neq 0$ . The sum of the first $20$ terms is four times the sum of the first $10$ terms.

(a) Show that $d = 2a$ .

(b) Given also that the $15$ th term is $87$ , find $a$ and $d$ .

Solution

(a)

S_{20} = 10(2a + 19d), \qquad S_{10} = 5(2a + 9d)

$S_{20} = 4S_{10}$ gives

\begin{aligned} 10(2a + 19d) &= 20(2a + 9d) \\ 20a + 190d &= 40a + 180d \\ 10d &= 20a \\ d &= 2a \end{aligned}

as required.

(b) $u_{15} = a + 14d = a + 28a = 29a = 87$ , so $a = 3$ and $d = 6$ .

Cutting a wire

A wire of length $5.8$ m is cut into pieces whose lengths, in centimetres, form an arithmetic progression. The shortest piece is $10$ cm and the longest is $48$ cm. Find the number of pieces and the common difference.

Solution

Work in centimetres: the total is $580$ cm. Using the first and last terms:

S_n = \frac{n}{2}(10 + 48) = 29n = 580 \quad\Rightarrow\quad n = 20

Then the last term gives $d$ :

48 = 10 + 19d \quad\Rightarrow\quad d = 2

There are $20$ pieces and the common difference is $2$ cm.

Watch out

Using $n$ instead of $n - 1$ . The $12$ th term is $a + 11d$ . Writing $a + 12d$ is the single most common error in this topic.

Mixing up a term and a sum. "She saves 380 dollars in the last month" is a term $u_n$ ; "she saves $10\,000$ dollars in total" is a sum $S_n$ . Read each sentence and decide before you write any equation.

Getting the sign of $d$ wrong. For $40, 37, 34, \ldots$ the difference is $-3$ . Always compute second term minus first term.

Giving a non-integer $n$ . $n$ counts terms, so it is a positive integer. If the inequality gives $n > 25.87$ , the answer is $26$ , not $25.87$ and not $25$ .

Miscounting terms between limits. From $105$ to $497$ in steps of $7$ there are $\frac{497 - 105}{7} + 1 = 57$ terms. Forgetting the $+1$ is a fence-post error.

Unit slips. If lengths are in centimetres and the total is given in metres, convert before forming the equation.

Exam tip

Show your equations. Marks are given for writing correct equations in $a$ and $d$ (often one mark each), then for solving. Write $a + 4d = 17$ , not just " $d = 4$ ".
"Show that" questions (like $d = 2a$ ) need every algebraic step; the final line must match the given result exactly.
Least or greatest $n$ . Solve the quadratic equation, then state the integer answer and, ideally, show the sums either side (as in the savings example). Examiners accept a solved inequality or a clear trial of values.
Context. Final answers need units and must answer the question asked ("in which month", "how many rows").
Combined questions. A common longer question links an AP to a geometric progression, for example "the first, second and fifth terms of an AP are the first three terms of a GP". These are covered in Problems combining progressions.
Formula list. Both sum formulas are in MF19, but you should be able to use them quickly from memory.

Summary

An AP has a constant difference $d$ between consecutive terms; find it as later term minus earlier term.
$u_n = a + (n - 1)d$ : there are $n - 1$ steps from the first term to the $n$ th.
$S_n = \tfrac{n}{2}(a + l) = \tfrac{n}{2}\left[2a + (n - 1)d\right]$ , proved by writing the sum forwards and backwards.
$p, q, r$ are consecutive terms of an AP exactly when $2q = p + r$ .
If $S_n$ is given, $u_n = S_n - S_{n-1}$ and $u_1 = S_1$ .
"Least $n$ " questions become quadratic inequalities; $n$ must be a positive integer.
Turn every sentence into an equation in $a$ and $d$ , then solve simultaneously.

Practice questions

Question

Find the $30$ th term and the sum of the first $30$ terms of the arithmetic progression $4, 11, 18, \ldots$
The sum of the first $8$ terms of an AP is $100$ and the sum of the first $16$ terms is $360$ . Find the first term and the common difference.
The numbers $x - 1$ , $2x + 1$ and $4x - 2$ are consecutive terms of an arithmetic progression. Find $x$ and the three terms.
An AP has first term $3$ and common difference $5$ . Find the least value of $n$ for which the sum of the first $n$ terms exceeds $1000$ .
Find the sum of all the integers from $1$ to $200$ inclusive that are not multiples of $3$ .
The sum of the first $n$ terms of a sequence is $S_n = 2n^2 - 5n$ . Find an expression for the $n$ th term, and find the first term of the sequence that is greater than $100$ .
A runner runs $5$ km on the first day of training and increases the distance by $0.4$ km each day. Find (a) the distance run on the $15$ th day, (b) the total distance run in the first $15$ days, (c) the day on which the total distance run first exceeds $300$ km.
The sum of the first ten terms of an AP is $145$ , and the sum of the next ten terms (the $11$ th to the $20$ th) is $445$ . Find the first term and the common difference.
The first term of an arithmetic progression is $6$ and the fifth term is $12$ . The progression has $n$ terms and the sum of all the terms is $90$ . Find the value of $n$ .
An arithmetic progression has first term $a$ and common difference $d$ , where $d \neq 0$ . For a particular value of $n$ , the sum of the first $2n$ terms is three times the sum of the first $n$ terms. (a) Show that $2a = (n + 1)d$ . (b) Given that $a = 12$ and the $n$ th term is $30$ , find $d$ and $n$ .

Answers

$a = 4$ , $d = 7$ . $u_{30} = 4 + 29 \times 7 = 207$ . $S_{30} = \tfrac{30}{2}(4 + 207) = 15 \times 211 = 3165$ .
$S_8 = 4(2a + 7d) = 100$ , so $2a + 7d = 25$ . $S_{16} = 8(2a + 15d) = 360$ , so $2a + 15d = 45$ . Subtracting, $8d = 20$ , so $d = 2.5$ and $2a = 25 - 17.5 = 7.5$ , giving $a = 3.75$ .
$2(2x + 1) = (x - 1) + (4x - 2)$ , so $4x + 2 = 5x - 3$ and $x = 5$ . The terms are $4$ , $11$ , $18$ (common difference $7$ ).
$S_n = \tfrac{n}{2}\left[6 + 5(n - 1)\right] = \tfrac{n(5n + 1)}{2} > 1000$ , so $5n^2 + n - 2000 > 0$ . The positive root is $n = \dfrac{-1 + \sqrt{1 + 40\,000}}{10} \approx 19.90$ , so $n = 20$ . Check: $S_{19} = 912$ , $S_{20} = 1010$ .
Sum of $1$ to $200$ : $\tfrac{200}{2}(1 + 200) = 20\,100$ . Multiples of $3$ : $3, 6, \ldots, 198$ , which is $66$ terms with sum $\tfrac{66}{2}(3 + 198) = 6633$ . Required sum: $20\,100 - 6633 = 13\,467$ .
$u_n = S_n - S_{n-1} = 2n^2 - 5n - \left[2(n - 1)^2 - 5(n - 1)\right] = 2n^2 - 5n - (2n^2 - 9n + 7) = 4n - 7$ . (Check: $u_1 = S_1 = -3 = 4 - 7$ .) We need $4n - 7 > 100$ , so $n > 26.75$ , giving $n = 27$ and $u_{27} = 101$ .
$a = 5$ , $d = 0.4$ . (a) $u_{15} = 5 + 14 \times 0.4 = 10.6$ km. (b) $S_{15} = \tfrac{15}{2}(5 + 10.6) = 117$ km. (c) $S_n = \tfrac{n}{2}\left[10 + 0.4(n - 1)\right] = 0.2n^2 + 4.8n > 300$ , so $n^2 + 24n - 1500 > 0$ . Positive root $n = \dfrac{-24 + \sqrt{576 + 6000}}{2} \approx 28.55$ , so the $29$ th day. Check: $S_{28} = 291.2$ km, $S_{29} = 307.4$ km.
$S_{10} = 5(2a + 9d) = 145$ , so $2a + 9d = 29$ . The first twenty terms sum to $145 + 445 = 590$ , so $10(2a + 19d) = 590$ and $2a + 19d = 59$ . Subtracting, $10d = 30$ , $d = 3$ , and $2a = 29 - 27 = 2$ , $a = 1$ .
$a + 4d = 12$ with $a = 6$ gives $d = 1.5$ . Then $\tfrac{n}{2}\left[12 + 1.5(n - 1)\right] = 90$ , so $n(10.5 + 1.5n) = 180$ , i.e. $1.5n^2 + 10.5n - 180 = 0$ , or $n^2 + 7n - 120 = 0$ . This factorises as $(n + 15)(n - 8) = 0$ , so $n = 8$ (reject $-15$ ). Check: $\tfrac{8}{2}(6 + 16.5) = 90$ .
(a) $S_{2n} = \tfrac{2n}{2}\left[2a + (2n - 1)d\right] = n\left[2a + (2n - 1)d\right]$ and $3S_n = \tfrac{3n}{2}\left[2a + (n - 1)d\right]$ . Setting them equal and dividing by $n$ (non-zero), then multiplying by $2$ :
$\begin{aligned} 4a + 2(2n - 1)d &= 6a + 3(n - 1)d \\ 4a + 4nd - 2d &= 6a + 3nd - 3d \\ nd + d &= 2a \end{aligned}$
so $2a = (n + 1)d$ . (b) $(n + 1)d = 24$ , and $u_n = 12 + (n - 1)d = 30$ gives $(n - 1)d = 18$ . Subtracting, $2d = 6$ , so $d = 3$ , and then $n + 1 = 8$ , so $n = 7$ . Check: $S_{14} = 7(24 + 39) = 441$ and $S_7 = \tfrac{7}{2}(24 + 18) = 147$ ; $3 \times 147 = 441$ .

Sequences, series and the language of APs

The nth term

Why the formula has n minus 1

How many terms?

The sum of the first n terms

The pairing argument

Getting terms back from a sum formula

Worked examples

Practice questions

Console