1.6 - Arithmetic progressions | Cambridge A-Level Mathematics

Adding the same amount each time creates a sequence with a simple structure. In this part of 1.6, you will recognise an arithmetic progression, find its terms and totals, and use equations to compare two progressions. The distinction between one term and a sum will guide every calculation.

Recognising a constant difference

Can two different increasing sequences have the same total after the same number of terms? Consider $5,8,11,\ldots$ and $8,10,12,\ldots$ . The second starts ahead, but the first grows faster. We will find when their totals become equal.

First, notice what stays constant: the first sequence adds $3$ at each step, while the second adds $2$ .

Arithmetic progression

An arithmetic progression, or AP, is a sequence in which each term is obtained by adding the same number, the common difference $d$ , to the previous term.

With first term $a$ , its terms are $a, a+d, a+2d,\ldots$ . The difference may be positive, negative or zero.

For three consecutive terms $a,b,c$ , equal differences mean

b-a=c-b\quad\Longleftrightarrow\quad2b=a+c.

Thus the middle term is the arithmetic mean of its neighbours. This is a useful equation when the terms contain an unknown. The letters in this three-term statement label the three terms; later, $a$ will denote the first term of a whole progression.

Worked example 1: a decreasing progression

Question. Show that $14,11,8$ are consecutive terms of an AP, and state its common difference.

Plan. Subtract each term from the next, preserving the order of subtraction.

11-14=-3, \qquad 8-11=-3.

The differences agree, so the terms form an AP.

Answer. Common difference $d=-3$ .

Check. The middle-term test gives $2(11)=14+8=22$ . Adding $-3$ also predicts the next term $5$ .

Keep the first term, but express the next two in terms of an unknown. The same equality of differences now finds that unknown.

Worked example 2: terms containing a parameter

Question. The first three terms of an AP are $14$ , $2p+1$ and $p+3$ . Find $p$ and the common difference.

Plan. Equate twice the middle term to the sum of its neighbours, then calculate a difference.

\begin{aligned} 2(2p+1)&=14+(p+3) &&\text{equal differences}\\ 4p+2&=p+17\\ 3p&=15\\ p&=5. \end{aligned}

The terms are therefore $14,11,8$ , giving $d=11-14=-3$ .

Answer. $p=5$ , $d=-3$ .

Check. Substitute directly into the two original differences: $(2p+1)-14=2p-13=-3$ and $(p+3)-(2p+1)=2-p=-3$ .

A descending AP is just as valid as an ascending one. A constant sequence such as $4,4,4,\ldots$ is also an AP, with $d=0$ .

Watch out

For $14,11,8$ , using $14-11=3$ reverses the direction of the sequence. Always calculate “next minus previous”, so $d=11-14=-3$ .

Use the middle-term equation in this check.

The nth term

To reach the fifth term from the first, make four additions of $d$ , not five. In general, the term with index $n$ is reached after $n-1$ steps. Write it as $u_n$ , where $n$ is a positive integer.

The nth term of an AP

For first term $a$ and common difference $d$ ,

u_n=a+(n-1)d, \qquad n=1,2,3,\ldots.

In the formula booklet.

This equation can find a term, identify its position, or recover $a$ and $d$ . Decide which quantity is unknown before substituting. For example, if $a=7$ and $d=4$ , then $u_4=7+3(4)=19$ .

Worked example 3: finding a term number

Question. An AP has first term $7$ and common difference $4$ . Which term is $55$ ?

Plan. Set the general term equal to $55$ and solve for its index.

\begin{aligned} 7+4(n-1)&=55\\ 4(n-1)&=48\\ n-1&=12\\ n&=13. \end{aligned}

Answer. $55$ is the thirteenth term.

Check. Twelve additions of $4$ take $7$ to $7+48=55$ . The adjacent terms are $51$ and $59$ , so the position is consistent.

Now hide the first term and common difference, giving two terms instead. Subtracting their equations removes the first term.

Worked example 4: recovering the progression

Question. The fourth and tenth terms of an AP are $19$ and $43$ . Find its first term and common difference, and determine which term is $55$ .

Plan. Use the six steps between the two known terms to find $d$ , then recover $a$ and the required index.

\begin{aligned} a+3d&=19\\ a+9d&=43\\ 6d&=24 &&\text{subtract the equations}\\ d&=4\\ a&=19-3(4)=7. \end{aligned}

For the required term, $7+4(n-1)=55$ , so $n=13$ .

Answer. $a=7$ , $d=4$ ; $55$ is the thirteenth term.

Check. Writing the first few terms gives $7,11,15,19$ . Six further steps from $19$ add $24$ , reaching $43$ at the tenth term. Three more reach $55$ .

An equation can have a real solution that is not a valid term number. In the same AP, asking for a term equal to $56$ gives $n=13.25$ . There is no term at that index, so $56$ is not in the progression. If $d=0$ , every term equals $a$ : a target equal to $a$ occurs at every positive integer index, and a different target never occurs.

Watch out

$u_n$ means one term; it is not the total of the first $n$ terms. Also count gaps rather than terms: the fourth and tenth terms are six steps apart, not seven.

Recover a new AP from two of its terms.

Summing an AP

Adding terms creates a series. Let $S_n$ be the sum of the first $n$ terms, and let the last of these terms be $l=u_n$ . For example, $S_3=5+8+11=24$ , while $u_3=11$ .

Write an AP forwards and backwards, then add corresponding terms:

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

There are $n$ equal pairs. Dividing by $2$ gives “number of terms times the average of the first and last”. This argument works for odd $n$ too: the middle term is counted twice when the two rows are added.

The sum of an AP

S_n=\frac n2(a+l) =\frac n2\{2a+(n-1)d\}, \qquad l=a+(n-1)d.

Here $n$ is a positive integer. Both forms work for positive, negative and zero $d$ .

In the formula booklet.

Choose the version matching your information. If a total is given instead of $n$ , substituting into the sum formula may produce a quadratic. Solve it, then check the roots against the meaning of $n$ .

Worked example 5: equal totals from different APs

Question. AP A begins $5,8,11,\ldots$ , and AP B begins $8,10,12,\ldots$ . Find the positive number $n$ for which their first $n$ terms have equal sums.

Plan. Express both sums in terms of the same $n$ , then equate them.

\begin{aligned} S_{A,n}&=\frac n2\{10+3(n-1)\} =\frac n2(3n+7)\\ S_{B,n}&=\frac n2\{16+2(n-1)\} =n(n+7). \end{aligned}

Equating sums and cancelling $n$ is valid because $n>0$ :

\begin{aligned} \frac n2(3n+7)&=n(n+7)\\ 3n+7&=2n+14\\ n&=7. \end{aligned}

Answer. The sums are equal after $7$ terms; each is $98$ .

Check. The seventh terms are $23$ and $20$ . Thus the sums are $7(5+23)/2=98$ and $7(8+20)/2=98$ , using the first-and-last form independently.

Keep both progressions, but ask for a specified difference between their sums. The equation now has a quadratic term to retain.

Worked example 6: one total overtakes another

Question. For the same APs, find $n$ when the sum of the first $n$ terms of A exceeds the sum of the first $n$ terms of B by $60$ .

Plan. Subtract the two sum expressions, set the result equal to $60$ , and retain only a positive integer root.

\begin{aligned} S_{A,n}-S_{B,n} &=\frac n2(3n+7)-n(n+7)\\ &=\frac n2(n-7)\\ \frac n2(n-7)&=60\\ n^2-7n-120&=0\\ (n-15)(n+8)&=0. \end{aligned}

The roots are $15$ and $-8$ . A negative number of terms is impossible.

Answer. $n=15$ .

Check. The fifteenth terms are $47$ and $36$ . Hence $S_{A,15}=15(5+47)/2=390$ and $S_{B,15}=15(8+36)/2=330$ ; their difference is $60$ .

This resolves the opening question: different APs can have equal totals after the same positive number of terms. Here, seven terms balance a higher starting value against a larger common difference.

Problem-solving

Translate “exceeds by $k$ ” as a difference equal to $k$ , with the larger quantity first. Keep the same index only when the question specifies the same number of terms, and reject roots that cannot count terms.

Compare a fresh pair of progressions.

Quick Recap

An AP adds a fixed difference; three consecutive terms satisfy $2b=a+c$ .
$u_n=a+(n-1)d$ finds a single term, with $n-1$ steps from the first.
$S_n=n(a+l)/2$ adds the first $n$ terms; substitute $l=a+(n-1)d$ when needed.
Equations from terms or sums can recover unknowns. A term count must be a positive integer.

The opening progressions have equal totals of $98$ after seven terms.

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1.6 - Arithmetic progressions

Recognising a constant difference

The nth term

Summing an AP

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