5A, 5B - Arithmetic sequences

How can you find a distant term without writing every term before it? In 5B, you will use constant differences to describe arithmetic sequences and recover their rules from known terms. In 5A, sigma notation gives you a compact way to describe a total of selected terms.

Common difference and nth term

The sequence starts $7,11,15,19,\ldots$ . What is its twenty-fifth term?

Each new term is $4$ more than the previous one. The first term is already there: reaching the fourth term takes three additions, so reaching the twenty-fifth takes twenty-four.

Arithmetic sequence

An arithmetic sequence has a constant common difference, $d$ , between consecutive terms. If its first term is $a$ , its terms begin $a,a+d,a+2d,\ldots$ .

For example, $12,9,6,3,\ldots$ is arithmetic with $d=-3$ . The sequence $2,4,8,16,\ldots$ is not arithmetic because its differences change.

Write $u_n$ for the term at position $n$ , where $n$ is a positive integer. There are $n-1$ steps from position $1$ to position $n$ .

Arithmetic nth term

u_n=a+(n-1)d

Here $a=u_1$ and $d=u_{n+1}-u_n$ . The difference may be positive, negative or zero.

Not on the formulae sheet: learn it.

Use the positions to count steps, rather than assuming that the position is the number of additions.

Worked example 1: a distant increasing term

Question. Find the twenty-fifth term of $7,11,15,19,\ldots$ .

Plan. Start at $7$ and add the common difference $24$ times.

\begin{aligned} d&=11-7=4 &&\text{consecutive difference}\\ u_{25}&=7+(25-1)4 &&\text{use }24\text{ steps}\\ &=103 &&\text{evaluate} \end{aligned}

Answer. The twenty-fifth term is $103$ .

Check. The formula gives $u_1=7$ and $u_2=11$ . Also $u_{24}=99$ , exactly $4$ below the answer.

Now keep the first term and position, but change the common difference to $-4$ . A sequence can pass through zero without changing its rule.

Worked example 2: a distant decreasing term

Question. Find the twenty-fifth term of $7,3,-1,-5,\ldots$ .

Plan. Use the signed difference $-4$ in the same nth-term formula.

\begin{aligned} d&=3-7=-4 &&\text{new minus previous}\\ u_{25}&=7+24(-4) &&\text{retain the sign}\\ &=-89 &&\text{subtract }96 \end{aligned}

Answer. The twenty-fifth term is $-89$ .

Check. $u_{24}=-85$ , and another decrease of $4$ gives $-89$ .

The same formula describes a constant sequence when $d=0$ : every term equals $a$ .

Watch out

$u_n=a+nd$ makes the first term $a+d$ . Use $a+(n-1)d$ , so substituting $n=1$ returns $a$ .

Use that first-term check on your own formula.

Finding the first term and difference

Two known positions let you count the steps between them, even when the first term is missing.

From position $4$ to position $9$ there are $9-4=5$ equal steps. In general, subtracting two term equations eliminates $a$ : $u_q-u_p=(q-p)d$ . Once $d$ is known, work backwards to position $1$ .

Worked example 3: recovering an increasing sequence

Question. An arithmetic sequence has $u_4=17$ and $u_9=37$ . Find $a$ and $d$ .

Plan. Translate the positions into equations, subtract, then substitute.

\begin{aligned} a+3d&=17 &&\text{fourth term}\\ a+8d&=37 &&\text{ninth term}\\ 5d&=20 &&\text{subtract the first equation}\\ d&=4 &&\text{divide by }5\\ a&=17-3(4)=5 &&\text{work back three steps} \end{aligned}

Answer. $a=5$ and $d=4$ .

Check. Starting $5,9,13,17$ puts $17$ fourth; five more steps give $17+5(4)=37$ .

Change only the ninth term. The later term is now smaller, so the recovered difference must be negative.

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5A, 5B - Arithmetic sequences

Common difference and nth term

Finding the first term and difference

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