3.1 Sequences

Learning objectives

3.1A Generating sequencesgenerate terms of a sequence using term-to-term and position-to-term definitions of the sequence Including odd, even, squares, multiples and powers 3.1B Subsequent terms of an integer sequence and the rulefind subsequent terms of an integer sequence and the rule for generating it 5, 9, 13, 17, … (add 4) 1, 2, 4, 8, … ( multiply by 2)3.1C nth terms of arithmetic sequencesuse linear expressions to describe the nth term of arithmetic sequences 1, 3, 5, 7, 9, … nth term is 2n – 1 nth term is 4n + 3, write down the first 3 terms of the sequence 3.1HA First term and common differenceunderstand and use common difference (d) and first term (a) in an arithmetic sequence e.g. given 2nd term is 7 and 5th term is 19, find a and d 3.1HB nth-term formula for arithmetic sequencesknow and use the nth-term formula a + (n − 1)d for an arithmetic sequence 3.1HC The sum of the first n terms of an arithmetic seriesfind the sum of the first n terms of an arithmetic series (Sn) e.g. given 4 + 7 + 10 + 13 + … find sum of first 50 terms

A sequence is an ordered list. A term-to-term rule tells how to get from one term to the next; a position-to-term rule gives any term directly from its position $n$ .

Rule type	Example	How to generate
term-to-term	start at 3, then multiply by 2	$3,6,12,24,\ldots$
position-to-term	$u_n=n^2$	$1,4,9,16,\ldots$
named pattern	odd numbers	$1,3,5,7,\ldots$
named pattern	powers of 2	$1,2,4,8,\ldots$

Identify the starting position, apply the stated rule exactly, and label the terms. For a position rule, substitute $n=1,2,3,\ldots$ rather than repeatedly changing the previous value.

Check that every generated term obeys the same definition. A list may look familiar but still follow a different rule after its displayed terms.

A finite list alone does not determine one unique sequence. Use the stated rule or context; do not invent a pattern only because it fits the first few terms.

To continue a sequence, compare consecutive terms and find an operation that is applied consistently. State both the operation and any starting information needed.

Consecutive comparison	Likely rule
constant difference	add or subtract that value
constant ratio	multiply or divide by that value
alternating changes	repeat the change cycle
changing first differences	inspect second differences or another structure

For $5,9,13,17,\ldots$ , each difference is $4$ , so the next terms are $21,25$ and the term-to-term rule is ‘add 4’.

For $1,2,4,8,\ldots$ , each term is twice the previous term, so the next terms are $16,32$ and the rule is ‘multiply by 2’.

Do not describe $1,2,4,8$ as ‘add 1, add 2, add 4’ if the intended invariant is multiply by 2. Prefer the simplest consistent rule supported by the task.

An arithmetic sequence has a constant difference $d$ . Its nth term is linear: $u_n=dn+c$ , where $c$ is chosen so that the formula reproduces the first term.

Step	Action
difference	calculate $d$ from consecutive terms
first draft	write $dn$
adjustment	compare $d$ with the first term to find $c=a-d$
verify	substitute $n=1,2,3$
use	substitute a position, or solve the formula for $n$

For $7,11,15,19,\ldots$ , $d=4$ . The sequence $4n$ begins $4,8,12,16$ , so add 3: $u_n=4n+3$ .

To test whether 83 is a term, solve $4n+3=83$ , giving $n=20$ . Because $20$ is a positive integer, 83 is a term.

The common difference is the coefficient of $n$ , not usually the complete nth term. Always include the adjustment and verify at least the first term.

Every arithmetic term satisfies $u_n=a+(n-1)d$ , where $a$ is the first term and $d$ is the common difference. Two independent term facts can determine both unknowns.

Given fact	Equation
second term is 7	$a+d=7$
fifth term is 19	$a+4d=19$
terms $u_p$ and $u_q$ known	subtract to get $(q-p)d=u_q-u_p$

If $u_2=7$ and $u_5=19$ , subtract $a+d=7$ from $a+4d=19$ : $3d=12$ , so $d=4$ and then $a=3$ .

Rebuild the stated terms: $u_2=3+4=7$ and $u_5=3+16=19$ . This catches an off-by-one error in the multiplier of $d$ .

The nth term contains $(n-1)d$ , not $nd$ , because the first term is reached before any common-difference steps have been taken.

The formula $u_n=a+(n-1)d$ gives any term of an arithmetic sequence directly from its first term $a$ , common difference $d$ , and position $n$ .

Goal	Set up
find a term	substitute its position for $n$
find a position	set $a+(n-1)d$ equal to the given value
compare terms	use $u_q-u_p=(q-p)d$
write linear form	expand to $u_n=dn+(a-d)$

For $a=4$ and $d=3$ , $u_{50}=4+49(3)=151$ . The multiplier is 49 because there are 49 steps from the first term to the 50th.

If $4+(n-1)3=100$ , then $3n+1=100$ and $n=33$ , so 100 is the 33rd term. A non-positive or non-integer result would mean it is not a term.

This formula applies to arithmetic sequences only. A constant ratio or changing difference requires a different model.

An arithmetic series is the sum of the terms of an arithmetic sequence. Pairing first and last terms gives $S_n=\frac n2(a+l)=\frac n2[2a+(n-1)d]$ .

Information available	Efficient form
first term $a$ and last term $l$	$S_n=\frac n2(a+l)$
first term $a$ and difference $d$	$S_n=\frac n2[2a+(n-1)d]$
a range of terms	subtract cumulative sums or re-index carefully

For $4+7+10+13+\cdots$ with 50 terms, $a=4$ , $d=3$ , so $S_{50}=\frac{50}{2}[2(4)+49(3)]=25(155)=3875$ .

The last term is $l=a+(n-1)d$ . Writing the series forwards and backwards makes each pair total $a+l$ ; there are $n$ such pairs across two copies of the series.

Do not substitute the last term as $n$ . First determine how many terms are being summed, especially when the sum starts after the first term.

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—3.2 Function notationRead

3.1 Sequences

Learning objectives

Generate sequence terms from two kinds of rule

Continue an integer sequence and state its rule

Find and use the nth term of an arithmetic sequence

Recover the first term and common difference

Use the arithmetic nth-term formula

Sum the first n terms of an arithmetic series

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