3.1 Sequences

Syllabus
2017
Topic
3.1
Level
Foundation

Learning objectives

Generate sequence terms from two kinds of rule

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 nn.

Rule type Example How to generate
term-to-term start at 3, then multiply by 2 3,6,12,24,3,6,12,24,\ldots
position-to-term un=n2u_n=n^2 1,4,9,16,1,4,9,16,\ldots
named pattern odd numbers 1,3,5,7,1,3,5,7,\ldots
named pattern powers of 2 1,2,4,8,1,2,4,8,\ldots

Identify the starting position, apply the stated rule exactly, and label the terms. For a position rule, substitute n=1,2,3,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.

Continue an integer sequence and state its rule

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 5,9,13,17,5,9,13,17,\ldots, each difference is 44, so the next terms are 21,2521,25 and the term-to-term rule is ‘add 4’.

For 1,2,4,8,1,2,4,8,\ldots, each term is twice the previous term, so the next terms are 16,3216,32 and the rule is ‘multiply by 2’.

Do not describe 1,2,4,81,2,4,8 as ‘add 1, add 2, add 4’ if the intended invariant is multiply by 2. Prefer the simplest consistent rule supported by the task.

Find and use the nth term of an arithmetic sequence

An arithmetic sequence has a constant difference dd. Its nth term is linear: un=dn+cu_n=dn+c, where cc is chosen so that the formula reproduces the first term.

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

For 7,11,15,19,7,11,15,19,\ldots, d=4d=4. The sequence 4n4n begins 4,8,12,164,8,12,16, so add 3: un=4n+3u_n=4n+3.

To test whether 83 is a term, solve 4n+3=834n+3=83, giving n=20n=20. Because 2020 is a positive integer, 83 is a term.

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

Recover the first term and common difference

Every arithmetic term satisfies un=a+(n1)du_n=a+(n-1)d, where aa is the first term and dd is the common difference. Two independent term facts can determine both unknowns.

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

If u2=7u_2=7 and u5=19u_5=19, subtract a+d=7a+d=7 from a+4d=19a+4d=19: 3d=123d=12, so d=4d=4 and then a=3a=3.

Rebuild the stated terms: u2=3+4=7u_2=3+4=7 and u5=3+16=19u_5=3+16=19. This catches an off-by-one error in the multiplier of dd.

The nth term contains (n1)d(n-1)d, not ndnd, because the first term is reached before any common-difference steps have been taken.

Use the arithmetic nth-term formula

The formula un=a+(n1)du_n=a+(n-1)d gives any term of an arithmetic sequence directly from its first term aa, common difference dd, and position nn.

Goal Set up
find a term substitute its position for nn
find a position set a+(n1)da+(n-1)d equal to the given value
compare terms use uqup=(qp)du_q-u_p=(q-p)d
write linear form expand to un=dn+(ad)u_n=dn+(a-d)

For a=4a=4 and d=3d=3, u50=4+49(3)=151u_{50}=4+49(3)=151. The multiplier is 49 because there are 49 steps from the first term to the 50th.

If 4+(n1)3=1004+(n-1)3=100, then 3n+1=1003n+1=100 and n=33n=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.

Sum the first n terms of an arithmetic series

An arithmetic series is the sum of the terms of an arithmetic sequence. Pairing first and last terms gives Sn=n2(a+l)=n2[2a+(n1)d]S_n=\frac n2(a+l)=\frac n2[2a+(n-1)d].

Information available Efficient form
first term aa and last term ll Sn=n2(a+l)S_n=\frac n2(a+l)
first term aa and difference dd Sn=n2[2a+(n1)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+4+7+10+13+\cdots with 50 terms, a=4a=4, d=3d=3, so S50=502[2(4)+49(3)]=25(155)=3875S_{50}=\frac{50}{2}[2(4)+49(3)]=25(155)=3875.

The last term is l=a+(n1)dl=a+(n-1)d. Writing the series forwards and backwards makes each pair total a+la+l; there are nn such pairs across two copies of the series.

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