FA19–FA21, HA23–HA25 - Linear sequences

FA19–FA21, HA23–HA25 - Linear sequences

Find distant terms without writing out an entire sequence. FA19 and HA23 introduce rules for generating terms; FA20 and HA24 identify arithmetic progressions; FA21 and HA25 let you describe a linear sequence using its position. You will also decide whether a given number can occur in a sequence.

Generating terms

A seating pattern has 3,7,11,15,…3,7,11,15,\ldots chairs in successive rows. How many chairs would row 5050 have? Adding 44 repeatedly works, but a rule using the row number will be quicker.

A sequence is an ordered list of terms. The first term has position 11, the second position 22, and so on. A term's position is different from its value: in 6,11,16,21,…6,11,16,21,\ldots, the term at position 33 has value 1616.

Two kinds of sequence rule

A term-to-term rule tells you how to get a new term from the previous term, so you also need a starting term. A position-to-term rule tells you a term directly from its position nn, where n=1,2,3,…n=1,2,3,\ldots.

For example, “start at 66, then add 55” and “the term at position nn is 5n+15n+1” generate the same sequence. The first rule moves along the list; the second jumps straight to a position. In an expression such as 5n+15n+1, multiplication happens before addition.

Worked example 1: using the previous term

Question. A sequence starts at 66. Its term-to-term rule is “add 55”. Write down the first four terms.

Plan. Include the starting term, then apply the rule three times.

6+5=11second term11+5=16third term16+5=21fourth term\begin{aligned} 6+5&=11 &&\text{second term}\\ 11+5&=16 &&\text{third term}\\ 16+5&=21 &&\text{fourth term} \end{aligned}

Answer. 6,11,16,216,11,16,21.

Check. Subtract 55 from each new term: 11−5=611-5=6, 16−5=1116-5=11, 21−5=1621-5=16. Each returns to the preceding term.

Keep the sequence, but change how its rule is given. You no longer need to generate earlier terms before finding a later one.

Worked example 2: using a position

Question. The term at position nn is 5n+15n+1. Write down the first four terms and find the tenth term.

Plan. Substitute each required position for nn.

n=1:5×1+1=6n=2:5×2+1=11n=3:5×3+1=16n=4:5×4+1=21n=10:5×10+1=51\begin{aligned} n=1:&\quad5\times1+1=6\\ n=2:&\quad5\times2+1=11\\ n=3:&\quad5\times3+1=16\\ n=4:&\quad5\times4+1=21\\ n=10:&\quad5\times10+1=51 \end{aligned}

Answer. The first four terms are 6,11,16,216,11,16,21; the tenth term is 5151.

Check. From position 11 to position 1010 there are nine increases of 55. Thus 6+9×5=516+9\times5=51, agreeing with direct substitution.

The expression is a rule for every allowed position, rather than one particular term.

Watch out

The first term of 5n+15n+1 is 66, not 11. Substitute n=1n=1; do not read the number added at the end as the first term.

Use a new position rule directly.

Arithmetic progressions

An arithmetic progression changes by the same amount at every step. Finding that change lets you continue the sequence in either direction.

Arithmetic progression

An arithmetic progression, also called an arithmetic sequence, has a constant common difference. Calculate it as the next term minus the previous term. A positive difference increases the terms; a negative difference decreases them; a zero difference repeats the same value.

For 4,10,16,22,…4,10,16,22,\ldots, the differences are 10−4=610-4=6, 16−10=616-10=6, 22−16=622-16=6. The rule is “add 66”. To move backwards, undo this by subtracting 66.

Check every supplied step. The list 2,5,8,122,5,8,12 starts with two increases of 33, but the final increase is 44, so it is not arithmetic. A few terms alone do not determine a unique infinite sequence; in these questions, use the stated rule or the intended consistent pattern.

Worked example 3: extending an increasing sequence

Question. Consecutive terms of an arithmetic progression are 14,20,2614,20,26. Find the term immediately before 1414 and the term immediately after 2626.

Plan. Find the common difference, then reverse it for the earlier term.

20−14=6difference26−20=6confirm the same difference14−6=8move one place backwards26+6=32move one place forwards\begin{aligned} 20-14&=6 &&\text{difference}\\ 26-20&=6 &&\text{confirm the same difference}\\ 14-6&=8 &&\text{move one place backwards}\\ 26+6&=32 &&\text{move one place forwards} \end{aligned}

Answer. The earlier term is 88; the later term is 3232.

Check. The extended list 8,14,20,26,328,14,20,26,32 has difference 66 at all four steps.

Now change the direction. “Difference” includes its sign.

Worked example 4: extending a decreasing sequence

Question. Consecutive terms of an arithmetic progression are 14,8,214,8,2. Find the preceding term and the next term.

Plan. Use next minus previous to find the signed difference.

8−14=−6difference2−8=−6confirm the difference14−(−6)=20undo adding negative six2+(−6)=−4continue forwards\begin{aligned} 8-14&=-6 &&\text{difference}\\ 2-8&=-6 &&\text{confirm the difference}\\ 14-(-6)&=20 &&\text{undo adding negative six}\\ 2+(-6)&=-4 &&\text{continue forwards} \end{aligned}

Answer. The preceding term is 2020; the next term is −4-4.

Check. Subtracting 66 repeatedly gives 20,14,8,2,−420,14,8,2,-4. Moving through zero has not changed the rule.

Continue this lesson.

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