A25 - The nth term of a linear sequence

A25 - The nth term of a linear sequence

A term-to-term rule gets you to the next term. An nth-term rule lets you reach any position in one calculation. In A25 you will build linear nth-term rules, including decreasing sequences and sequences with missing terms, then use your rule to decide whether a number appears.

Building a rule from the common difference

Will the sequence 5,8,11,14,…5, 8, 11, 14, \ldots ever reach 20272027? Listing hundreds of terms would work, but a rule can settle it much faster.

You already know

An arithmetic, or linear, sequence has a constant difference. The position nn normally starts at 11. Substituting a position into an nth-term rule gives the value at that position.

The differences in 5,8,11,14,…5, 8, 11, 14, \ldots are all 33. Compare those terms with 3,6,9,12,…3, 6, 9, 12, \ldots, the multiples of 33. Every term is 22 bigger than the matching multiple. That fixed extra amount is what the common difference alone does not tell you.

[DIAGRAM: asset_name: Multiples and a fixed extra amount; asset_slug: aqa8300_a16_linear_groups; caption: Each pattern has n groups of three tiles and two extra tiles, giving the rule 3n plus 2; description: Pattern 1 has a column of three blue tiles and two green tiles, Pattern 2 has two columns of three blue tiles and two green tiles, and Pattern 3 has three columns of three blue tiles and two green tiles. Below each is an equation: 3 times 1 plus 2 equals 5, 3 times 2 plus 2 equals 8, and 3 times 3 plus 2 equals 11. The final line reads n groups of 3 plus 2 fixed tiles.]
Each pattern has n groups of three tiles and two extra tiles, giving the rule 3n plus 2

The picture shows why the rule has two jobs: 3n3n counts the growing part; +2+2 supplies the same correction at every position.

The nth term of a linear sequence

For positions n=1,2,3,…n=1, 2, 3, \ldots, a linear sequence with first term aa and common difference dd has

nth term=a+(n−1)d=dn+(a−d)\begin{aligned} \text{nth term} &= a+(n-1)d \\ &= dn+(a-d) \end{aligned}

Equivalently, write dn+cdn+c, where c=first term−dc=\text{first term}-d.

Not on the formulae sheet: learn it.

Why n−1n-1 differences? To reach the first term you make no moves. To reach the second you make one move. To reach the nth you make n−1n-1 moves. You can use that route or compare the sequence with the multiples of dd.

Worked example 1: a positive correction

Question. A linear sequence begins 5,8,11,145, 8, 11, 14. Work out an expression for the nth term. Non-calculator.

Plan. Find the common difference, use its multiples, then work out the fixed correction.

d=8−5=3also 11−8=14−11=3start with 3nmultiples increase by 3c=5−3=2compare the first termsnth term=3n+2add 2 at every position\begin{aligned} d &= 8-5=3 && \text{also }11-8=14-11=3 \\ \text{start with }&3n && \text{multiples increase by }3 \\ c &= 5-3=2 && \text{compare the first terms} \\ \text{nth term} &= 3n+2 && \text{add }2\text{ at every position} \end{aligned}

Why this step? The difference 33 is the coefficient of nn, because increasing nn by 11 increases 3n3n by 33.

Answer. 3n+23n+2.

Check. At n=1n=1, 3(1)+2=53(1)+2=5. At n=4n=4, 3(4)+2=143(4)+2=14, matching both ends of the given list.

Now move the whole sequence down by 44. The difference stays 33, but the fixed correction changes sign.

Worked example 2: a negative correction

Question. A linear sequence begins 1,4,7,101, 4, 7, 10. Work out an expression for the nth term. Non-calculator.

Plan. Compare the terms with 3n3n, then use the first term to find the correction.

d=4−1=3check the constant differenceterms of 3n:3,6,9,12positions 1,2,3,4c=1−3=−2subtract 2 from each multiplenth term=3n−2combine the two pieces\begin{aligned} d &= 4-1=3 && \text{check the constant difference} \\ \text{terms of }3n &: 3, 6, 9, 12 && \text{positions }1, 2, 3, 4 \\ c &= 1-3=-2 && \text{subtract }2\text{ from each multiple} \\ \text{nth term} &= 3n-2 && \text{combine the two pieces} \end{aligned}

Why this step? The first term is 11, but the constant correction is −2-2. They are different because nn starts at 11, not 00.

Answer. 3n−23n-2.

Check. 3(1)−2=13(1)-2=1 and 3(4)−2=103(4)-2=10.

There is a pleasing balance in an arithmetic sequence: terms equally far from opposite ends of a finite list have the same sum. Here 1+10=4+7=111+10=4+7=11. Moving one term up by 33 and its partner down by 33 leaves the total unchanged.

Watch out

Wrong: write n+3n+3 because the rule is “add 33”. That expression gives 4,5,6,…4, 5, 6, \ldots, increasing by 11. Right: start with 3n3n, then find the fixed correction.

Change the difference, while keeping the idea of a negative correction.

Decreasing sequences and fractional differences

A negative common difference gives a negative coefficient of nn.

For 10,8,6,4,…10, 8, 6, 4, \ldots, calculate 8−10=−28-10=-2. Using +2+2 would describe growth in the opposite direction. The same rule dn+cdn+c still works; the sign belongs to the difference.

Signed common difference

The common difference is the next term minus the previous term. For a decreasing linear sequence it is negative. It may also be a fraction or decimal: 10,8.5,7,5.5,…10, 8.5, 7, 5.5, \ldots has common difference −1.5-1.5.

Use the first term to find cc after choosing the signed coefficient. Subtracting a negative number increases the correction.

Worked example 3: a negative coefficient

Question. A linear sequence begins 10,8,6,410, 8, 6, 4. Work out an expression for the nth term. Non-calculator.

Plan. Use difference −2-2, then choose the constant that gives 1010 when n=1n=1.

d=8−10=−2next minus previousnth term=−2n+cuse the signed difference−2(1)+c=10match the first termc=12add 2 to both sides\begin{aligned} d &= 8-10=-2 && \text{next minus previous} \\ \text{nth term} &= -2n+c && \text{use the signed difference} \\ -2(1)+c &= 10 && \text{match the first term} \\ c &= 12 && \text{add }2\text{ to both sides} \end{aligned}

Why this step? −2n-2n begins at −2-2. Adding 1212, rather than 1010, brings its first term to 1010.

Answer. 12−2n12-2n.

Check. The first and fourth values are 12−2(1)=1012-2(1)=10 and 12−2(4)=412-2(4)=4.

Now change only the size of the decrease, from 22 to 1.51.5, keeping the first term at 1010. A decimal difference needs no new method.

Worked example 4: a decimal decrease

Question. A linear sequence begins 10,8.5,7,5.510, 8.5, 7, 5.5. Work out an expression for the nth term. Non-calculator.

Plan. Put the difference −1.5-1.5 in front of nn, then match the first term.

d=8.5−10=−1.5the constant decreasenth term=−1.5n+cuse that coefficient−1.5+c=10substitute n=1c=11.5add 1.5\begin{aligned} d &= 8.5-10=-1.5 && \text{the constant decrease} \\ \text{nth term} &= -1.5n+c && \text{use that coefficient} \\ -1.5+c &= 10 && \text{substitute }n=1 \\ c &= 11.5 && \text{add }1.5 \end{aligned}

Why this step? A half-unit in the difference must stay in the rule. Rounding −1.5-1.5 to −2-2 changes the sequence.

Answer. 11.5−1.5n11.5-1.5n, or 232−32n\frac{23}{2}-\frac32 n.

Check. 11.5−1.5(1)=1011.5-1.5(1)=10 and 11.5−1.5(4)=5.511.5-1.5(4)=5.5.

A constant sequence is the boundary case: 6,6,6,…6, 6, 6, \ldots has difference 00, so its nth term is simply 66. A decreasing number sequence can continue below zero, unless its context restricts the number of terms. These are applications of the linear rule, which is not on the formulae sheet: learn it.

Watch out

Wrong: 10−2n10-2n for 10,8,6,4,…10, 8, 6, 4, \ldots. At n=1n=1 this gives 88, already one term too far along. Right: 10−2(n−1)=12−2n10-2(n-1)=12-2n.

Keep the first term fixed again, but change the size of the decrease.

Continue this lesson.

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