3.1 - Arithmetic sequences and nth terms

3.1 - Arithmetic sequences and nth terms

Learn to turn a constant-step pattern into an nth-term rule and decide whether a number belongs to it. Higher students will also use first-term notation and recover an arithmetic sequence from two separated terms.

Finding a linear nth term

Can you locate 119119 in 5,11,17,23,…5, 11, 17, 23, \ldots without writing every earlier term? A rule using the position will turn a long list into a short calculation.

You already know

A position-to-term rule uses n=1,2,3,…n=1, 2, 3, \ldots. Substitute the position into the rule, and use brackets when multiplying negative numbers.

First notice what makes 7,11,15,19,…7, 11, 15, 19, \ldots special: every jump is 44. The terms 4n4n also jump by 44, so compare the two lists.

Arithmetic sequence

An arithmetic sequence has the same difference between consecutive terms. For example, 7,11,15,19,…7, 11, 15, 19, \ldots is arithmetic. The squares 1,4,9,16,…1, 4, 9, 16, \ldots are not: their differences change.

The multiplier of nn controls the jump. Adding a fixed number shifts every term by the same amount without changing that jump. This is why arithmetic sequences have linear nth-term expressions.

Worked example 1: compare with multiples

Question. Find the nth term of 7,11,15,19,…7, 11, 15, 19, \ldots, then find the twentieth term.

Plan. Compare the sequence with multiples of its constant difference.

11−7=4constant step4n:4,8,12,16,…multiples of four7−4=3adjustment at position oneun=4n+3same adjustment throughoutu20=4(20)+3=83substitute the position\begin{aligned} 11-7&=4 &&\text{constant step}\\ 4n&: 4, 8, 12, 16, \ldots &&\text{multiples of four}\\ 7-4&=3 &&\text{adjustment at position one}\\ u_n&=4n+3 &&\text{same adjustment throughout}\\ u_{20}&=4(20)+3=83 &&\text{substitute the position} \end{aligned}

Why add 33? Each multiple is 33 below the corresponding term, not 77 below it.

Answer. un=4n+3u_n=4n+3; the twentieth term is 8383.

Check. u1=7u_1=7 and u4=19u_4=19, matching the given sequence.

Now change only the direction of the step: start at 77 again, but subtract 44. A negative coefficient records this decrease.

Worked example 2: a negative step

Question. Find the nth term of 7,3,−1,−5,…7, 3, -1, -5, \ldots, then find the twentieth term.

Plan. Compare with −4n-4n, keeping its negative sign.

3−7=−4constant step−4n:−4,−8,−12,−16,…matching negative step7−(−4)=11adjustment at position oneun=−4n+11add the adjustmentu20=−4(20)+11=−69substitute twenty\begin{aligned} 3-7&=-4 &&\text{constant step}\\ -4n&: -4, -8, -12, -16, \ldots &&\text{matching negative step}\\ 7-(-4)&=11 &&\text{adjustment at position one}\\ u_n&=-4n+11 &&\text{add the adjustment}\\ u_{20}&=-4(20)+11=-69 &&\text{substitute twenty} \end{aligned}

Why use −4n-4n? A rule with coefficient 44 would increase, whatever constant you added.

Answer. un=−4n+11u_n=-4n+11; the twentieth term is −69-69.

Check. The first four outputs are 7,3,−1,−57, 3, -1, -5, as required.

Always test the first position before trusting your rule.

Watch out

The wrong rule 4n+74n+7 gives first term 1111, not 77. The first term is not automatically the constant. Comparing with 4n4n gives 4n+34n+3.

Neither the Foundation nor Higher formulae sheet supplies an arithmetic nth-term rule. Learn the constant-step comparison; it also checks rules written in another form.

Deciding whether a number is a term

An equation can have a fractional solution, but a sequence cannot have a “twentieth-and-a-bit” position. Set the nth-term rule equal to the proposed value, solve, then interpret the position.

Allowed position

For a sequence beginning at its first term, an allowed position nn is a positive integer. A fractional, zero or negative value of nn does not name a term in that sequence.

For example, the rule 6n−16n-1 gives −1-1 at n=0n=0, but that does not make −1-1 a term when positions start at 11. Solving is only part of a membership decision.

Worked example 3: find an integer position

Question. Is 119119 a term of the sequence with nth term 6n−16n-1? Give a reason.

Plan. Solve 6n−1=1196n-1=119 and check whether the result is an allowed position.

6n−1=119set the term equal to the target6n=120add onen=20divide by six\begin{aligned} 6n-1&=119 &&\text{set the term equal to the target}\\ 6n&=120 &&\text{add one}\\ n&=20 &&\text{divide by six} \end{aligned}

Why interpret 2020? Membership depends on reaching the value at a positive integer position.

Answer. Yes. 119119 is the twentieth term.

Check. 6(20)−1=1196(20)-1=119. This answers the opening question for 5,11,17,23,…5, 11, 17, 23, \ldots.

Change only the target number to 120120. Being close to a term does not guarantee being a term.

Worked example 4: reject a fractional position

Question. Is 120120 a term of the sequence with nth term 6n−16n-1? Give a reason.

Plan. Solve the same equation with the new target, then apply the position restriction.

6n−1=120new target6n=121add onen=1216divide by six\begin{aligned} 6n-1&=120 &&\text{new target}\\ 6n&=121 &&\text{add one}\\ n&=\frac{121}{6} &&\text{divide by six} \end{aligned}

Why reject this solution? 121121 is not divisible by 66, so the position is not an integer.

Answer. No. 120120 does not occur in the sequence.

Check. The neighbouring terms are u20=119u_{20}=119 and u21=125u_{21}=125; the sequence jumps past 120120.

When the question asks for a reason, include the restriction in words.

Watch out

The wrong conclusion “n≈20n\approx20, so yes” rounds a position. The equation must give a positive integer exactly. Here n=121/6n=121/6, so the correct conclusion is no.

For Pearson's “Give a reason” questions, the calculation and what it means belong together. A bare yes or no does not explain membership.

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