2.21 - nth terms of linear sequences

An nth-term rule lets you go straight to any position in a sequence. You will build rules for numerical and algebraic linear sequences, then use them to decide whether a value occurs and which term first passes a boundary.

From a constant difference to a rule

The sequence $143, 137, 131, \ldots$ keeps falling by $6$ . Will it ever land exactly on $-1000$ , or will it step straight past? A rule will answer that without listing nearly two hundred terms.

Start with $7, 11, 15, 19, \ldots$ . To reach position $4$ from position $1$ , you make three jumps of $4$ , giving $7 + 3 \times 4 = 19$ . Position $n$ is reached after $n - 1$ jumps.

Linear sequence

A linear sequence has a constant difference between consecutive terms. Its common difference is the next term minus the current term. For example, $7, 11, 15, \ldots$ has common difference $4$ ; $7, 3, -1, \ldots$ has common difference $-4$ . The sequence $1, 4, 9, \ldots$ is not linear because its differences change.

Write $T_n$ for the term in position $n$ . Unless told otherwise, positions are the positive integers $1, 2, 3, \ldots$ : a negative term is allowed, but a negative position is not.

[DIAGRAM: asset_name: Counting jumps in a linear sequence; asset_slug: l2fm-linear-sequence-jumps; caption: Position n is reached by n minus 1 equal jumps from the first term; description: Two rows show the terms in positions 1 to 5. The increasing row is 7, 11, 15, 19, 23 with arrows labelled plus 4. The decreasing row is 7, 3, minus 1, minus 5, minus 9 with arrows labelled minus 4. A bracket under the first four positions shows three jumps from position 1 to position 4.]
Position n is reached by n minus 1 equal jumps from the first term

The first term supplies the starting value; the common difference supplies the size and direction of every jump.

Nth term of a linear sequence

If the first term is $a$ and the common difference is $d$ , then

T_n = a + (n - 1)d = dn + (a - d), \qquad n = 1, 2, 3, \ldots

The difference $d$ may be positive, negative or zero. When $d = 0$ , all the terms are equal.
Not on the formulae sheet: learn it.

Use the jump form to build the rule, then simplify if you want the form $dn + c$ .

Worked example 1: an increasing sequence

Question. Work out an expression for the nth term of $7, 11, 15, 19, \ldots$ .

Plan. Start at $7$ and add $n - 1$ jumps of $4$ .

\begin{aligned} d &= 11 - 7 = 4 && \text{next term minus current term} \\ T_n &= 7 + 4(n - 1) && \text{start plus the jumps} \\ &= 4n + 3 && \text{expand and collect} \end{aligned}

Why this step? There are $n - 1$ jumps, because position $1$ already contains the first term.

Answer. $T_n = 4n + 3$ .

Check. At $n = 1$ , the rule gives $7$ ; at $n = 4$ , it gives $19$ . Increasing $n$ by $1$ adds $4$ .

Change only the common difference from $4$ to $-4$ . The first term stays $7$ , so the starting value is unchanged.

Worked example 2: a decreasing sequence

Question. Work out an expression for the nth term of $7, 3, -1, -5, \ldots$ .

Plan. Start at $7$ and make $n - 1$ jumps of $-4$ .

\begin{aligned} d &= 3 - 7 = -4 && \text{preserve the direction} \\ T_n &= 7 - 4(n - 1) && \text{substitute the negative difference} \\ &= 11 - 4n && \text{expand the whole bracket} \end{aligned}

Why this step? The common difference is negative, so the coefficient of $n$ must be negative too.

Answer. $T_n = 11 - 4n$ .

Check. $T_1 = 7$ and $T_4 = -5$ , matching the given terms.

Decimals and fractions work in the same way. For $2, 4.5, 7, \ldots$ , the difference is $2.5$ , so $T_n = 2 + 2.5(n - 1) = 2.5n - 0.5$ . The positions remain integers even when the terms are not.

Watch out

Wrong: $T_n = 7 - 4n$ for the decreasing example. It gives $3$ at $n = 1$ , missing the first term. Right: $T_n = 7 - 4(n - 1) = 11 - 4n$ . Always test position $1$ .

Try a different negative difference. Keep the starting value and the jump count separate.

Linear sequences with algebraic terms

Letters in the terms do not change the method: subtract consecutive terms to identify the common difference, then count the jumps.

In a sequence with terms involving $r$ and $s$ , these letters stand for fixed values throughout the sequence. Only the position $n$ changes. Keep brackets when subtracting a complete term; the repeated $r$ often cancels.

Worked example 3: an algebraic common difference

Question. Work out the nth term of the linear sequence $r + s, r + 3s, r + 5s, \ldots$ .

Plan. Subtract the first term from the second, then use the first-term rule.

\begin{aligned} d &= (r + 3s) - (r + s) = 2s && \text{the }r\text{ terms cancel} \\ T_n &= (r + s) + (n - 1)(2s) && \text{first term plus jumps} \\ &= r + (2n - 1)s && \text{collect the multiples of }s \end{aligned}

Why this step? The common difference is $2s$ , not $s$ : the coefficients of $s$ go $1, 3, 5, \ldots$ .

Answer. $T_n = r + (2n - 1)s$ .

Check. Positions $1$ and $3$ give $r + s$ and $r + 5s$ . If $s = 0$ , every term becomes $r$ , as the rule predicts.

Change only the common difference from $2s$ to $3s$ , keeping the first term $r + s$ . This changes the coefficient of $n$ , as well as the constant adjustment.

Worked example 4: changing the algebraic step

Question. Work out the nth term of the linear sequence $r + s, r + 4s, r + 7s, \ldots$ .

Plan. Use first term $r + s$ and common difference $3s$ .

\begin{aligned} d &= (r + 4s) - (r + s) = 3s && \text{subtract whole terms} \\ T_n &= r + s + 3s(n - 1) && \text{count }n - 1\text{ jumps} \\ &= r + (3n - 2)s && \text{collect the }s\text{ terms} \end{aligned}

Why this step? Expanding $3s(n - 1)$ contributes $-3s$ , which combines with the starting $+s$ to give $-2s$ .

Answer. $T_n = r + (3n - 2)s$ .

Check. At $n = 1, 2, 3$ , the coefficient of $s$ is $1, 4, 7$ , respectively.

There is a neat balance in every linear sequence: terms equally far either side of a middle term add to twice that middle term. Here, $(r + s) + (r + 7s) = 2(r + 4s)$ . In general, the equal backwards and forwards jumps cancel. This holds even if the difference is negative or zero.

Watch out

Wrong: $r + 3ns$ for Worked example 4. At $n = 1$ , it gives $r + 3s$ . Right: $r + (3n - 2)s$ . The first term fixes the adjustment after you choose the coefficient of $n$ .

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2.21 - nth terms of linear sequences

From a constant difference to a rule

Linear sequences with algebraic terms

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