D4 - Arithmetic sequences and series | AQA A-Level Mathematics

Use a constant difference to locate terms, recover unknown parameters and derive arithmetic sums. You will solve equations and inequalities for term counts, checking their integer domain and any extra conditions.

The common difference and nth term

Add the terms $12,10,8,\ldots$ in order. Can the running total equal $40$ at two different term counts? A total need not keep growing if the later terms become negative. We will first learn how to locate those terms, then return to the total.

A simpler increasing list is $7,10,13,\ldots$ . To reach its fourth term, make three jumps of $3$ : $7+3(3)=16$ . The number of jumps is one less than the term's position.

Arithmetic sequence

An arithmetic sequence has a constant common difference $d=u_{n+1}-u_n$ . With first term $a$ , its terms are $a,a+d,a+2d,\ldots$ . For example, $7,10,13,\ldots$ has $d=3$ ; $2,4,8,\ldots$ is not arithmetic because its differences change.

Repeated addition produces the nth-term formula. A positive, zero or negative difference gives an increasing, constant or decreasing sequence respectively; term values need not be integers.

Arithmetic nth term

u_n=a+(n-1)d,\qquad n\in\mathbb N

If the last included term is $l$ , then $l=a+(n-1)d$ .
Not in the formula booklet: learn it.

Use the formula for a term and an inequality for a boundary. Keep these two jobs distinct.

Worked example 1: a term and a strict boundary

Question. An arithmetic sequence has $a=7,d=3$ . Find $u_{20}$ and the number of terms less than $100$ .

Plan. Substitute $n=20$ , then solve the strict inequality for positive integer indices.

\begin{aligned} u_{20}&=7+19(3)=64 && \text{nineteen increments}\\ 7+3(n-1)&<100 && \text{translate the boundary}\\ 3n+4&<100\ \Longrightarrow\ n<32 && \text{solve for the index} \end{aligned}

Why this step? The index must be an integer, and a strict upper bound $n<32$ stops at $31$ .

Answer. $u_{20}=64$ , and $31$ terms are less than $100$ .

Check. $u_{31}=97<100$ , whereas $u_{32}=100$ is excluded.

Change only the boundary wording from “less than” to “at most”. Equality now includes one more index.

Worked example 2: including the boundary

Question. For the same sequence $a=7,d=3$ , how many terms are at most $100$ ?

Plan. Use a non-strict inequality and retain every permitted positive integer index.

\begin{aligned} 7+3(n-1)&\le100 && \text{at most includes equality}\\ 3n+4&\le100 && \text{simplify}\\ n&\le32 && \text{divide by positive }3 \end{aligned}

Why this step? A term count changes when the boundary value itself is a term of the sequence.

Answer. $32$ terms.

Check. $u_{32}=100$ qualifies and $u_{33}=103$ does not.

A negative difference uses the same formula: with $a=18,d=-3/2$ , the eighth term is $18+7(-3/2)=15/2$ . The sequence decreases because each update subtracts $3/2$ , even while its early terms are positive.

Watch out

Wrong: $u_{20}=a+20d$ . The first term already counts as one position. Right: $u_{20}=a+19d$ . Also preserve the exact inequality symbol when selecting an integer boundary.

Change the first term and boundary value while keeping the non-strict task.

Recovering a sequence from conditions

From the fourth term to the tenth term there are six jumps, not seven. Subtracting the corresponding term equations removes the first term and isolates those jumps.

You already know

The recalled formula $u_n=a+(n-1)d$ turns each known term into a linear equation in $a$ and $d$ . Subtracting simultaneous equations can eliminate a shared unknown.

In general, $u_k-u_j=(k-j)d$ , obtained by subtracting the two nth-term formulae. This derived relationship is not separately in the formula booklet; count the gaps rather than memorising another formula.

Worked example 3: two separated terms

Question. An arithmetic sequence has $u_4=11,u_{10}=29$ . Find $a,d$ and $u_{20}$ .

Plan. Subtract the term equations to find $d$ , then substitute to find $a$ .

\begin{aligned} a+3d&=11,\qquad a+9d=29 && \text{use the supplied positions}\\ 6d&=18\ \Longrightarrow\ d=3 && \text{subtract the equations}\\ a&=11-3(3)=2 && \text{substitute into the first equation}\\ u_{20}&=2+19(3)=59 && \text{use the recovered parameters} \end{aligned}

Why this step? The same $a$ appears in both equations, so subtraction reveals the total change across six intervals.

Answer. $a=2,d=3,u_{20}=59$ .

Check. $2+3(3)=11$ and $2+9(3)=29$ , so both conditions hold.

Change only the later known term from $29$ to $-7$ . The sign of the difference is now determined by the data rather than guessed.

Worked example 4: recovering a decreasing sequence

Question. An arithmetic sequence has $u_4=11,u_{10}=-7$ . Find $a,d$ and $u_{20}$ .

Plan. Use the same subtraction route, retaining the negative change.

\begin{aligned} a+3d&=11,\qquad a+9d=-7 && \text{form term equations}\\ 6d&=-18\ \Longrightarrow\ d=-3 && \text{subtract in the same order}\\ a&=11-3(-3)=20 && \text{substitute}\\ u_{20}&=20+19(-3)=-37 && \text{calculate the requested term} \end{aligned}

Why this step? Subtracting $u_{10}-u_4$ must be paired with $10-4$ ; reversing only one subtraction changes the sign incorrectly.

Answer. $a=20,d=-3,u_{20}=-37$ .

Check. $20+9(-3)=-7$ , and directly $u_{20}=u_{10}+10d=-7-30=-37$ .

If only a later term is requested, moving from a known term by the right number of gaps may avoid finding $a$ . That is the same arithmetic relationship, not a different sequence model.

Watch out

Wrong: divide $u_{10}-u_4$ by $7$ , counting the endpoints. Right: divide by $10-4=6$ , the number of changes between them.

Change the later known value again, leaving the positions fixed.

Deriving the sum of an arithmetic series

For $5+9+13+\cdots+41$ , the endpoints add to $46$ , as does every inward pair. Reversing the whole list makes that symmetry work for any positive integer number of terms, including an odd one.

Write $l=a+(n-1)d$ . Then

\begin{aligned} S_n&=a+(a+d)+\cdots+l\\ S_n&=l+(l-d)+\cdots+a\\ 2S_n&=(a+l)+(a+l)+\cdots+(a+l) && \text{add the }n\text{ columns}\\ &=n(a+l) && \text{count equal column sums} \end{aligned}

The odd middle term is counted once in each line and pairs with itself. No assumption that $n/2$ is an integer is needed.

Arithmetic series sum

S_n=\frac n2(a+l)=\frac n2\bigl[2a+(n-1)d\bigr],\qquad n\in\mathbb N

These apply to positive, negative or zero common difference. The average term is $(a+l)/2$ , as the derivation shows.
In the formula booklet.

Choose the version matching the information you have. For a consecutive block, use that block's first value, last value and inclusive term count.

Worked example 5: summing the first ten terms

Question. An arithmetic sequence has $a=5,d=4$ . Find $S_{10}$ .

Plan. Find the last included term, then multiply the average endpoint value by the count.

\begin{aligned} u_{10}&=5+9(4)=41 && \text{last included term}\\ S_{10}&=\frac{10}{2}(5+41) && \text{supplied endpoint sum formula}\\ &=230 && \text{evaluate} \end{aligned}

Why this step? Ten terms have average $(5+41)/2=23$ , so the total is ten copies of that average.

Answer. $230$ .

Check. The other form gives $5[2(5)+9(4)]=230$ .

Change only the selected block: sum terms eleven to twenty of the same sequence. The original last index is no longer the number of terms being added.

Worked example 6: a later block of ten terms

Question. For $a=5,d=4$ , find $u_{11}+u_{12}+\cdots+u_{20}$ .

Plan. Treat this block as its own arithmetic list.

\begin{aligned} N&=20-11+1=10 && \text{inclusive term count}\\ u_{11}&=5+10(4)=45 && \text{first selected value}\\ u_{20}&=5+19(4)=81 && \text{last selected value}\\ \text{sum}&=\frac{10}{2}(45+81)=630 && \text{average endpoints times count} \end{aligned}

Why this step? The block still has common difference $4$ , so the supplied arithmetic sum formula applies to its ten values.

Answer. $630$ .

Check. $S_{20}-S_{10}=860-230=630$ . Subtracting $S_{11}$ would remove a term that is required.

The pairing argument has a pleasing consequence: the first $n$ positive odd integers sum to $n^2$ , because their first and last terms are $1$ and $2n-1$ . Thus the first $100$ odd integers add to exactly $10000$ . This identity is not separately in the formula booklet; it follows from the supplied sum.

Watch out

Wrong: terms eleven to twenty make twenty terms. Right: there are $20-11+1=10$ . In $n(a+l)/2$ , the symbols represent the selected list, not automatically the original sequence.

Move the selected block on by ten positions.

Solving conditions on a sum

When the unknown is the count, the sum formula usually creates a quadratic. Solve it, then ask what each root means as a positive integer number of terms.

Worked example 7: one admissible term count

Question. An arithmetic sequence has $a=3,d=4$ . Its first $n$ terms sum to $210$ . Find $n$ .

Plan. Form the sum equation, solve the quadratic and apply the count domain.

\begin{aligned} \frac n2[6+4(n-1)]&=210 && \text{use the supplied sum formula}\\ 2n^2+n-210&=0 && \text{expand and rearrange}\\ (n-10)(2n+21)&=0 && \text{factorise}\\ n&=10\text{ or }-\frac{21}{2} && \text{solve} \end{aligned}

Why this step? The formula is algebraic, but the count must be a positive integer; the second root is inadmissible.

Answer. $n=10$ .

Check. $S_{10}=10(2(10)+1)=210$ .

Now allow a negative common difference. Later terms can decrease the running total, so do not assume that a sum has only one admissible count.

Worked example 8: the same total twice

Question. An arithmetic sequence has $a=12,d=-2$ . Find all positive integers $n$ for which $S_n=40$ . No positivity condition is imposed on the terms.

Plan. Solve the quadratic and check every positive integer root in the original sum.

\begin{aligned} S_n&=\frac n2[24-2(n-1)]=n(13-n) && \text{simplify the sum}\\ n(13-n)&=40 && \text{apply the condition}\\ (n-5)(n-8)&=0 && \text{rearrange and factorise}\\ n&=5\text{ or }8 && \text{both are positive integers} \end{aligned}

Why this step? Negative later terms are allowed here, so they cannot be used to reject a valid count.

Answer. $n=5$ and $n=8$ .

Check. $S_5=5(8)=40$ and $S_8=8(5)=40$ . The added terms $u_6,u_7,u_8$ are $2,0,-2$ , summing to zero.

This resolves the opening question. The same total appears twice because the extra terms cancel. These example-specific sum expressions and roots are not supplied in the formula booklet; derive them using the supplied general sum. If a context requires every term to be positive, apply that condition separately.

Watch out

Wrong: always keep only the smaller positive root. Right: check every root against the actual domain and context. Two positive integer roots can both be valid when later terms become negative.

Change only the target sum in the decreasing sequence.

Quick Recap

Reach term $n$ using $n-1$ common differences; this nth-term formula must be recalled.
Subtract known term equations using the number of gaps between their indices.
Derive the supplied sum by adding the same list forwards and backwards.
For a block of terms, use its own endpoints and inclusive count.
Solve sum conditions and retain every root permitted by the integer domain and context.

The opening sequence reaches total $40$ after both five and eight terms: the three additional terms are $2,0,-2$ , which cancel.

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D4 - Arithmetic sequences and series

The common difference and nth term

Recovering a sequence from conditions

Deriving the sum of an arithmetic series

Solving conditions on a sum

Quick Recap

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