3.1 - Sums of arithmetic series

Learn to add an arithmetic sequence efficiently and explain why its sum formula works. You will use totals to solve a saving-budget problem. This lesson is for Higher students.

Pairing the first and last terms (Higher)

You plan to save $£4$ in week one, then $£2$ more each week. How many complete weeks can you fund from $£100$ ? The amount in one week and the total across several weeks are different questions.

You already know

An arithmetic sequence has first term $a$ , common difference $d$ and nth term $a+(n-1)d$ . From the first to the nth term there are $n-1$ jumps.

Adding $1$ through $100$ looks like a hundred calculations, but a useful surprise makes it one multiplication: $1+100=2+99=101$ . Every opposite-end pair has the same total, so the whole sum is $50(101)=5050$ .

Arithmetic series

An arithmetic series adds consecutive terms of an arithmetic sequence. The symbol $u_n$ names one term; $S_n$ names the sum of the first $n$ terms.

Suppose the first term is $a$ and the last is $l$ . The second is $a+d$ , while the second-last is $l-d$ ; their sum is still $a+l$ . Write the series once forwards and once backwards. Every column then totals $a+l$ .

[DIAGRAM: asset_slug: m044_pairing_terms; caption: A forward and reversed arithmetic series make equal column sums; description: The two copies of 4 plus 7 plus 10 through 31 are aligned in opposite orders; each of ten columns totals 35, so two sums total 350; file: m044_pairing_terms.png; recommended_method: matplotlib]

The picture gives $2S_n=n(a+l)$ . Divide by $2$ because you added two copies, not because the original series must have an even number of terms.

Arithmetic sum from endpoints

For an arithmetic series of $n$ terms, first term $a$ and last term $l$ ,

S_n=\frac{n}{2}(a+l)

This endpoint form is not printed on the Higher formulae sheet; it follows from the supplied arithmetic-series formula. Not in the formulae sheet: learn it. Foundation does not require arithmetic-series sums and has neither form on its sheet.

Count terms carefully when only the endpoints and difference are given.

Worked example 1: count the terms before adding

Question. Work out $4+7+10+\cdots+31$ .

Plan. Count the equal jumps, add one to count terms, then use the endpoints.

\begin{aligned} \frac{31-4}{3}&=9 &&\text{number of jumps}\\ n&=9+1=10 &&\text{include the first term}\\ S_{10}&=\frac{10}{2}(4+31) &&\text{endpoint formula}\\ &=5(35)=175 &&\text{five equal pairs} \end{aligned}

Why add $1$ to the jump count? Nine gaps join ten terms.

Answer. The sum is $175$ .

Check. Two reversed copies total $10(35)=350$ ; half is $175$ .

Now change only the final term from $31$ to $28$ . There are nine terms, so the two-copy argument is especially helpful.

Worked example 2: an odd number of terms

Question. Work out $4+7+10+\cdots+28$ .

Plan. Count the terms and use the same endpoint formula.

\begin{aligned} \frac{28-4}{3}&=8 &&\text{number of jumps}\\ n&=8+1=9 &&\text{number of terms}\\ S_9&=\frac{9}{2}(4+28) &&\text{endpoint formula}\\ &=\frac{9}{2}(32)=144 &&\text{half of two copies} \end{aligned}

Why is $9/2$ allowed? It comes from halving nine pairs across two copies; it need not count whole pairs in one copy.

Answer. The sum is $144$ .

Check. Four outer pairs total $4(32)=128$ ; the middle term is $16$ , giving $128+16=144$ .

Use the same count for $n$ everywhere in the calculation.

Watch out

The wrong line $S=9(4+31)/2$ uses nine jumps as nine terms. There are ten terms, so the correct line is $S=10(4+31)/2=175$ .

The next sum changes only the final term once more.

Using first term and common difference (Higher)

Usually a question gives the first term and common difference rather than the last term. Substitute $l=a+(n-1)d$ into the endpoint formula:

S_n=\frac{n}{2}\bigl[a+a+(n-1)d\bigr]

Combining the two copies of $a$ produces the form printed on the sheet.

Arithmetic-series sum

For constant difference $d$ , first term $a$ , and positive integer number of terms $n$ ,

S_n=\frac{n}{2}\left[2a+(n-1)d\right]

Higher: In the formulae sheet. The Foundation sheet does not supply this result, and arithmetic-series sums are Higher content.

Identify $a$ , $d$ and $n$ before substituting. Keep a negative difference in brackets so its sign survives the multiplication.

Worked example 3: substitute three quantities

Question. Find the sum of the first $12$ terms when $a=18$ and $d=2$ .

Plan. Substitute into the supplied sum formula, then simplify its bracket first.

\begin{aligned} S_{12}&=\frac{12}{2}\left[2(18)+(12-1)(2)\right] &&\text{substitute }a,d,n\\ &=6(36+22) &&\text{eleven jumps}\\ &=6(58)=348 &&\text{multiply} \end{aligned}

Why use $11d$ inside the bracket? The last term is eleven jumps beyond the first.

Answer. $S_{12}=348$ .

Check. The last term is $18+11(2)=40$ , so $12(18+40)/2=348$ .

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Handling data

3.1 - Sums of arithmetic series

Pairing the first and last terms (Higher)

Using first term and common difference (Higher)

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