5A, 5B - Arithmetic series

5A, 5B - Arithmetic series

An arithmetic sequence is easy to continue, but adding many terms one by one is slow. In 5B, you will calculate arithmetic totals, recover an unknown number of terms, and find when a total reaches a target. The sigma notation from 5A lets you specify exactly which consecutive terms to add.

Adding an arithmetic sequence

How could you add the first twenty terms of 3,7,11,15,…3,7,11,15,\ldots without twenty separate additions?

Start with five terms: 3+7+11+15+193+7+11+15+19. Write them backwards underneath. Every vertical pair adds to 2222, so two copies of the sum total 5×225\times22. One copy is 5555. An odd number of terms causes no difficulty because we are pairing two copies, rather than trying to split one list into whole pairs.

[DIAGRAM: asset_name: Pairing an arithmetic sum with its reverse; asset_slug: arithmetic-pairing; caption: Five equal pair totals give twice the arithmetic sum; description: The top row is 3, 7, 11, 15, 19 and the second row is 19, 15, 11, 7, 3. Each column totals 22. Two copies total 5 times 22 equals 110, so one sum is 55.]
Five equal pair totals give twice the arithmetic sum

Arithmetic series

An arithmetic series is the sum of terms of an arithmetic sequence. Write SnS_n for the sum of its first nn terms, while unu_n denotes just its nth term.

If the first term is aa and last term is l=a+(n−1)dl=a+(n-1)d, each forward-and-backward pair totals a+la+l. This gives the useful formula without needing a formal proof.

Arithmetic sum

For a positive integer nn,

Sn=n2(a+l)=n2[2a+(n−1)d]S_n=\frac n2(a+l)=\frac n2\bigl[2a+(n-1)d\bigr]

The second form is on the formulae sheet. The first form follows by substituting l=a+(n−1)dl=a+(n-1)d.

Choose the form that fits the information you have.

Worked example 1: adding from the first term

Question. Find the sum of the first twenty terms of 3,7,11,15,…3,7,11,15,\ldots.

Plan. Use a=3a=3, d=4d=4 and n=20n=20 in the sum formula.

S20=202[2(3)+19(4)]substitute=10(82)simplify the bracket=820multiply\begin{aligned} S_{20}&=\frac{20}{2}\bigl[2(3)+19(4)\bigr] &&\text{substitute}\\ &=10(82) &&\text{simplify the bracket}\\ &=820 &&\text{multiply} \end{aligned}

Answer. The total is 820820.

Check. The twentieth term is 3+19(4)=793+19(4)=79. The mean of the first and last terms is 4141, and 20×41=82020\times41=820.

Now add a block that starts later in the same sequence. The sum formula still works, but its first term must be the first term being added.

Worked example 2: adding between two positions

Question. For uk=3+4(k−1)u_k=3+4(k-1), find ∑k=620uk\sum_{k=6}^{20}u_k.

Plan. Find the endpoints and count the included terms before using the first-plus-last form.

u6=3+5(4)=23first included termu20=79last included termN=20−6+1=15inclusive term count∑k=620uk=152(23+79)=765sum this block\begin{aligned} u_6&=3+5(4)=23 &&\text{first included term}\\ u_{20}&=79 &&\text{last included term}\\ N&=20-6+1=15 &&\text{inclusive term count}\\ \sum_{k=6}^{20}u_k&=\frac{15}{2}(23+79)=765 &&\text{sum this block} \end{aligned}

Answer. 765765.

Check. The first five terms sum to 5555, so the same total is S20−S5=820−55=765S_{20}-S_5=820-55=765.

Subtracting S6S_6 would remove the sixth term as well. To keep terms 66 onwards, remove only terms 11 to 55.

Finding an unknown number of terms

A stated total becomes an equation in nn, often a quadratic because both the number of terms and the last term depend on nn.

The algebra treats nn as a real number; a term count must be a positive integer. Check that condition after solving, then substitute a retained value into the original information. A decreasing series can sometimes have two valid term counts, so do not discard a solution merely because there are two.

Worked example 3: recovering a term count from a total

Question. An arithmetic series has first term 44, common difference 33, and sum Sn=116S_n=116. Find nn.

Plan. Write the sum formula, solve the quadratic, then apply the term-count condition.

n2[8+3(n−1)]=116translate the total3n2+5n−232=0multiply by 2 and rearrange(n−8)(3n+29)=0factorisen=8orn=−293solve\begin{aligned} \frac n2\bigl[8+3(n-1)\bigr]&=116 &&\text{translate the total}\\ 3n^2+5n-232&=0 &&\text{multiply by }2\text{ and rearrange}\\ (n-8)(3n+29)&=0 &&\text{factorise}\\ n&=8\quad\text{or}\quad n=-\frac{29}{3} &&\text{solve} \end{aligned}

Answer. n=8n=8; the other root is not a positive integer.

Check. The eighth term is 2525, so S8=8(4+25)/2=116S_8=8(4+25)/2=116.

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