1.04h - Arithmetic sequences and sums

1.04h - Arithmetic sequences and sums

Find terms and totals when each step adds the same amount. You will derive the arithmetic sum formula, recover missing information and decide whether a proposed term or total is actually possible.

A constant difference gives the term formula

A theatre has 1212 seats in its first row and 33 more seats in each successive row. How many seats are there in 2020 rows, without counting them individually? The first rows contain 12,15,18,21,…12,15,18,21,\ldots: the change, rather than the size, stays fixed.

You already know

A sequence is an ordered list; unu_n means its term at position nn. Subtract consecutive terms to find their difference. A position is a positive integer unless another starting index is stated.

You reach the twentieth row by making nineteen changes. That small counting detail drives the whole method.

Arithmetic sequence

An arithmetic sequence, also called an arithmetic progression or AP, has a constant difference dd between successive terms. For example, 7,4,1,−2,…7,4,1,-2,\ldots has d=−3d=-3. The list 2,4,8,…2,4,8,\ldots is not arithmetic: its differences are 22 and 44.

Starting at aa, the terms are a,a+d,a+2d,…a,a+d,a+2d,\ldots. Positive, negative and zero differences give increasing, decreasing and constant sequences respectively.

Arithmetic term

For first term aa, common difference dd and positive integer nn,

un=a+(n−1)du_n=a+(n-1)d

Not in the formula sheet: learn it.

Use the position to count changes before substituting.

Worked example 1: reaching a later term

Question. Find the tenth term of 7,4,1,−2,…7,4,1,-2,\ldots.

Plan. Identify the first term and difference, then make nine changes.

a=7,d=4−7=−3read the patternu10=7+(10−1)(−3)nine changes=−20simplify\begin{aligned} a&=7,\quad d=4-7=-3 && \text{read the pattern} \\ u_{10}&=7+(10-1)(-3) && \text{nine changes} \\ &=-20 && \text{simplify} \end{aligned}

Why nine changes? The first term is already present at position one.

Answer. The tenth term is −20-20.

Check. The ninth term is −17-17; subtracting 33 once more gives −20-20.

Now change only what is unknown: the term value is supplied and its position must be found.

Worked example 2: testing membership

Question. Does −22-22 occur in 7,4,1,−2,…7,4,1,-2,\ldots?

Plan. Set the term formula equal to −22-22, then check the resulting position.

7−3(n−1)=−22term equation−3(n−1)=−29subtract 7n=323solve\begin{aligned} 7-3(n-1)&=-22 && \text{term equation} \\ -3(n-1)&=-29 && \text{subtract }7 \\ n&=\frac{32}{3} && \text{solve} \end{aligned}

Why inspect the answer for nn? Solving an equation does not guarantee a whole-number position.

Answer. No: 32/332/3 is not an integer.

Check. The tenth and eleventh terms are −20-20 and −23-23; the sequence skips −22-22.

The algebra describes a line through possible positions, but the sequence uses only the permitted integer inputs.

Watch out

Wrong: un=a+ndu_n=a+nd. That gives u1=a+du_1=a+d, already one change too far. Right: un=a+(n−1)du_n=a+(n-1)d. A non-integer solution for nn is not a term position.

Try the membership decision with a fresh target in the same decreasing sequence.

Pair the terms to derive the sum

A total becomes much easier when you pair small terms with large ones. For 1+2+⋯+1001+2+\cdots+100, the pairs 1+1001+100, 2+992+99 and so on all equal 101101. There are 5050 pairs, giving 50505050. More surprisingly, the same idea works for odd numbers of terms: use two complete reversed rows, then halve.

Write l=a+(n−1)dl=a+(n-1)d for the last term. Then

Sn=a+(a+d)+⋯+lSn=l+(l−d)+⋯+a2Sn=n(a+l)add matching columns\begin{aligned} S_n&=a+(a+d)+\cdots+l \\ S_n&=l+(l-d)+\cdots+a \\ 2S_n&=n(a+l) && \text{add matching columns} \end{aligned}

Each column has the same sum. This proves the formula even when a single row cannot be divided into whole pairs.

Arithmetic sum

For nn terms of an AP, with first term aa, last term ll and difference dd,

Sn=n2(a+l)=n2{2a+(n−1)d}S_n=\frac n2(a+l)=\frac n2\{2a+(n-1)d\}

In the formula sheet.

Choose the version that uses the information you have. The sum of the first nn natural numbers follows by taking a=1a=1, d=1d=1, l=nl=n: ∑k=1nk=n(n+1)/2\sum_{k=1}^n k=n(n+1)/2. This special case is derived from the sheet rather than separately printed on it.

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