Ms12, Ms13 - Arithmetic sequences and sums | OCR A-Level Mathematics B (MEI)

Use equal differences to find terms, totals and missing information. You will explain the arithmetic sum formula and read an unfamiliar endpoint estimate, deciding when a correction makes it exact.

A constant difference gives the term formula

A light display has $8$ lamps in its first strip and $4$ more in every successive strip. Could you find the total in $25$ strips without adding every strip separately? Start with $8,12,16,20,\ldots$ . The number changes, but each step has the same size.

You already know

The term $u_n$ is the value at position $n$ , not the sum of the first $n$ values. Subtract consecutive values to compare their changes. Positions here are positive integers.

There are $24$ steps from strip one to strip twenty-five: count the gaps, not the strips.

Arithmetic sequence

An arithmetic sequence, or arithmetic progression (AP), has the same difference $d$ between every pair of successive terms. For example, $8,12,16,\ldots$ has $d=4$ ; $8,12,18,\ldots$ is not arithmetic. A constant sequence is an AP with $d=0$ .

With first term $a$ , the first few terms are $a,a+d,a+2d,\ldots$ . A negative difference makes the values decrease.

Arithmetic term

For positive integer $n$ , first term $a$ and difference $d$ ,

u_n=a+(n-1)d

Not in the formula sheet: learn it.

Finding a later value is one substitution; deciding whether a value occurs also needs a check on the position.

Worked example 1: find a term

Question. Find the twelfth term of $8,12,16,\ldots$ .

Plan. Make eleven additions of $4$ after the first term.

\begin{aligned} a&=8,\quad d=12-8=4 && \text{identify the AP} \\ u_{12}&=8+11(4)=52 && \text{eleven changes} \end{aligned}

Why eleven? The first value needs no addition.

Answer. $u_{12}=52$ .

Check. $u_{11}=48$ , and adding $4$ gives $52$ .

Change only the unknown: supply a value and recover its possible position.

Worked example 2: find a position

Question. Determine whether $54$ occurs in $8,12,16,\ldots$ .

Plan. Solve the term equation and inspect its solution.

\begin{aligned} 8+4(n-1)&=54 && \text{target value} \\ 4(n-1)&=46 && \text{subtract }8 \\ n&=\frac{25}{2} && \text{solve} \end{aligned}

Why not round the position? Rounding changes the term being considered.

Answer. $54$ does not occur, because $25/2$ is not an integer.

Check. The twelfth and thirteenth terms are $52$ and $56$ , on opposite sides of $54$ .

A term equation can have a real solution even when the discrete sequence has no such term.

Watch out

Wrong: $u_n=a+nd$ . Substituting $n=1$ exposes the extra step. Right: $u_n=a+(n-1)d$ , with a permitted integer position.

The next target changes the value again; keep the position exact.

Pair the terms to derive the sum

An arithmetic total is its number of terms multiplied by their mean. The mean is halfway between the endpoints, even if no actual term equals it.

For example, the mean of $1,2,\ldots,50$ is $25.5$ , so the total is $1275$ . Here is why that shortcut works. Write the $n$ terms forwards and backwards, with last term $l$ :

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

This uses two complete rows, so it works for odd as well as even $n$ .

Arithmetic sum

For $n$ terms of an AP,

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

In the formula sheet.

Putting $a=d=1$ gives $\sum_{r=1}^n r=n(n+1)/2$ , the sum of the first $n$ natural numbers. That special case is derived from the printed formula. A pleasing consequence is that two consecutive triangular totals make a square: $n(n+1)/2+n(n-1)/2=n^2$ .

Worked example 3: sum a known count

Question. Find the total lamps in $25$ strips, starting with $8$ and adding $4$ per strip.

Plan. Calculate the last value, then use the endpoint average.

\begin{aligned} l&=8+24(4)=104 && \text{last strip} \\ S_{25}&=\frac{25}{2}(8+104) && \text{sum formula} \\ &=1400 && \text{evaluate} \end{aligned}

Why use the average? Reversing the strips pairs every value with another whose total is $112$ .

Answer. There are $1400$ lamps.

Check. The middle, thirteenth strip has $56$ lamps, and $25(56)=1400$ .

Change only the stopping information: a last value replaces the term count.

Worked example 4: recover the count first

Question. Find $8+12+16+\cdots+72$ .

Plan. Count the gaps, include both endpoints, then sum.

\begin{aligned} 72&=8+4(n-1) && \text{last-term equation} \\ n&=17 && \text{sixteen gaps} \\ S_{17}&=\frac{17}{2}(8+72)=680 && \text{seventeen terms} \end{aligned}

Why add one to the gap count? The first value is included before the first gap.

Answer. The sum is $680$ .

Check. Eight pairs give $8(80)$ , and the middle value is $40$ : the total is $680$ .

Write the count clearly before the sum; a wrong endpoint count can leave otherwise tidy algebra answering a different question.

Watch out

Wrong: $(72-8)/4=16$ terms. Right: that is $16$ gaps, hence $17$ terms. Always include the initial term.

Try a fresh endpoint in the same AP.

Recover unknown terms and counts

Known terms at two positions determine the change per gap. Their equations also reveal the first term, which may be quite different from either given value.

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Ms12, Ms13 - Arithmetic sequences and sums

A constant difference gives the term formula

Pair the terms to derive the sum

Recover unknown terms and counts

Continue this lesson.

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