P2 4.2 - Arithmetic sequences and sums | Edexcel A-Level International Mathematics

Find terms and finite sums of arithmetic sequences, and understand the proof behind the sum formula. You will use sigma notation, sum the first natural numbers and recover unknown sequence parameters from given information.

Nth terms

How could you add $3+7+11+\cdots+43$ without writing every addition? First you need to know how many terms there are. Counting the gaps between terms is useful, but the first term must be counted too.

In $7,11,15,19,\ldots$ , every next term is $4$ more than the previous one. Starting at $a=7$ , the second term is $a+d$ , the third is $a+2d$ , and the fourth is $a+3d$ , with $d=4$ . There is one fewer gap than there are terms.

Arithmetic sequence

An arithmetic sequence, or arithmetic progression, has a constant common difference $d=u_{n+1}-u_n$ . Write its first term as $a$ . A positive $d$ gives increase, a negative $d$ gives decrease, and $d=0$ gives a constant sequence.

That repeated addition gives the explicit formula.

Arithmetic nth term

For a positive integer $n$ ,

u_n=a+(n-1)d.

In the formula booklet.

To decide whether a number is a term, set $u_n$ equal to that number and solve for $n$ . A term position must be a positive integer; do not round a non-integer solution to create a term.

Worked example 1: finding a term and its position

Question. The arithmetic sequence $7,11,15,\ldots$ continues. Find its fifteenth term and determine whether $99$ is a term.

Plan. Identify $a,d$ , then use the nth-term formula for both questions.

\begin{aligned} u_{15}&=7+(15-1)4=63\\ 99&=7+(n-1)4\\ 92&=4(n-1)\\ n&=24. \end{aligned}

Answer. $u_{15}=63$ , and $99$ is the twenty-fourth term.

Check. $7+23\times4=99$ . Also, $u_{14}=59$ and adding one common difference gives $u_{15}=63$ .

Change only the first term from $7$ to $9$ , keeping the common difference and the target $99$ . The same target may no longer occupy an integer position.

Worked example 2: a target between terms

Question. The arithmetic sequence $9,13,17,\ldots$ continues. Find its fifteenth term and determine whether $99$ is a term.

Plan. Use $a=9,d=4$ and test the calculated position of $99$ .

\begin{aligned} u_{15}&=9+14(4)=65\\ 99&=9+(n-1)4\\ n-1&=\frac{90}{4}=22.5\\ n&=23.5. \end{aligned}

Answer. $u_{15}=65$ . The number $99$ is not a term, because $23.5$ is not an integer position.

Check. The neighbouring terms are $u_{23}=97$ and $u_{24}=101$ . The target $99$ lies strictly between them.

Count the initial position as well as the gaps.

Watch out

For a final term $l$ and nonzero $d$ , $(l-a)/d$ counts the gaps. The number of terms is $1+(l-a)/d$ , provided that result is a positive integer.

Use a new first term with the same common difference.

Deriving the finite-sum formula

Adding an arithmetic sequence forwards and backwards produces equal column totals, which makes the entire sum easy to find.

For $S=3+7+11+15+19$ , reverse the terms underneath:

\begin{array}{rccccc} S=&3+&7+&11+&15+&19\\ S=&19+&15+&11+&7+&3\\\hline 2S=&22+&22+&22+&22+&22. \end{array}

There are five equal column totals, so $2S=5\times22$ and $S=55$ . This is two complete copies of the sum, rather than a requirement to divide the original terms into pairs.

[DIAGRAM: asset_name: An arithmetic sum written twice; asset_slug: arithmetic-sum-pairs; caption: Reversing a second copy makes every column sum equal to the first term plus the last.; description: Five columns show forward terms three seven eleven fifteen nineteen and reverse terms nineteen fifteen eleven seven three. Each column totals twenty-two; the five column totals add to one hundred and ten, representing twice the original sum of fifty-five.]
Reversing a second copy makes every column sum equal to the first term plus the last.

For a general arithmetic series with first term $a$ , common difference $d$ and last term $l=a+(n-1)d$ , write

\begin{aligned} S_n&=a+(a+d)+\cdots+(l-d)+l\\ S_n&=l+(l-d)+\cdots+(a+d)+a. \end{aligned}

In every column, the $d$ -steps cancel: $(a+kd)+(l-kd)=a+l$ . There are $n$ columns. Adding the two rows therefore gives $2S_n=n(a+l)$ . Dividing by two proves the formula.

Finite arithmetic sum

For $n$ terms, with $l=a+(n-1)d$ ,

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

Both forms are in the formula booklet. Know the proof by reversing a second copy and adding the two rows.

A series is a sum of sequence terms. Thus $u_n$ is one term, while $S_n$ is the sum of the first $n$ terms. Choose the first sum form when you know the last term; choose the second when you know $a,d,n$ .

Worked example 3: summing an even number of terms

Question. Find the sum of the first eight terms of $3,7,11,\ldots$ .

Plan. Find the eighth term, then multiply the average of the first and last terms by the number of terms.

\begin{aligned} l&=3+(8-1)4=31\\ S_8&=\frac82(3+31)=4\times34=136. \end{aligned}

Answer. $136$ .

Check. Four endpoint pairs each sum to $34$ : $3+31$ , $7+27$ , $11+23$ , $15+19$ . Their total is $4\times34=136$ .

Change only the number of terms from eight to nine. The formula still works when the number is odd.

Worked example 4: summing an odd number of terms

Question. Find the sum of the first nine terms of $3,7,11,\ldots$ .

Plan. Use the same sum formula, now with $n=9$ .

\begin{aligned} l&=3+8(4)=35\\ S_9&=\frac92(3+35)=\frac92\times38=171. \end{aligned}

Answer. $171$ .

Check. Add the ninth term to the previous sum: $S_9=S_8+u_9=136+35=171$ . Also, the middle term is $19$ , the average of the endpoints, so nine terms have total $9\times19=171$ .

The derivation used two rows of $n$ terms, so it did not assume that $n$ was even.

Watch out

The factor $n/2$ does not mean “take half a term”. It comes from halving two copies of the sum, so the formula works for odd $n$ as well as even $n$ .

The opening sum ends at $43$ , which is $3+10\times4$ , the eleventh term. Use that count to finish the calculation.

Sigma notation and natural-number sums

Sigma notation states exactly which terms are being added, so its lower and upper limits must both be read.

For example,

\sum_{r=1}^{4}(3r+2)=(3\times1+2)+(3\times2+2)+(3\times3+2)+(3\times4+2) =5+8+11+14.

Here $r$ takes every integer value from $1$ to $4$ , inclusive. It is a counter within the sum, not a factor multiplying the whole sum. You could call it $k$ instead without changing the result.

Sigma notation

The symbol $\sum$ , capital Greek sigma, means “add”. In $\sum_{r=m}^{n}f(r)$ , add the values of $f(r)$ for $r=m,m+1,\ldots,n$ . There are $n-m+1$ terms, for integer limits $m\le n$ .

The first $n$ natural numbers $1,2,\ldots,n$ form an arithmetic sequence with $a=1,d=1,l=n$ . Substituting into the proved sum formula gives

Sum of the first natural numbers

For a positive integer $n$ ,

\sum_{r=1}^{n}r=1+2+\cdots+n=\frac{n(n+1)}2.

This follows from the arithmetic sum formula in the booklet; this separate identity is not printed in the P2 section.

You can separate a sum of two expressions and take a constant factor outside:

\sum_{r=1}^{n}(Ar+B) =A\sum_{r=1}^{n}r+\sum_{r=1}^{n}B =A\frac{n(n+1)}2+Bn.

The constant $B$ is added once for each of the $n$ terms. For a later lower limit, either subtract the unwanted starting terms or identify the first term, last term and number of terms directly.

Worked example 5: using the natural-number sum

Question. Evaluate $\sum_{r=1}^{10}(3r+2)$ .

Plan. Separate the variable contribution from ten copies of the constant.

\begin{aligned} \sum_{r=1}^{10}(3r+2) &=3\sum_{r=1}^{10}r+2(10)\\ &=3\left(\frac{10\times11}{2}\right)+20\\ &=185. \end{aligned}

Answer. $185$ .

Check. The terms form an arithmetic series with endpoints $5$ and $32$ and ten terms. Its sum is $(10/2)(5+32)=185$ .

Change only the lower limit from $1$ to $4$ . Three unwanted terms must now be removed.

Worked example 6: a later starting index

Question. Evaluate $\sum_{r=4}^{10}(3r+2)$ .

Plan. Subtract the terms for $r=1,2,3$ from the complete sum in Example 5.

\begin{aligned} \sum_{r=4}^{10}(3r+2) &=185-\bigl[(3+2)+(6+2)+(9+2)\bigr]\\ &=185-(5+8+11)\\ &=161. \end{aligned}

Answer. $161$ .

Check. There are $10-4+1=7$ terms, starting at $14$ and ending at $32$ . The arithmetic sum is $(7/2)(14+32)=161$ .

The limits tell you the number of constant contributions too.

Watch out

$\sum_{r=4}^{10}2=7\times2=14$ , not $2$ or $20$ . Also, $\sum_{r=4}^{10}r$ is not $10\times11/2$ ; that latter sum begins at $1$ .

Move the starting limit by one and check the inclusive count.

Solving for unknown sequence parameters

Information about two terms gives two equations for the first term and common difference; the distance between their positions tells you how many differences separate them.

For instance, from $u_4$ to $u_{10}$ there are six steps, so $u_{10}-u_4=6d$ . Subtracting the two nth-term equations eliminates the first term. Once $d$ is known, substitute back to find $a$ , then use the finite sum formula if required.

Problem-solving

Translate each term value into $a+(n-1)d$ before solving. If a sum is given instead, translate it into $n[2a+(n-1)d]/2$ . The data become equations in the same unknowns.

Use the given positions as well as the given values.

Worked example 7: recovering parameters from two terms

Question. An arithmetic sequence has $u_4=17$ and $u_{10}=41$ . Find $a$ , $d$ and $S_{12}$ .

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

\begin{aligned} a+3d&=17\\ a+9d&=41\\ 6d&=24 &&\text{subtract}\\ d&=4\\ a&=17-3(4)=5\\ S_{12}&=\frac{12}{2}\bigl[2(5)+11(4)\bigr]=324. \end{aligned}

Answer. $a=5$ , $d=4$ , $S_{12}=324$ .

Check. $u_4=5+12=17$ and $u_{10}=5+36=41$ , as supplied. The twelfth term is $49$ , so the other sum form gives $(12/2)(5+49)=324$ .

Change only the tenth term from $41$ to $29$ , keeping $u_4=17$ . Both parameters must be reconsidered.

Worked example 8: changing one piece of term data

Question. An arithmetic sequence has $u_4=17$ and $u_{10}=29$ . Find $a$ , $d$ and $S_{12}$ .

Plan. Use the same six-step gap, then substitute into the fourth-term equation.

\begin{aligned} a+3d&=17\\ a+9d&=29\\ 6d&=12\\ d&=2,\qquad a=17-3(2)=11\\ S_{12}&=\frac{12}{2}\bigl[22+11(2)\bigr]=264. \end{aligned}

Answer. $a=11$ , $d=2$ , $S_{12}=264$ .

Check. $11+3(2)=17$ and $11+9(2)=29$ . The last term is $33$ , giving the alternative sum $(12/2)(11+33)=264$ .

Term data are not the only way to find $a,d$ . If $u_4=17$ and $S_{10}=230$ , then

a+3d=17,\qquad 5(2a+9d)=230.

The second equation reduces to $2a+9d=46$ . Subtracting twice the first equation gives $3d=12$ , so $d=4,a=5$ . The sum check is $5(10+36)=230$ . The method remains “translate, solve, check”.

When $d<0$ , adding terms increases the running sum while the new term is positive, leaves it unchanged for a zero term, and decreases it once the new terms are negative. For $a=20,d=-3$ , $u_n=23-3n$ ; $u_7=2>0$ and $u_8=-1<0$ . The largest partial sum is therefore $S_7=(7/2)(20+2)=77$ . Check: $S_6=75$ and $S_8=76$ , both smaller. If a zero term occurs, the sums immediately before and after adding that term tie. Term counts remain integers throughout.

The opening sum $3+7+\cdots+43=253$ is now explained by both its count and its endpoint average, rather than a long list of additions.

Watch out

$u_{10}-u_4$ spans $10-4=6$ common differences, not ten or seven. Subtract the positions to count gaps; use a positive integer $n$ when interpreting a term or partial sum.

Change the tenth-term value once more and solve the two equations afresh.

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P2 4.2 - Arithmetic sequences and sums

Nth terms

Deriving the finite-sum formula

Sigma notation and natural-number sums

Solving for unknown sequence parameters

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