Arithmetic Sequences and Series

Learning goals

Test for a constant difference $d$ throughout
Apply $a_n = a_1 + (n-1)d$ , counting steps not terms
Recover $d$ from two terms by dividing rise by steps
Count terms with $n = \tfrac{L - a_1}{d} + 1$
Prove $S_n = \tfrac{n}{2}(a_1 + a_n)$ by reversing and adding
Read a quadratic $S_n$ backwards to recover $d$

What makes a sequence arithmetic

A sequence is a function on the positive integers, so $a_n$ is the output at input $n$ . That function can do anything at all. An arithmetic sequence is one that does the simplest possible thing: it adds the same number every time you advance the input by one.

A sequence $a_1, a_2, a_3, \ldots$ is arithmetic when there is a fixed number $d$ , the common difference, with

a_{n+1} - a_n = d \qquad \text{for every } n \ge 1,

which is the same as the recursive rule $a_{n+1} = a_n + d$ . The word common is doing real work here: the difference must be the same for every pair of neighbours, not just the first pair you happen to check.

Three quick examples and two non-examples make the definition concrete:

$4, 9, 14, 19, \ldots$ is arithmetic with $d = 5$ (each term is five more than the one before).
$20, 13, 6, -1, \ldots$ is arithmetic with $d = -7$ . A negative $d$ is perfectly legal; the sequence simply runs downhill.
$3, 3, 3, 3, \ldots$ is arithmetic with $d = 0$ . It is a dull sequence, but it satisfies the definition, and the formulas below for the $n$ th term and for the sum will handle it without complaint. (One later formula, the one that counts terms, will need $d \neq 0$ , and the reason is worth seeing when you get there.)
$1, 2, 4, 8, \ldots$ is not arithmetic. Its differences are $1$ , then $2$ , then $4$ , and they never settle down.
$1, 4, 9, 16, \ldots$ (the squares) is not arithmetic either. Its differences are $3$ , then $5$ , then $7$ . They change by a constant amount, which is interesting, but they are not constant.

To test a sequence, subtract each term from the one after it and compare all the results. Two matching differences prove nothing: $2, 5, 8, 13$ starts with $3$ and $3$ and then jumps by $5$ .

The nth term, by counting steps

You know $a_1$ and you know $d$ . How do you reach $a_{50}$ without writing down forty-nine additions?

Every move to the next term costs one $d$ . So the question is not “how many terms are there” but “how many steps do I take”, and those are two different counts.

Five terms, but only four steps between them. Each step adds d, so getting from the 1st term to the 5th costs four lots of d, not five. That is where the n minus 1 in the formula comes from.

Five terms sit on that line, and four arrows join them. The terms are the places; $d$ is the toll you pay to move between places. Nobody pays a toll to arrive at the place they are already standing on, so the first term costs nothing. Because that first place is free, the count of tolls is one less than the count of places.

The nth term is $a_n = a_1 + (n-1)d$ #

Take any $n \ge 2$ (the case $n = 1$ just says $a_1 = a_1$ and needs no argument).

The definition hands you one equation per step. Stepping from term $1$ to term $2$ gives $a_2 - a_1 = d$ , stepping from term $2$ to term $3$ gives $a_3 - a_2 = d$ , and so on. The steps run from $1$ to $2$ , from $2$ to $3$ , all the way up to the step from $n-1$ to $n$ . So there are exactly $n - 1$ of them, one for each starting index between $1$ and $n-1$ .

Add all $n - 1$ of those equations together. Every right side is $d$ , so the right sides total $(n-1)d$ :

\sum_{k=1}^{n-1} \big(a_{k+1} - a_k\big) = \sum_{k=1}^{n-1} d = (n-1)d.

The left side collapses. Written out, it is $(a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3) + \cdots + (a_n - a_{n-1})$ , and every interior term appears twice, once with a plus sign and once with a minus sign. Those pairs annihilate each other and only the two ends survive, so the left side is $a_n - a_1$ . Therefore

a_n - a_1 = (n-1)d, \qquad \text{so} \qquad a_n = a_1 + (n-1)d.

The whole content of the formula is that count $n - 1$ . It is the number of steps, and the number of steps between $n$ terms is always one fewer than $n$ . Fence posts show the same pattern: five posts have four gaps between them.

Two rearrangements of that formula earn their keep. Solving for $d$ gives

d = \frac{a_n - a_1}{n - 1},

and the same argument run between any two terms (there are $m - k$ steps from term $k$ to term $m$ ) gives the more flexible version

a_m = a_k + (m - k)d, \qquad \text{so} \qquad d = \frac{a_m - a_k}{m - k}.

Both denominators count steps, never terms. From $a_3$ to $a_{11}$ there are $8$ steps, so you divide by $8$ , not by $11$ and not by $9$ .

The fence-post count is the kind of thing that stays slippery until you have done it with your hands, so the figure below hands you one step at a time. You set the number the step starts from and the number added to it, and the arrow shows that single step and where it lands.

Set the start to $-4$ and the number added to $2$ . The arrow lands on $-2$ , so a sequence with $a_1 = -4$ and $d = 2$ has $a_2 = -2$ . Move the start to $-2$ , where you just landed, and take the same step again to reach $0$ , which is $a_3$ . Carry on from $0$ and then from $2$ . You end on $4$ , and $4$ is $a_5$ . Now count what you did: five terms named, four steps taken, which is the fence-post rule above with the posts in your hands. That gap of one is the whole of $n - 1$ , and it is not a convention anyone chose. The first term is where you begin, so it costs nothing, and only the terms after it have to be walked to. Set the number added to $-2$ and run the route backwards from $4$ to confirm it from the other side. You take four steps again, and land back on $-4$ , because a negative common difference changes the direction of the walk and nothing at all about the counting.

Number line walk

Start at Then add

-4 + 2 = -2. Adding a positive number moves 2 units to the right.

A single step along a number line, sized and positioned by you. Because the step can be retaken from the point the last one reached, an arithmetic sequence can be walked out term by term and its steps counted against the terms they produce.

An arithmetic sequence is a linear function of n

Expand the formula and something familiar appears:

a_n = a_1 + (n-1)d = dn + (a_1 - d).

The right side is $dn + b$ with $b = a_1 - d$ , so $a_n$ is a linear function of $n$ with slope $d$ . The converse is true as well. If a sequence obeys $a_n = mn + b$ for every $n$ , then

a_{n+1} - a_n = \big(m(n+1) + b\big) - \big(mn + b\big) = m,

a constant, so the sequence is arithmetic with $d = m$ . Both directions hold, so “arithmetic” and “linear in $n$ ” are two names for the same family of sequences. That is a genuinely useful equivalence: whenever you meet a rule like $a_k = 4k + 1$ , you may read off $d = 4$ on sight, with no subtraction at all.

It also tells you what these sequences look like. Plot the points $(n, a_n)$ and they land on a straight line of slope $d$ , evenly spaced, one point per positive integer. The intercept $a_1 - d$ is the “zeroth term”, the value the line would take at $n = 0$ , a term the sequence itself never has.

How many terms are there

Turn the formula around once more. If a sequence starts at $a_1$ , steps by $d$ , and its last term is $L$ , then $L = a_1 + (n-1)d$ , so

n = \frac{L - a_1}{d} + 1.

The division tells you how many steps separate the first term from the last, and then the $+1$ converts steps back into terms. It is the fence-post count again, run backwards, and the missing $+1$ is one of the most common errors in this entire topic.

This is the one formula in the lesson that demands $d \neq 0$ , since it divides by $d$ . The restriction is not a flaw in the formula but a fact about the question: if $d = 0$ then every term equals $L$ . In that case $3, 3, 3$ and $3, 3, 3, 3, 3$ look identical from their values alone, so no rule could recover the count from them. The $n$ th-term and sum formulas never divide by $d$ , so they stay valid even in that degenerate case.

Worked example 1 Find the 30th term of $4, 11, 18, 25, \ldots$

The differences are $11 - 4 = 7$ , $18 - 11 = 7$ , and $25 - 18 = 7$ , so the sequence is arithmetic with $a_1 = 4$ and $d = 7$ .

Getting from the 1st term to the 30th takes $29$ steps, not $30$ :

a_{30} = a_1 + (30 - 1)d = 4 + 29 \cdot 7 = 4 + 203 = 207.

Had you multiplied by $30$ instead, you would have got $4 + 210 = 214$ , which is the 31st term. The formula never says “multiply $d$ by the term number”; it says “multiply $d$ by the number of steps”.

Worked example 2 An arithmetic sequence has $a_6 = 23$ and $a_{14} = 55$ . Find $a_1$ and a formula for $a_n$ .

From term $6$ to term $14$ is $14 - 6 = 8$ steps, and those steps carry the value from $23$ up to $55$ , a rise of $32$ . Each step is worth the same amount, so

d = \frac{a_{14} - a_6}{14 - 6} = \frac{55 - 23}{8} = \frac{32}{8} = 4.

Now walk backwards from $a_6$ to $a_1$ , which is $5$ steps down:

a_1 = a_6 - 5d = 23 - 5 \cdot 4 = 3.

That gives the explicit rule, which you can simplify:

a_n = 3 + (n-1)4 = 4n - 1.

Check it against the data you were given: $a_6 = 24 - 1 = 23$ and $a_{14} = 56 - 1 = 55$ . Both match. Notice the check is worth doing, because dividing by $9$ (the number of terms from the 6th to the 14th) instead of $8$ would have produced $d = \tfrac{32}{9}$ . With that value of $d$ , nothing would have come out whole.

Check your understanding

An arithmetic sequence has $a_1 = 6$ and $d = 4$ . What is $a_{12}$ ?

50

54

46

48

Worked example 3 How many terms are in $8, 15, 22, \ldots, 148$ ?

The common difference is $d = 7$ , and the last term is $L = 148$ .

First count the steps from the first term to the last. Each step adds $7$ , and the total climb is $148 - 8 = 140$ , so

\frac{L - a_1}{d} = \frac{148 - 8}{7} = \frac{140}{7} = 20 \text{ steps}.

Twenty steps land you on the last term, but you were standing on a term before you took any step at all. So the number of terms is

n = 20 + 1 = 21.

Confirm it forwards: $a_{21} = 8 + 20 \cdot 7 = 148$ , exactly the last term.

Adding the terms, by writing the sum twice

A series is what you get when you add up the terms of a sequence. Write $S_n$ for the sum of the first $n$ terms, which in the sigma notation from the last lesson is

S_n = \sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n.

Adding a hundred terms one at a time is hopeless, and the trick that rescues you is one of the most famous in mathematics. Write the sum out forwards. Then, underneath it, write the very same sum backwards:

\begin{aligned} S_n &= a_1 + (a_1 + d) + (a_1 + 2d) + \cdots + a_n, \\ S_n &= a_n + (a_n - d) + (a_n - 2d) + \cdots + a_1. \end{aligned}

Now add the two lines column by column. The first column gives $a_1 + a_n$ . The second gives $(a_1 + d) + (a_n - d)$ , and the $d$ ‘s cancel, so it is $a_1 + a_n$ again. The third gives $(a_1 + 2d) + (a_n - 2d) = a_1 + a_n$ . Every column produces the same total, and the reason is easiest to see by walking along the two lines. As you walk right along the top line you gain $d$ at each step, while the bottom line loses $d$ at each step, and the two motions cancel exactly.

The sum of the first n terms is $S_n = \frac{n}{2}(a_1 + a_n)$ #

Let the sequence be arithmetic with common difference $d$ . The $k$ th column of the layout above pairs the $k$ th term counted from the front with the $k$ th term counted from the back. In symbols, that column pairs $a_k$ with $a_{n+1-k}$ , since counting back $k - 1$ steps from $a_n$ lands on index $n+1-k$ . Use the formula for the $n$ th term on both:

a_k + a_{n+1-k} = \big(a_1 + (k-1)d\big) + \big(a_1 + (n-k)d\big) = 2a_1 + (n-1)d.

The $k$ has vanished, because $(k-1) + (n-k) = n-1$ no matter what $k$ is. So every column has the same total, and since $a_n = a_1 + (n-1)d$ , that common total is exactly

2a_1 + (n-1)d = a_1 + a_n.

There are $n$ columns, one for each term, so adding the two copies of the sum gives

2S_n = n(a_1 + a_n), \qquad \text{hence} \qquad S_n = \frac{n}{2}\,(a_1 + a_n).

The doubling is what makes the argument airtight. Suppose a student instead tries to fold the sum in half and pair the terms off physically, first with last, second with second-last. That student immediately has to ask whether $n$ is even (so the pairs come out whole) or odd (so one lonely middle term is stranded). Writing a second, reversed copy of the whole sum never asks that question: every one of the $n$ terms gets a partner, whether $n$ is $100$ or $101$ . When $n$ is odd the middle term is simply paired with itself, and that is not a special case at all. That is because $a_m + a_m = a_1 + a_n$ holds for the middle index $m$ , just as the column identity above says it must. No case split, no leftovers.

The doubling trick in pictures, for 2 + 5 + 8 + 11 + 14. The dark bars are the sum written forwards; the accented bars stacked on top are the same sum written backwards. Every column now stands 16 units tall, the first term plus the last, and there are 5 columns, so twice the sum is 5 times 16, which is 80. The sum itself is 40.

The same formula, wearing different clothes

The formula $S_n = \tfrac{n}{2}(a_1 + a_n)$ wants the last term. Often you know $d$ instead, and substituting $a_n = a_1 + (n-1)d$ turns it into

S_n = \frac{n}{2}\Big(a_1 + a_1 + (n-1)d\Big) = \frac{n}{2}\big(2a_1 + (n-1)d\big).

These are not two formulas to memorize. They are one statement written with different inputs: use the first when the last term is handed to you, the second when only $a_1$ and $d$ are. Both come from the single fact $2S_n = n(a_1 + a_n)$ .

Read $\tfrac{n}{2}(a_1 + a_n)$ once more as $n \cdot \tfrac{a_1 + a_n}{2}$ , and the sum tells you something you can feel. An arithmetic series is just its number of terms times the average of its first and last term. The terms are evenly spread, so the small ones at one end compensate exactly for the large ones at the other. When $n$ is odd the middle term is that average, since the middle index is $m = \tfrac{n+1}{2}$ and

a_m = a_1 + \frac{n-1}{2}\,d = \frac{a_1 + a_n}{2},

so a sum of $9$ terms is $9$ times its 5th term, and a sum of $11$ terms is $11$ times its 6th. That shortcut works only because the sequence is arithmetic. It is not a fact about sums in general.

One last observation, which the quadratics chapter has already prepared you for. Expanding the second form gives

S_n = \frac{d}{2}n^2 + \Big(a_1 - \frac{d}{2}\Big)n,

a quadratic in $n$ with no constant term.

The converse holds too, and it costs one line, so do not take it on trust. Suppose all you are told is that some sequence has running total $S_n = An^2 + Bn$ for every $n$ . Recover its $n$ th term as the jump the total makes when that term arrives:

a_n = S_n - S_{n-1} = \big(An^2 + Bn\big) - \big(A(n-1)^2 + B(n-1)\big) = 2An + (B - A).

That is linear in $n$ , and a sequence linear in $n$ is arithmetic, by the equivalence proved earlier. So the sequence is arithmetic with $d = 2A$ , whatever $A$ and $B$ are. If a problem hands you $S_n = 2n^2 + 3n$ , you may read off $d = 4$ and $a_n = 4n + 1$ without ever seeing a single term of the sequence.

Worked example 4 Add the whole numbers from $1$ to $100$

The sequence $1, 2, 3, \ldots, 100$ is arithmetic with $a_1 = 1$ , $d = 1$ , and $a_{100} = 100$ , so there is nothing to compute before applying the sum.

S_{100} = \frac{100}{2}\,(1 + 100) = 50 \cdot 101 = 5050.

The doubling argument is even easier to see here: written forwards and backwards, the columns are $1 + 100$ , $2 + 99$ , $3 + 98$ , and so on, one hundred columns each totalling $101$ , giving $2S = 100 \cdot 101 = 10100$ .

Worked example 5 Evaluate $\sum_{k=1}^{20} (3k - 2)$

The rule $a_k = 3k - 2$ is linear in $k$ , so by the equivalence proved above the terms form an arithmetic sequence with $d = 3$ . You need only its two ends:

a_1 = 3(1) - 2 = 1, \qquad a_{20} = 3(20) - 2 = 58.

There are $20$ terms, so apply the sum directly:

S_{20} = \frac{20}{2}\,(1 + 58) = 10 \cdot 59 = 590.

The second form is a useful check, since it never mentions $a_{20}$ :

S_{20} = \frac{20}{2}\big(2(1) + 19 \cdot 3\big) = 10(2 + 57) = 590. \checkmark

Check your understanding

An arithmetic sequence has $a_1 = 3$ and $a_{20} = 60$ . What is $S_{20}$ , the sum of its first $20$ terms?

1{,}260

63

630

600

Worked example 6 A stadium section has $24$ seats in the first row and $4$ more in each row after it. How many seats are in the first $15$ rows?

Row by row the seat counts are $24, 28, 32, \ldots$ , an arithmetic sequence with $a_1 = 24$ and $d = 4$ . The total you want is $S_{15}$ .

Take the route through the last row first. Reaching row $15$ from row $1$ takes $14$ steps:

a_{15} = 24 + 14 \cdot 4 = 80 \text{ seats in row } 15.

Now sum the fifteen rows:

S_{15} = \frac{15}{2}\,(24 + 80) = \frac{15}{2} \cdot 104 = 15 \cdot 52 = 780.

The second form skips the intermediate step and must agree:

S_{15} = \frac{15}{2}\big(2 \cdot 24 + 14 \cdot 4\big) = \frac{15}{2}(48 + 56) = \frac{15}{2} \cdot 104 = 780. \checkmark

The section holds $780$ seats. Sanity-check the size: fifteen rows averaging $52$ seats is about $780$ , and $52$ is indeed the average of the first row ( $24$ ) and the last ( $80$ ).

Check your understanding

How many terms are in the sequence $7, 12, 17, \ldots, 97$ ?

18

19

20

90

Common mistakes

Multiplying $d$ by $n$ instead of by $n - 1$ . This is the single most common error in the topic. Getting from $a_1$ to $a_n$ takes $n - 1$ steps, not $n$ , so $a_{10} = a_1 + 9d$ . Writing $a_1 + 10d$ silently produces the eleventh term.
Forgetting the $+1$ when counting terms. The division $(L - a_1) \div d$ counts steps. The number of terms is one more. From $3$ to $43$ by fours is $10$ steps but $11$ terms.
Dividing by the wrong count when recovering $d$ from two terms. Between $a_4$ and $a_9$ there are $9 - 4 = 5$ steps, so divide the rise by $5$ , never by $9$ and never by $6$ .
Testing only one difference. A sequence is arithmetic only if every consecutive difference agrees. In $2, 5, 8, 13$ the first two differences match and the sequence still is not arithmetic.
Dropping the half in the sum. The doubling produces $2S_n = n(a_1 + a_n)$ , so $n(a_1 + a_n)$ is twice the answer. The $\tfrac{n}{2}$ is the whole point of the trick.
Using the sum formula on a sequence that is not arithmetic. For $1, 2, 4, 8$ the formula would claim $\tfrac{4}{2}(1 + 8) = 18$ , while the true sum is $15$ . The columns are only constant when the differences are.
Blending the two forms. $\tfrac{n}{2}(a_1 + a_n)$ needs the last term; $\tfrac{n}{2}(2a_1 + (n-1)d)$ needs the common difference. Something like $\tfrac{n}{2}(a_1 + d)$ is neither of them.
Losing the sign of a negative $d$ . In $20, 17, 14, \ldots$ the common difference is $-3$ , so $a_{12} = 20 + 11(-3) = -13$ . Subtracting $d$ “because the sequence goes down” double-counts the sign.

Key takeaways

A sequence is arithmetic when $a_{n+1} - a_n = d$ for every $n$ , with the same $d$ throughout. A negative $d$ runs the sequence downhill; $d = 0$ makes it constant.
The nth term: $a_n = a_1 + (n-1)d$ . The $n - 1$ counts the steps from the first term to the $n$ th, and a step is what costs you a $d$ .
Between any two terms, $a_m = a_k + (m-k)d$ , so $d = \dfrac{a_m - a_k}{m - k}$ : divide the rise by the number of steps.
Counting terms: $n = \dfrac{L - a_1}{d} + 1$ . The division gives steps, and the $+1$ turns steps back into terms. This is the one formula that needs $d \neq 0$ , because it divides by $d$ .
The sum (arithmetic series): write it forwards and backwards and add columnwise. Every column totals $a_1 + a_n$ and there are $n$ of them, so $2S_n = n(a_1 + a_n)$ and $S_n = \dfrac{n}{2}(a_1 + a_n) = \dfrac{n}{2}\big(2a_1 + (n-1)d\big)$ .
The doubling works for odd and even $n$ alike, because a reversed copy gives every term a partner, including the middle one, which partners itself.
A sum is the number of terms times the average of the first and last term. The terms are linear in $n$ ; their running total $S_n$ is quadratic in $n$ with no constant term. That runs both ways: a sequence with $S_n = An^2 + Bn$ has $a_n = 2An + (B - A)$ , so it is arithmetic with $d = 2A$ .

Arithmetic Sequences and Series

Learning goals

What makes a sequence arithmetic

The nth term, by counting steps

An arithmetic sequence is a linear function of n

How many terms are there

Adding the terms, by writing the sum twice

The same formula, wearing different clothes

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)