Arithmetic Sequences and Series

What makes a sequence arithmetic

A sequence is a function whose domain is a run of consecutive 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

which is the same recurrence as $a_{n+1} = a_n + d$; paired with a seed $a_1$, that recurrence becomes a complete recursive rule for the sequence. The word common is doing real work here: the difference must be the same for every pair of neighbors, 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 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.

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

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

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.

An arithmetic sequence is a linear function of n

Expand the formula and something familiar appears:

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 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

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.

That formula also assumes $L$ really is a term, not just some number you are curious about. Take $6, 10, 14, \ldots$ and ask how many terms run up to $31$. The division gives $(31 - 6) \div 4 = 6.25$, and a fractional step count is the formula telling you $31$ is never landed on; it falls between two terms. A question like “how many terms up to the first one past $31$” is not asking for that division at all. It is asking where the sequence first clears $31$, so solve the inequality instead: $6 + 4(n - 1) > 31$ gives $n > 7.25$, so the first term past $31$ is the eighth one, $a_8 = 34$.

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 constant case.

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

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:

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 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

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

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

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:

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.

A sum written in sigma notation does not have to start at $k = 1$. For $\sum_{k=5}^{20} (3k - 2)$, the terms still form the same arithmetic sequence with $d = 3$, but the sum now starts partway through it: the first included term is at $k = 5$, not $k = 1$. Evaluate both ends directly from the rule, $3(5) - 2 = 13$ and $3(20) - 2 = 58$, then count terms the same way as always, between the two indices that bound the sum: from $5$ to $20$ is $15$ steps, so there are $15 + 1 = 16$ terms, and $S = \tfrac{16}{2}(13 + 58) = 8 \cdot 71 = 568$.

Common mistakes