# Formulas for arithmetic and geometric sequences (Part 1)

I’m not particularly a fan of memorizing formulas. Apparently, most college students aren’t fans either, because they often don’t have immediate recall of certain formulas from high school when they’re needed in the collegiate curriculum.

While I’m not a fan of making students memorize formulas, I am a fan of teaching students how to derive formulas. Speaking for myself, if I ever need to use a formula that I know exists but have long since forgotten, the ability to derive the formula allows me to get it again.

Which leads me to today’s post: the derivation of the formulas for the $n$th term of an arithmetic sequence and of a geometric sequence. This topic is commonly taught in Precalculus but, in my experience, is often forgotten by students years later when needed in later classes.

An arithmetic sequence is specified by two numbers: the first term and the common difference between terms. For example, if the first term is $16$ and the common difference is $3$, then the sequence begins as

$16, 19, 22, 25, 28, 31, 34, \dots$

If the first term is $29$ and the common difference is $-4$, then the sequence begins as

$29, 25, 21, 17, 13, 9, 5, 1, -3, \dots$

For those of us old enough to remember, our favorite arithmetic sequences came from Schoolhouse Rock:

Let’s discuss the first arithmetic sequence, whose first seven terms are:

$16, 19, 22, 25, 28, 31, 34, \dots$

How do we get the $8$th term? That’s easy: we just add $3$ to $34$ to get $37$.

How to we get the $100$th term. That’s easy: we just add $3$ to the $99$th term.

Oops. We don’t know the $99$th term. To get the $99th$ term, we need the $98$th term, which in turn requires the $97$th term. Et cetera, et cetera, et cetera.

The trouble (so far) is that an arithmetic sequence is recursively defined: to get one term, I add something to the previous term. Mathematically, the arithmetic sequence is defined by

$a_n = a_{n-1} + d$,

where $d$ is the common difference. This can be very intimidating to students when seeing it for the first time. So, to make this formula less intimidating, I usually read this equation as “Each next term in the sequence is equal to the previous term in the sequence plus the common difference.”

It would be far better to have a closed-form formula, where I could just plug in $100$ to get the $100$th term, without first figuring out the previous $99$ terms.

To this end, we notice the following pattern:

• Second term: $19 = 16 + 3$
• Third term: $22 = 19 + 3 = 16 + 3 + 3 = 16 + 2 \times 3$
• Fourth term: $25 = 22+ 3 = 16 + (2 \times 3) + 3 = 16 + 3 \times 3$
• Fifth term: $28 = 25+ 3 = 16 + (3 \times 3) + 3 = 16 + 4 \times 3$
• Sixth term: $31 = 28+ 3 = 16+ (4 \times 3) + 3 = 16 + 5 \times 3$
• Seventh term: $34 = 31 + 3 = 16 + (5 \times 3) + 3 = 16 + 6 \times 3$

It looks like we have a pattern, so we can guess that:

• One hundredth term = $16 + (100-1) \times 3 = 313$

In general, we have justified the closed-form formula

$a_n = a_1 + (n-1)d$,

where $a_1$ is the first term, and $d$ is the common difference.  In words: to get the $n$th term of an arithmetic sequence, we add $d$ to the first term $n-1$ times. (This may be formally proven using mathematical induction, though I won’t do so here.)

A closed-form formula for a geometric sequence is similarly obtained. In a geometric sequence, each term is equal to the previous term multiplied by a common ratio. Mathematically, the geometric sequence is recursively defined by

$a_n = a_{n-1}r$,

where $r$ is the common ratio. For example, if the first term is $3$ and the common ratio is $2$, then the first few terms of the sequence are

$3, 6, 12, 24, 48, dots$

By the same logic used above, to get the $n$th term of an geometric sequence, we multiply $r$ to the first term $n-1$ times. Thus justifies the formula

$a_n = a_1 r^{n-1}$,

which may be formally proven using mathematical induction.