Distinguish arithmetic from geometric sequences and use the explicit nth-term formulas to find terms and relate sequences to linear and exponential functions (A.F.2).

What this topic is asking

This part of A.F.2 asks you to tell arithmetic from geometric sequences and use the explicit nth-term formulas to find terms, connecting sequences to linear and exponential functions. On the Virginia Algebra I SOL these are items where the sequence formulas (which are on the formula sheet) get applied: find a specific term, identify the type, or write a rule. They appear as fill-in-the-blank and multiple choice.

Arithmetic sequences

An arithmetic sequence has a common difference $d$: you add the same number to get from one term to the next. For $7, 11, 15, 19, \dots$, the common difference is $d = 4$. The explicit formula for the nth term is

where $a_1$ is the first term. The $(n - 1)$ matters: to reach the $n$th term you take $(n - 1)$ steps from the first.

Geometric sequences

A geometric sequence has a common ratio $r$: you multiply by the same number each step. For $3, 6, 12, 24, \dots$, the common ratio is $r = 2$. The explicit formula is

Again the exponent is $(n - 1)$, the number of multiplications needed to reach the $n$th term from the first.

Telling them apart

Check what gets you from term to term:

• Constant difference (you add the same number): arithmetic.
• Constant ratio (you multiply by the same number): geometric.

For $3, 6, 9, 12$ the difference is a constant $3$ (arithmetic). For $3, 6, 12, 24$ the ratio is a constant $2$ (geometric). If neither is constant, it is neither type.

The link to linear and exponential functions

Sequences are functions whose inputs are the term numbers $1, 2, 3, \dots$:

• An arithmetic sequence is a linear function in disguise. The common difference $d$ plays the role of the slope, and $a_n = a_1 + (n - 1)d$ rearranges to a line in $n$.
• A geometric sequence is an exponential function in disguise. The common ratio $r$ plays the role of the base, and $a_n = a_1 \cdot r^{\,n-1}$ has the same multiply-each-step structure as $ab^x$.

Why the exponent and multiplier are (n - 1)

The $(n - 1)$ in both formulas trips students up, but it is exactly right because the first term needs no steps. The first term $a_1$ is where you start, so reaching it requires zero additions or multiplications: plugging $n = 1$ gives $a_1 + (1 - 1)d = a_1$ and $a_1 \cdot r^{1-1} = a_1 \cdot r^0 = a_1$, the first term, as it must. The second term takes one step from the first, the third takes two, and the $n$th takes $(n - 1)$. So the difference is applied $(n - 1)$ times and the ratio is raised to the power $(n - 1)$. If the formula used $n$ instead of $(n - 1)$, it would overshoot by one full step, the most common error on these items. Counting "how many steps from the first term?" rather than "what is the term number?" keeps the index straight.

How the SOL examines this topic

• Fill-in-the-blank. Use the formula to find a specific term and type it.
• Multiple choice. Identify a sequence as arithmetic or geometric, or pick a missing term.
• Table items. Extend a sequence or match it to its rule.

Try this

Q1. Find the $10$th term of $2, 5, 8, 11, \dots$. [2 points]

• Cue. $a_1 = 2$, $d = 3$: $a_{10} = 2 + 9(3) = 29$.

Q2. Is $4, 12, 36, 108, \dots$ arithmetic or geometric? [1 point]

• Cue. Constant ratio $3$, so geometric.