Identify terms of arithmetic and geometric sequences when given in recursive form, and write a formula for the $n$th term of arithmetic and geometric sequences given several of their terms (TEKS A.12C, A.12D).

What this topic is asking

Sequences sit in the Number and Algebraic Methods category. TEKS A.12C asks you to read terms of arithmetic and geometric sequences given in recursive form, and A.12D asks you to write a formula for the $n$th term. The key fact for STAAR is that the sequence formulas are not on the reference sheet, so you must memorize them. Sequences also connect forward: arithmetic sequences are linear (constant difference) and geometric sequences are exponential (constant ratio).

Arithmetic sequences: a common difference

In an arithmetic sequence, consecutive terms differ by the same amount, the common difference $d = a_n - a_{n-1}$. The sequence $3, 7, 11, 15, \dots$ has $d = 4$.

• Recursive form. $a_1 = 3$ and $a_n = a_{n-1} + 4$. Each term is the previous one plus $d$.
• Explicit form. $a_n = a_1 + (n - 1)d = 3 + (n - 1)(4)$. This gives any term directly; $a_{10} = 3 + 9(4) = 39$.

Because the change is a constant addition, an arithmetic sequence is a linear pattern, and $d$ plays the role of slope.

Geometric sequences: a common ratio

In a geometric sequence, consecutive terms have the same ratio, the common ratio $r = \frac{a_n}{a_{n-1}}$. The sequence $2, 6, 18, 54, \dots$ has $r = 3$.

• Recursive form. $a_1 = 2$ and $a_n = a_{n-1} \cdot 3$.
• Explicit form. $a_n = a_1 \cdot r^{n-1} = 2 \cdot 3^{n-1}$. Then $a_5 = 2 \cdot 3^4 = 162$.

Because the change is a constant multiplication, a geometric sequence is an exponential pattern, and $r$ plays the role of the base.

Finding the rule from terms

Given several terms, decide the type first, then extract $d$ or $r$.

How STAAR examines sequences

• Multiple choice. Find a specific term from a recursive rule (watch the off-by-one), or identify the correct $n$th-term formula.
• Equation editor. Build the explicit formula such as $a_n = 5 \cdot 3^{n-1}$. Exact-match scoring means the $n - 1$ must be correct.
• Inline choice. Classify a sequence as arithmetic or geometric from a dropdown.

A clarifying idea is that "$n - 1$" appears because the first term has not yet been changed: at $n = 1$, $a_1 = a_1 + (1 - 1)d = a_1$ and $a_1 = a_1 \cdot r^{1-1} = a_1$. That is the test of whether your formula starts in the right place.

Recursive versus explicit, and which to use

A recursive rule has two parts: a starting value and a step that depends on the previous term, such as $a_1 = 7$ with $a_n = a_{n-1} + 4$. It is efficient for finding the next term, but slow for a far-off term because you must build every term in between. An explicit rule, such as $a_n = 7 + (n - 1)(4)$, computes any term directly from its position $n$, which is why STAAR favors it when a question asks for the 20th or 50th term. Being able to translate between the two forms is itself an assessed skill: read off the start and the constant change from the recursive rule, then drop them into the matching $n$th-term formula. The constant change is $d$ for an arithmetic sequence (added) and $r$ for a geometric sequence (multiplied), and spotting which one is constant, the difference or the ratio, is the decision that selects the formula.

Try this

Q1. An arithmetic sequence has $a_1 = 10$ and $d = -3$. Find $a_6$. [1 point]

• Cue. $a_6 = 10 + (6 - 1)(-3) = 10 - 15 = -5$.

Q2. Write the $n$th term of $5, 10, 20, 40, \dots$. [2 points]

• Cue. Geometric, $r = 2$: $a_n = 5 \cdot 2^{n-1}$.