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How do arithmetic and geometric sequences work, and how do you find a term using the reference-sheet formulas?

Recognize arithmetic and geometric sequences and find a term using the explicit formulas, connecting them to linear and exponential functions (LA A1: F-BF.A.2, F-IF.A.3).

A Louisiana LEAP 2025 Algebra I answer on sequences (LA A1: F-BF.A.2): common difference and common ratio, the explicit term formulas from the reference sheet, and the link to linear and exponential functions.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Arithmetic sequences
  3. Geometric sequences
  4. Telling them apart
  5. How LEAP examines this topic
  6. Why sequences are functions of their position
  7. Try this

What this topic is asking

Standards A1: F-BF.A.2 and F-IF.A.3 ask you to recognize arithmetic and geometric sequences and find a term using the explicit formulas. On LEAP 2025 these are Type I items, and the sequence formulas are on the reference sheet, so the credit is for identifying the type, finding the difference or ratio, and substituting correctly.

Arithmetic sequences

In an arithmetic sequence each term is the previous term plus a fixed number, the common difference dd. Find dd by subtracting any term from the next.

Geometric sequences

In a geometric sequence each term is the previous term times a fixed number, the common ratio rr. Find rr by dividing any term by the previous one.

Telling them apart

Check the consecutive terms:

  • Constant difference (you subtract to a fixed number): arithmetic.
  • Constant ratio (you divide to a fixed number): geometric.

For 4,7,10,134, 7, 10, 13: differences are all 33, arithmetic. For 3,6,12,243, 6, 12, 24: ratios are all 22, geometric.

How LEAP examines this topic

  • Equation response. Find a specified term using the reference-sheet formula.
  • Multiple choice. Identify the sequence type, the common difference or ratio, or a missing term.
  • Drag and drop. Match sequences to their explicit rules.

A clarifying idea: the formulas use nβˆ’1n - 1 because the first term (n=1n = 1) has had no step applied: a1=a1+0β‹…da_1 = a_1 + 0 \cdot d and a1=a1β‹…r0a_1 = a_1 \cdot r^0. The first step happens between term 1 and term 2.

Why sequences are functions of their position

A sequence is really a function whose input is the term number, which is the conceptual link F-IF.A.3 draws and the reason sequences sit in the functions module. Writing ana_n is the same as writing f(n)f(n): you feed in a position nn (a positive integer) and the formula returns the term at that position. Seen this way, an arithmetic sequence is a linear function in disguise, an=a1+(nβˆ’1)da_n = a_1 + (n - 1)d has a constant rate of change dd per step, exactly like slope, so its terms fall on a straight line. A geometric sequence is an exponential function in disguise, an=a1r nβˆ’1a_n = a_1 r^{\,n-1} multiplies by the constant factor rr per step, so its terms grow or decay exponentially. This is why the "constant difference versus constant ratio" test for sequences mirrors the "common difference versus common ratio" test for linear versus exponential functions, they are the same idea applied to integer inputs. Recognizing the connection lets you carry everything you know about lines and exponentials over to sequences, and it explains why the reference sheet groups the sequence formulas near the other growth formulas.

Try this

Q1. Find the 10th term of the arithmetic sequence 2,5,8,…2, 5, 8, \ldots. [2 points]

  • Cue. a10=2+(10βˆ’1)(3)=2+27=29a_{10} = 2 + (10 - 1)(3) = 2 + 27 = 29.

Q2. Find the 4th term of the geometric sequence 5,10,20,…5, 10, 20, \ldots. [2 points]

  • Cue. a4=5β‹…24βˆ’1=5β‹…8=40a_4 = 5 \cdot 2^{4-1} = 5 \cdot 8 = 40.

Exam-style practice questions

Practice questions written in the style of LDOE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

LA LEAP 2025 Math (style)2 marksEquation response. An arithmetic sequence is 4,7,10,13,…4, 7, 10, 13, \ldots. Find the 20th term.
Show worked answer β†’

The 20th term is 6161.

The sequence has first term a1=4a_1 = 4 and common difference d=3d = 3 (each term adds 33). The reference sheet gives an=a1+(nβˆ’1)da_n = a_1 + (n - 1)d. So a20=4+(20βˆ’1)(3)=4+57=61a_{20} = 4 + (20 - 1)(3) = 4 + 57 = 61. The common slip is using nn instead of nβˆ’1n - 1; the formula multiplies the difference by nβˆ’1n - 1 because the first term has no difference added yet.

LA LEAP 2025 Math (style)2 marksEquation response. A geometric sequence is 3,6,12,24,…3, 6, 12, 24, \ldots. Find the 6th term.
Show worked answer β†’

The 6th term is 9696.

The first term is a1=3a_1 = 3 and the common ratio is r=2r = 2 (each term is multiplied by 22). The reference sheet gives an=a1r nβˆ’1a_n = a_1 r^{\,n-1}. So a6=3β‹…26βˆ’1=3β‹…25=3β‹…32=96a_6 = 3 \cdot 2^{6-1} = 3 \cdot 2^5 = 3 \cdot 32 = 96. Again the exponent is nβˆ’1n - 1, not nn, because the first term is multiplied by r0=1r^0 = 1.

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