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How do you write and use explicit and recursive formulas for arithmetic and geometric sequences, and connect them to linear and exponential functions?

Write and evaluate explicit and recursive formulas for arithmetic and geometric sequences, and relate arithmetic sequences to linear functions and geometric sequences to exponential functions (MA.912.AR.5 and MA.912.F).

A B.E.S.T. Algebra 1 EOC answer on sequences, the explicit and recursive formulas on the reference sheet, finding the common difference or ratio, and linking arithmetic to linear and geometric to exponential growth.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. Telling them apart
  3. The explicit formulas (on the reference sheet)
  4. The recursive formulas
  5. How the B.E.S.T. EOC examines this topic
  6. Why arithmetic is linear and geometric is exponential
  7. Try this

What this topic is asking

A sequence is an ordered list of numbers. On the B.E.S.T. Algebra 1 EOC you identify whether a sequence is arithmetic (constant difference) or geometric (constant ratio), write its explicit and recursive formula, and connect it to a function: arithmetic sequences are linear and geometric sequences are exponential. Both nnth-term formulas are on the reference sheet, so the credit is for using them correctly.

Telling them apart

Check consecutive terms. If you add the same number each time, it is arithmetic (find dd by subtracting any term from the next). If you multiply by the same number each time, it is geometric (find rr by dividing any term by the previous one).

  • 3,7,11,15,…3, 7, 11, 15, \dots is arithmetic with d=4d = 4.
  • 3,6,12,24,…3, 6, 12, 24, \dots is geometric with r=2r = 2.

The explicit formulas (on the reference sheet)

The explicit formula gives any term directly from nn, without listing the ones before it. The key detail is the nβˆ’1n - 1: the first term has had the difference or ratio applied zero times, so the multiplier or exponent is one less than the term number.

The recursive formulas

A recursive formula defines each term from the previous one and requires a starting value:

  • Arithmetic: a1=a_1 = (given), an=anβˆ’1+da_n = a_{n-1} + d.
  • Geometric: a1=a_1 = (given), an=anβˆ’1β‹…ra_n = a_{n-1} \cdot r.

Recursive form is handy for generating the next term quickly, but to jump straight to the 50th term you want the explicit formula.

How the B.E.S.T. EOC examines this topic

  • Equation editor. Write the explicit or recursive formula for a given sequence.
  • Multiple choice. Find a specific term, with off-by-one (nn vs nβˆ’1n - 1) distractors.
  • Editing task choice or matching. Match a sequence to its formula or to linear vs exponential.

A clarifying idea: the simplified explicit arithmetic formula an=3n+2a_n = 3n + 2 above is a linear function of nn (slope 33, intercept 22). That is the deep link the standard wants, an arithmetic sequence is just a line evaluated at the whole numbers.

Why arithmetic is linear and geometric is exponential

The connection to functions is exact, not loose. In an arithmetic sequence you add dd each step, so the term grows by the same amount per unit increase in nn, which is precisely the definition of constant slope, a linear function an=a1+d(nβˆ’1)a_n = a_1 + d(n - 1). In a geometric sequence you multiply by rr each step, so the term is scaled by the same factor per unit increase in nn, which is the defining behavior of an exponential function an=a1β‹…rnβˆ’1a_n = a_1 \cdot r^{n-1}. This is why the EOC can ask the same idea two ways: a table that grows by adding is linear, a table that grows by multiplying is exponential, and the sequence formulas are those two functions restricted to the counting numbers 1,2,3,…1, 2, 3, \dots.

Try this

Q1. Find a8a_8 for the geometric sequence with a1=2a_1 = 2 and r=3r = 3. [2 points]

  • Cue. a8=2β‹…37=2β‹…2187=4374a_8 = 2 \cdot 3^{7} = 2 \cdot 2187 = 4374.

Q2. Write the recursive formula for 7,4,1,βˆ’2,…7, 4, 1, -2, \dots. [1 point]

  • Cue. a1=7a_1 = 7, an=anβˆ’1βˆ’3a_n = a_{n-1} - 3.

Exam-style practice questions

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

B.E.S.T. (style)1 marksMultiple choice. An arithmetic sequence has first term a1=6a_1 = 6 and common difference d=4d = 4. What is the value of a10a_{10}? (A) 4040 (B) 4242 (C) 4646 (D) 3636
Show worked answer β†’

The correct answer is (B).

The reference sheet gives an=a1+(nβˆ’1)da_n = a_1 + (n - 1)d. With a1=6a_1 = 6, d=4d = 4, n=10n = 10: a10=6+(10βˆ’1)(4)=6+36=42a_{10} = 6 + (10 - 1)(4) = 6 + 36 = 42. The classic error is using nn instead of nβˆ’1n - 1, giving 6+40=466 + 40 = 46 (choice C); the multiplier is one less than the term number because the first term has no difference added yet.

B.E.S.T. (style)2 marksEquation editor. A geometric sequence has first term a1=3a_1 = 3 and common ratio r=2r = 2. Write the explicit formula for the nnth term.
Show worked answer β†’

Enter an=3β‹…2nβˆ’1a_n = 3 \cdot 2^{n-1} (or an=3(2)nβˆ’1a_n = 3(2)^{n-1}).

The reference sheet gives the geometric formula an=a1β‹…r nβˆ’1a_n = a_1 \cdot r^{\,n-1}. With a1=3a_1 = 3 and r=2r = 2, that is an=3β‹…2nβˆ’1a_n = 3 \cdot 2^{n-1}. The exponent is nβˆ’1n - 1 so the first term comes out right: a1=3β‹…20=3a_1 = 3 \cdot 2^0 = 3. Writing 3β‹…2n3 \cdot 2^n gives 66 for the first term, which is wrong.

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