Algebra 2 Quiz: Arithmetic And Geometric Sequences As Functions

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Arithmetic And Geometric Sequences As FunctionsQuestion 1 of 20

A student deposits $40 into a jar each week and does not remove any money. Let $a_n$ be the total amount in the jar after $n$ weeks, with $a_1=40$ . Write both recursive and explicit formulas (domain: positive integers).

Recursive:

a_1=40,\ a_{n+1}=40a_n

; Explicit:

a_n=40\cdot 40^{n-1}

Recursive:

a_1=0,\ a_{n+1}=a_n+40

; Explicit:

a_n=40n

Recursive:

a_1=40,\ a_{n+1}=a_n+40

; Explicit:

a_n=40+40(n-1)

Recursive:

a_1=40,\ a_{n+1}=a_n+4

; Explicit:

a_n=40+4(n-1)

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Algebra 2 Quiz

Algebra 2 Quiz: Arithmetic And Geometric Sequences As Functions

Practice Arithmetic And Geometric Sequences As Functions in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Arithmetic And Geometric Sequences As Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Recursive: $a_1=40,\ a_{n+1}=40a_n$ ; Explicit: $a_n=40\cdot 40^{n-1}$
Recursive: $a_1=0,\ a_{n+1}=a_n+40$ ; Explicit: $a_n=40n$
Recursive: $a_1=40,\ a_{n+1}=a_n+40$ ; Explicit: $a_n=40+40(n-1)$ (correct answer)
Recursive: $a_1=40,\ a_{n+1}=a_n+4$ ; Explicit: $a_n=40+4(n-1)$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. Geometric sequences have constant ratio r (multiply by same factor): recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the multiplying pattern, while explicit form aₙ = a₁·r^(n-1) gives direct calculation. Each form has advantages! For depositing $40 each week with a₁=40 after 1 week, it's arithmetic with d=40, so recursive a₁=40, aₙ₊₁=aₙ+40 and explicit aₙ=40+40(n-1). Choice C correctly writes both formulas for this arithmetic sequence. A distractor like choice A might confuse it with geometric growth, but since it's fixed additions, not multiplications, check for constant differences to confirm. Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 2

The sequence $5,\ 15,\ 45,\ 135,\dots$ is used to model the number of views a video gets each day, where day 1 has 5 views and each day has the same multiplicative change from the previous day. Write both a recursive and an explicit formula for the sequence (integer domain $n\ge 1$ ).

Recursive: $a_1=5,\ a_{n+1}=a_n+10$ ; Explicit: $a_n=5+10(n-1)$
Recursive: $a_1=15,\ a_{n+1}=3a_n$ ; Explicit: $a_n=15\cdot 3^{n-1}$
Recursive: $a_1=5,\ a_{n+1}=3a_n$ ; Explicit: $a_n=5\cdot 3^{n-1}$ (correct answer)
Recursive: $a_1=5,\ a_{n+1}=a_n+3$ ; Explicit: $a_n=5+3(n-1)$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 15, 45, 135: differences are 10, 30, 90 (not constant), but ratios are 15/5 = 3, 45/15 = 3, 135/45 = 3 (constant!), so this is geometric with r = 3. With first term a₁ = 5 and ratio r = 3, we get recursive: a₁ = 5, aₙ₊₁ = 3aₙ and explicit: aₙ = 5 · 3^(n-1). Choice C correctly identifies this as a geometric sequence and provides both formulas. Choice D incorrectly treats it as arithmetic with d = 3, which would give the sequence 5, 8, 11, 14,... instead of 5, 15, 45, 135,.... Always check both differences and ratios to determine sequence type. If ratios are constant, it's geometric; if differences are constant, it's arithmetic.

Question 3

A bookstore's reward points increase by 15 points each week. In week 1, a customer has 40 points. Model the total points after $n$ weeks as a function with integer domain $n \ge 1$ by writing both a recursive formula and an explicit formula for the sequence.

Recursive: $a_1=40,\ a_{n+1}=15a_n$ ; Explicit: $a_n=40\cdot 15^{(n-1)}$
Recursive: $a_1=40,\ a_{n+1}=a_n+15$ ; Explicit: $a_n=40+15(n-1)$ (correct answer)
Recursive: $a_1=55,\ a_{n+1}=a_n+15$ ; Explicit: $a_n=55+15(n-1)$
Recursive: $a_1=15,\ a_{n+1}=a_n+40$ ; Explicit: $a_n=15+40(n-1)$

Question 4

A town has a population of 50,000 and grows by 3% each year. Model the population after $n$ years with both a recursive and an explicit formula, where $a_1$ is the population after 1 year (domain: positive integers).

Recursive: $a_1=51{,}500,\ a_{n+1}=1.03a_n$ ; Explicit: $a_n=51{,}500\cdot(1.03)^{n-1}$ (correct answer)
Recursive: $a_1=50{,}000,\ a_{n+1}=1.03a_n$ ; Explicit: $a_n=50{,}000\cdot(1.03)^{n-1}$
Recursive: $a_1=50{,}000,\ a_{n+1}=a_n+1{,}500$ ; Explicit: $a_n=50{,}000+1{,}500(n-1)$
Recursive: $a_1=50{,}000,\ a_{n+1}=a_n+0.03$ ; Explicit: $a_n=50{,}000+0.03(n-1)$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. Geometric sequences have constant ratio r (multiply by same factor): recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the multiplying pattern, while explicit form aₙ = a₁·r^(n-1) gives direct calculation. Each form has advantages! For this population growth at 3% per year starting from 50,000, with a₁ after 1 year as 50,000×1.03=51,500, it's geometric with r=1.03, so recursive a₁=51,500, aₙ₊₁=1.03aₙ and explicit aₙ=51,500·(1.03)^{n-1}. Choice B correctly writes both formulas, aligning a₁ with the value after the first year. A distractor like choice C might use the initial population as a₁, but carefully read the problem to identify what a₁ represents—here it's after 1 year, so adjust accordingly. Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 5

A phone battery is at 90% at the start of the day and then retains 80% of its charge at the end of each hour (due to constant use). Let $a_n$ be the battery percentage after $n-1$ hours (so $a_1=90$ ), with domain $n\in\{1,2,3,\dots\}$ . Write both a recursive definition and an explicit formula.

Recursive: $a_1=90,\ a_{n+1}=a_n-20$ ; Explicit: $a_n=90-20(n-1)$
Recursive: $a_1=90,\ a_{n+1}=0.8a_n$ ; Explicit: $a_n=90\cdot(0.8)^{n-1}$ (correct answer)
Recursive: $a_1=90,\ a_{n+1}=a_n\cdot 90$ ; Explicit: $a_n=0.8\cdot 90^{n-1}$
Recursive: $a_1=72,\ a_{n+1}=0.8a_n$ ; Explicit: $a_n=72\cdot(0.8)^{n-1}$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. Geometric sequences have constant ratio r (multiply by same factor): recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the multiplying pattern, while explicit form aₙ = a₁·r^(n-1) gives direct calculation. Each form has advantages! For the battery model starting at 90% and retaining 80% (r=0.8) each hour, it's geometric with a₁=90, so recursive a₁=90, aₙ₊₁=0.8aₙ and explicit aₙ=90·(0.8)^{n-1}. Choice B correctly writes both formulas, modeling the exponential decay. Choices like A use arithmetic subtraction, but multiplication by 0.8 makes it geometric—fantastic application to real life! Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 6

A car worth $\$ 24{,}000 $depreciates by 10% each year. Let$ V_n $be the value after$ n $years, where$ n $is an integer and$ V_0$ is the initial value. Which choice correctly models this situation with both recursive and explicit formulas?

Recursive: $V_0=24000,\ V_{n+1}=V_n-0.10$ ; Explicit: $V_n=24000-0.10n$
Recursive: $V_0=24000,\ V_{n+1}=0.90V_n$ ; Explicit: $V_n=24000\cdot 0.90^{n}$ (correct answer)
Recursive: $V_1=24000,\ V_{n+1}=0.90V_n$ ; Explicit: $V_n=24000\cdot 0.90^{n-1}$
Recursive: $V_0=24000,\ V_{n+1}=1.10V_n$ ; Explicit: $V_n=24000\cdot 1.10^{n}$

Question 7

A car loses $12\%$ of its value each year. If its value is $18{,}000$ dollars at year 0, let $V(n)$ be its value after $n$ years (domain: nonnegative integers). Which choice gives both a correct recursive and explicit model for $V(n)$ ?

Recursive: $V(0)=18000,\ V(n+1)=V(n)-0.12$ ; Explicit: $V(n)=18000-0.12n$
Recursive: $V(0)=18000,\ V(n+1)=0.88V(n)$ ; Explicit: $V(n)=18000\cdot 0.88^{n}$ (correct answer)
Recursive: $V(1)=18000,\ V(n+1)=0.88V(n)$ ; Explicit: $V(n)=18000\cdot 0.88^{n-1}$
Recursive: $V(0)=18000,\ V(n+1)=1.12V(n)$ ; Explicit: $V(n)=18000\cdot 1.12^{n}$

Question 8

A video gets 300 views on day 1, and then gains 40 more views each day. Let $v(n)$ be the number of views on day $n$ , with domain $n \in \{1,2,3,\dots\}$ . Write both a recursive formula and an explicit formula for $v(n)$ .

Recursive: $v(1)=300,\ v(n+1)=1.4v(n)$ ; Explicit: $v(n)=300\cdot 1.4^{n-1}$
Recursive: $v(1)=340,\ v(n+1)=v(n)+40$ ; Explicit: $v(n)=340+40(n-1)$
Recursive: $v(1)=300,\ v(n+1)=v(n)+40$ ; Explicit: $v(n)=300+40(n-1)$ (correct answer)
Recursive: $v(1)=300,\ v(n+1)=v(n)-40$ ; Explicit: $v(n)=300-40(n-1)$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. For this video views problem: day 1 has 300 views (so v(1) = 300), and gains 40 more views each day (so d = 40), making this arithmetic with recursive v(1) = 300, v(n+1) = v(n) + 40 and explicit v(n) = 300 + 40(n - 1). Choice C correctly writes both formulas with v(1) = 300 as the starting views and d = 40 as the daily increase. Choice A incorrectly treats this as geometric (multiplying by 1.4), B has wrong starting value 340, and D subtracts instead of adds the daily views. Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1).

Question 9

A bacteria culture starts with 80 bacteria and triples every hour. Let $a_n$ be the number of bacteria after $n-1$ hours (so $a_1=80$ ), with domain $n\in\{1,2,3,\dots\}$ . Model this situation with both a recursive definition and an explicit formula.

Recursive: $a_1=80,\ a_{n+1}=a_n+3$ ; Explicit: $a_n=80+3(n-1)$
Recursive: $a_1=80,\ a_{n+1}=3a_n$ ; Explicit: $a_n=80\cdot 3^{n-1}$ (correct answer)
Recursive: $a_1=80,\ a_{n+1}=a_n\cdot 80$ ; Explicit: $a_n=3\cdot 80^{n-1}$
Recursive: $a_1=240,\ a_{n+1}=3a_n$ ; Explicit: $a_n=240\cdot 3^{n-1}$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. Geometric sequences have constant ratio r (multiply by same factor): recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the multiplying pattern, while explicit form aₙ = a₁·r^(n-1) gives direct calculation. Each form has advantages! In this bacteria growth model starting at 80 and tripling each hour, it's geometric with a₁=80 and r=3, so recursive a₁=80, aₙ₊₁=3aₙ and explicit aₙ=80·3^(n-1). Choice B correctly writes both formulas, capturing the multiplicative growth accurately. Choices like A model it as arithmetic addition, but since it triples (multiplies), it's geometric—great job recognizing real-world patterns like this! Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 10

A bacterial culture starts with 200 bacteria and triples every 4 hours. If $b_n$ represents the number of bacteria after $4n$ hours, which explicit formula represents this growth, and approximately how many bacteria will there be after 16 hours?

$b_n = 200 \cdot 4^n$ ; After 16 hours: approximately 51,200 bacteria
$b_n = 200 \cdot 3^{n-1}$ ; After 16 hours: approximately 5,400 bacteria
$b_n = 200 + 3n$ ; After 16 hours: approximately 212 bacteria
$b_n = 200 \cdot 3^n$ ; After 16 hours: approximately 16,200 bacteria (correct answer)

Explanation: When you encounter bacterial growth problems, you're dealing with exponential growth where the population multiplies by a constant factor at regular intervals. The key is identifying the growth factor and how time relates to the variable. Here, bacteria triple every 4 hours, starting with 200. Since

b_n

represents bacteria after

4n

hours, when

n = 1

, we have 4 hours elapsed; when

n = 2

, we have 8 hours elapsed, and so on. After each 4-hour period, the population multiplies by 3. The correct formula is

b_n = 200 \cdot 3^n

. At

n = 0

(start):

b_0 = 200 \cdot 3^0 = 200

. At

n = 1

(4 hours):

b_1 = 200 \cdot 3^1 = 600

. At

n = 4

(16 hours):

b_4 = 200 \cdot 3^4 = 200 \cdot 81 = 16,200

. Choice A uses base 4 instead of 3, confusing the time interval (4 hours) with the growth factor (tripling). Choice B uses

3^{n-1}

, which would give the wrong starting value—this shifts the entire sequence incorrectly. Choice C shows linear growth (

200 + 3n

), completely missing that this is exponential growth where populations multiply rather than add. Remember: in exponential growth problems, identify what multiplies (the base) versus what measures time intervals. The population multiplies by the growth factor, while time intervals just help you determine the exponent. Always verify your formula works at

n = 0

and

n = 1

Question 11

A theater charges $15 for admission and plans to increase the price by $2 each month. If $ $a_n$ $ represents the admission price in the nth month, which recursive formula correctly models this situation, and what will be the price in the 8th month?

$a_1 = 15, a_n = a_{n-1} + 2$ for $n \geq 2$ ; Price in 8th month: $29 (correct answer)
$a_1 = 17, a_n = a_{n-1} + 2$ for $n \geq 2$ ; Price in 8th month: $31
$a_0 = 15, a_n = a_{n-1} + 2$ for $n \geq 1$ ; Price in 8th month: $31
$a_1 = 15, a_n = a_{n-1} + 15$ for $n \geq 2$ ; Price in 8th month: $120

Explanation: The initial price is $15, so $

a_1 = 15

. Each month the price increases by

2, so

a_n = a_{n-1} + 2

. For the 8th month:

a_8 = 15 + 7(2) = 29

. Choice B incorrectly starts at

17. Choice C uses

a_0

which would make the 8th month

a_8 = 15 + 8(2) = 31

$. Choice D incorrectly adds $15 each month instead of $2.

Question 12

A sequence follows the pattern: 5, 20, 80, 320, ... . If this pattern continues, what are both the recursive and explicit formulas for the nth term, where $a_1 = 5$ ?

Recursive: $a_n = a_{n-1} + 15$ ; Explicit: $a_n = 5 + 15(n-1)$
Recursive: $a_n = 4a_{n-1}$ ; Explicit: $a_n = 5 \cdot 4^{n-1}$ (correct answer)
Recursive: $a_n = 4a_{n-1} + 5$ ; Explicit: $a_n = 5 \cdot 4^n$
Recursive: $a_n = 4a_{n-1}$ ; Explicit: $a_n = 5 + 4^{n-1}$

Explanation: When you encounter a sequence problem, your first step is identifying the pattern. Look at how each term relates to the previous one to determine if it's arithmetic (constant difference) or geometric (constant ratio). Let's examine the given sequence: 5, 20, 80, 320, ... Check the ratios between consecutive terms:

$\frac{20}{5} = 4$
$\frac{80}{20} = 4$
$\frac{320}{80} = 4$

Since each term is 4 times the previous term, this is a geometric sequence with common ratio

r = 4

. For geometric sequences, the recursive formula is

a_n = r \cdot a_{n-1}

, so here it's

a_n = 4a_{n-1}

. The explicit formula is

a_n = a_1 \cdot r^{n-1}

, which gives us

a_n = 5 \cdot 4^{n-1}

. This confirms answer choice B is correct. Let's check why the other options fail: Choice A represents an arithmetic sequence (adding 15 each time), but

5 + 15 = 20

, then

20 + 15 = 35 ≠ 80

. Choice C has the wrong explicit formula:

a_n = 5 \cdot 4^n

would give

a_1 = 5 \cdot 4^1 = 20

, not 5. Choice D combines the correct recursive formula with an incorrect explicit formula that represents addition rather than multiplication:

a_1 = 5 + 4^0 = 6 ≠ 5

. Study tip: Always verify your formulas by plugging in the first few values of

n

to ensure they match the given terms. For geometric sequences, remember the explicit formula uses

r^{n-1}

, not

r^n

Question 13

Given the recursive formula $a_1 = 7$ and $a_n = 2a_{n-1} - 3$ for $n \geq 2$ , what type of sequence is this, and what is the explicit formula?

Geometric sequence; $a_n = 7 \cdot 2^{n-1} - 3(n-1)$
Arithmetic sequence; $a_n = 7 + (n-1)(4)$
Neither arithmetic nor geometric; $a_n = 4 \cdot 2^{n-1} + 3$ (correct answer)
Geometric sequence; $a_n = 4 + 3 \cdot 2^{n-1}$

Explanation: Computing terms:

a_1 = 7

a_2 = 11

a_3 = 19

a_4 = 35

. Differences: 4, 8, 16 (not constant, so not arithmetic). Ratios: 11/7, 19/11, 35/19 (not constant, so not geometric). For

a_n = 2a_{n-1} - 3

, let

a_n = A \cdot 2^{n-1} + B

. Substituting:

A \cdot 2^{n-1} + B = 2(A \cdot 2^{n-2} + B) - 3 = A \cdot 2^{n-1} + 2B - 3

. So

B = 2B - 3

, giving

B = 3

. With

a_1 = 7

A + 3 = 7

, so

A = 4

. Thus

a_n = 4 \cdot 2^{n-1} + 3

Question 14

Translate the explicit formula $a_n=3\cdot 2^{n-1}$ into an equivalent recursive definition (domain: positive integers).

$a_1=3,\ a_{n+1}=a_n+2$
$a_1=6,\ a_{n+1}=2a_n$
$a_1=3,\ a_{n+1}=2a_n$ (correct answer)
$a_1=3,\ a_{n+1}=a_n\cdot 2^{n-1}$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Translating between forms: if you have recursive (like a₁ = 3, aₙ₊₁ = aₙ + 5), extract a₁ = 3 and d = 5 (the amount added), then write explicit aₙ = 3 + 5(n - 1). If you have explicit (like aₙ = 2·3^(n-1)), extract a₁ = 2 (when n = 1) and r = 3 (the base), then write recursive a₁ = 2, aₙ₊₁ = 3aₙ. The parameters connect the two forms! For the explicit aₙ=3·2^{n-1}, evaluate at n=1 to get a₁=3, and recognize r=2 from the base, so recursive is a₁=3, aₙ₊₁=2aₙ. Choice C correctly translates to the recursive definition for this geometric sequence. A distractor like choice A might swap a₁, but always verify by computing the first few terms from the explicit formula to match. Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 15

A car worth $\$ 18{,}000 $depreciates by 10% each year. Let$ a_n $be the value after$ n $years, with$ a_1$ the value after 1 year. Write both recursive and explicit formulas (domain: positive integers).

Recursive: $a_1=18{,}000,\ a_{n+1}=0.9a_n$ ; Explicit: $a_n=18{,}000\cdot(0.9)^{n-1}$
Recursive: $a_1=16{,}200,\ a_{n+1}=0.9a_n$ ; Explicit: $a_n=16{,}200\cdot(0.9)^{n-1}$ (correct answer)
Recursive: $a_1=16{,}200,\ a_{n+1}=a_n-1{,}800$ ; Explicit: $a_n=18{,}000-1{,}800(n-1)$
Recursive: $a_1=18{,}000,\ a_{n+1}=1.1a_n$ ; Explicit: $a_n=18{,}000\cdot(1.1)^{n-1}$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. Geometric sequences have constant ratio r (multiply by same factor): recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the multiplying pattern, while explicit form aₙ = a₁·r^(n-1) gives direct calculation. Each form has advantages! For this car depreciating 10% per year from $18,000, with a₁ after 1 year as 18,000×0.9=16,200, it's geometric with r=0.9, so recursive a₁=16,200, aₙ₊₁=0.9aₙ and explicit aₙ=16,200·(0.9)^{n-1}. Choice B correctly writes both formulas, ensuring a₁ matches the value after the first year. A distractor like choice A might start with the initial value as a₁, but pay close attention to the problem's definition of a₁ to set it correctly. Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 16

Write both recursive and explicit formulas for the sequence $5,\ 15,\ 45,\ 135,\ \dots$ (domain: positive integers).

Recursive: $a_1=15,\ a_{n+1}=3a_n$ ; Explicit: $a_n=15\cdot 3^{n-1}$
Recursive: $a_1=5,\ a_{n+1}=a_n+10$ ; Explicit: $a_n=5+10(n-1)$
Recursive: $a_1=5,\ a_{n+1}=3a_n$ ; Explicit: $a_n=5\cdot 3^{n-1}$ (correct answer)
Recursive: $a_1=5,\ a_{n+1}=a_n\cdot n$ ; Explicit: $a_n=5n$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. Geometric sequences have constant ratio r (multiply by same factor): recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the multiplying pattern, while explicit form aₙ = a₁·r^(n-1) gives direct calculation. Each form has advantages! For the sequence 5, 15, 45, 135, ..., the ratios are 3 each time, so it's geometric with a₁=5 and r=3, leading to recursive a₁=5, aₙ₊₁=3aₙ and explicit aₙ=5·3^{n-1}. Choice B correctly writes both formulas for this geometric sequence. A distractor like choice A might mistake it for arithmetic, but since ratios are constant and differences increase, it's geometric—great job checking both! Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 17

Given the explicit formula $a_n=9+4(n-1)$ , write an equivalent recursive definition (domain: positive integers).

$a_1=9,\ a_{n+1}=a_n-4$
$a_1=9,\ a_{n+1}=4a_n$
$a_1=9,\ a_{n+1}=a_n+4$ (correct answer)
$a_1=13,\ a_{n+1}=a_n+4$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Translating between forms: if you have recursive (like a₁ = 3, aₙ₊₁ = aₙ + 5), extract a₁ = 3 and d = 5 (the amount added), then write explicit aₙ = 3 + 5(n - 1). If you have explicit (like aₙ = 2·3^(n-1)), extract a₁ = 2 (when n = 1) and r = 3 (the base), then write recursive a₁ = 2, aₙ₊₁ = 3aₙ. The parameters connect the two forms! For the explicit aₙ=9+4(n-1), simplify to see it's arithmetic: a₁=9 (at n=1), and d=4 (coefficient of n term), so recursive is a₁=9, aₙ₊₁=aₙ+4. Choice C correctly translates to the recursive definition for this arithmetic sequence. A distractor like choice A might misidentify it as geometric, but check if differences are constant (yes) versus ratios (no) to confirm the type. Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 18

A student deposits $25 into a jar every week and does not remove any money. Let $a_n$ be the total amount of money in the jar after $n$ weeks, with $a_0=0$ and domain $n\in\{0,1,2,\dots\}$ . Write both a recursive definition and an explicit formula for $a_n$ .

Recursive: $a_0=0,\ a_{n+1}=25a_n$ ; Explicit: $a_n=25^n$
Recursive: $a_0=25,\ a_{n+1}=a_n+25$ ; Explicit: $a_n=25+25n$
Recursive: $a_0=0,\ a_{n+1}=a_n+25$ ; Explicit: $a_n=25n$ (correct answer)
Recursive: $a_0=0,\ a_{n+1}=a_n-25$ ; Explicit: $a_n=-25n$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Arithmetic sequences have constant difference d (add same amount each step): recursive form a₁ = [value], aₙ₊₁ = aₙ + d shows the stepping pattern, while explicit form aₙ = a₁ + (n - 1)d lets you jump directly to any term. Geometric sequences have constant ratio r (multiply by same factor): recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the multiplying pattern, while explicit form aₙ = a₁·r^(n-1) gives direct calculation. Each form has advantages! In this savings model with $25 added weekly from a₀=0, it's arithmetic with d=25, so recursive a₀=0, aₙ₊₁=aₙ+25 and explicit aₙ=25n (note the indexing starts at 0). Choice C correctly writes both formulas, fitting a₁=25, a₂=50, etc. Choices like A model geometric multiplication, but constant addition makes it arithmetic—keep up the excellent modeling! Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 19

Translate the recursive definition to an explicit formula (with integer domain $n\ge 1$ ):

Recursive: $a_1=6,\ a_{n+1}=3a_n$ .

Which explicit formula matches this sequence?

$a_n=6\cdot 3^{n-1}$ (correct answer)
$a_n=6+3(n-1)$
$a_n=3\cdot 6^{n-1}$
$a_n=6\cdot 3^{n}$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Translating between forms: if you have recursive (like a₁ = 3, aₙ₊₁ = aₙ + 5), extract a₁ = 3 and d = 5 (the amount added), then write explicit aₙ = 3 + 5(n - 1). If you have recursive (like a₁ = 2, aₙ₊₁ = 3aₙ), extract a₁ = 2 (when n = 1) and r = 3 (the base), then write explicit aₙ = 2·3^(n-1). From the recursive formula a₁ = 6, aₙ₊₁ = 3aₙ, we identify this as geometric with first term a₁ = 6 and ratio r = 3, so the explicit formula is aₙ = 6 · 3^(n-1). Choice A correctly translates the recursive definition to explicit form. Choice D incorrectly uses exponent n instead of n-1, which would give a₁ = 6 · 3¹ = 18 instead of 6. When translating geometric sequences from recursive to explicit, remember that the exponent is (n-1) when the sequence starts at a₁. This ensures that when n = 1, we get a₁ = a₁ · r⁰ = a₁ · 1 = a₁.

Question 20

A bacteria culture starts with 80 bacteria and increases by 10% each hour. Let $a_n$ be the number of bacteria after $n$ hours, where $n$ is a nonnegative integer and $a_0$ is the initial amount. Which choice gives both a recursive and an explicit model?

Recursive: $a_1=80,\ a_{n+1}=1.10a_n$ ; Explicit: $a_n=80\cdot 1.10^{n}$
Recursive: $a_0=80,\ a_{n+1}=1.10a_n$ ; Explicit: $a_n=80\cdot 1.10^{n}$ (correct answer)
Recursive: $a_0=80,\ a_{n+1}=a_n+10$ ; Explicit: $a_n=80+10n$
Recursive: $a_0=80,\ a_{n+1}=0.90a_n$ ; Explicit: $a_n=80\cdot 0.90^{n}$