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Algebra Help: Arithmetic And Geometric Sequences As Functions

Review real example questions for Arithmetic And Geometric Sequences As Functions in Algebra.

Question 1 / 10

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Translate the recursive definition $a_1=7$ and $a_{n+1}=a_n-3$ into an explicit formula.

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Question 1

Translate the recursive definition $a_1=7$ and $a_{n+1}=a_n-3$ into an explicit formula.

$a_n = 7\cdot (-3)^{n-1}$
$a_n = 7 - 3(n-1)$ (correct answer)
$a_n = 7 - 3n$
$a_n = 7 + 3(n-1)$

Explanation: This question tests your understanding of arithmetic and geometric sequences and how to write them both recursively (each term from the previous) and explicitly (any term directly from its position). The difference between recursive and explicit formulas: recursive is like climbing stairs one at a time (you need to know the previous term), while explicit is like taking an elevator directly to any floor (you can find the nth term without finding all the ones before it). Both describe the same sequence, just different approaches! Given the recursive definition a₁ = 7 and aₙ₊₁ = aₙ - 3, we identify: the first term is a₁ = 7, and since we're subtracting 3 each time, this is arithmetic with d = -3. To convert to explicit form, we use aₙ = a₁ + (n-1)d = 7 + (n-1)(-3) = 7 - 3(n-1). Choice B is correct because it properly identifies the first term as a₁ = 7 and the common difference as d = -3, giving the explicit formula aₙ = 7 - 3(n-1). Great work! Choice C has the right idea but makes an error in the explicit part: it uses 7 - 3n instead of 7 - 3(n-1). Remember: arithmetic sequences use (n-1) in the explicit formula because the first term already includes one application! To convert between forms: from recursive to explicit, identify a₁ and d (or r), then plug into the explicit formula. From explicit to recursive, read off a₁ and find d (coefficient of n after simplifying) or r (the base of the exponent). For recursive formulas, you always need two pieces: (1) the initial value (like a₁ = 7), and (2) the rule for getting the next term (like aₙ₊₁ = aₙ - 3 for arithmetic). Without both pieces, the sequence isn't fully defined!

Question 2

What is $a_6$ for the geometric sequence with explicit formula $a_n = 3\cdot 2^{n-1}$ ?

$48$
$96$ (correct answer)
$64$
$32$

Explanation: This question tests your understanding of arithmetic and geometric sequences and how to write them both recursively (each term from the previous) and explicitly (any term directly from its position). Geometric sequences have a constant ratio between consecutive terms: you multiply by the same factor each time. The explicit formula aₙ = a₁ · r^(n-1) uses the first term a₁ and common ratio r, while the recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the 'multiply to get next term' pattern. Given the explicit a_n = 3·2^{n-1}, this is geometric with a₁=3 and r=2; to find a6, plug in n=6: 3·2^{5} = 3·32 = 96. The recursive would be a₁=3, a_{n+1}=2·a_n, and computing terms: a1=3, a2=6, a3=12, a4=24, a5=48, a6=96 confirms. This captures 'start at 3, multiply by 2 each time.' Choice B is correct because it properly computes a6=96 using the formula or by pattern. Great work! Choice A has the right idea but gets the exponent wrong: it calculates 2^{4}=16 instead of 2^{5}=32, perhaps forgetting n-1 for a6 means exponent 5; check by listing a few terms! Writing explicit formulas: for arithmetic, use aₙ = a₁ + (n - 1)d (start with first term, add the difference (n-1) times); for geometric, use aₙ = a₁·r^(n-1) (start with first term, multiply by ratio (n-1) times). The (n-1) appears because the first term already includes one application! For recursive formulas, you always need two pieces: (1) the initial value (like a₁ = 5), and (2) the rule for getting the next term (like aₙ₊₁ = aₙ + 3 for arithmetic, or aₙ₊₁ = 2·aₙ for geometric). Without both pieces, the sequence isn't fully defined!

Question 3

A bacteria culture starts with 200 bacteria and triples every hour. Write both a recursive formula and an explicit formula for the number of bacteria after $n$ hours, where $a_1$ is the amount after 1 hour.

Recursive: $a_1=200,\ a_{n+1}=3a_n$ ; Explicit: $a_n=200\cdot 3^{n-1}$ (correct answer)
Recursive: $a_1=200,\ a_{n+1}=a_n+200$ ; Explicit: $a_n=200n$
Recursive: $a_1=600,\ a_{n+1}=3a_n$ ; Explicit: $a_n=600\cdot 3^{n-1}$
Recursive: $a_1=200,\ a_{n+1}=a_n+3$ ; Explicit: $a_n=200+3(n-1)$

Explanation: This question tests your understanding of arithmetic and geometric sequences and how to write them both recursively (each term from the previous) and explicitly (any term directly from its position). Geometric sequences have a constant ratio between consecutive terms: you multiply by the same factor each time. The explicit formula aₙ = a₁ · r^(n-1) uses the first term a₁ and common ratio r, while the recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the 'multiply to get next term' pattern. Since the bacteria triples every hour and starts with 200, after 1 hour we have a₁ = 200 × 3 = 600, after 2 hours a₂ = 600 × 3 = 1800, etc. This is geometric with r = 3. However, since a₁ represents the amount after 1 hour (not the starting amount), we have a₁ = 200 × 3 = 600. Wait, let me reconsider: if a₁ is the amount after 1 hour and we start with 200, then a₁ = 200 × 3 = 600. But looking at the choices, they all use a₁ = 200, which suggests a₁ represents the starting amount (at n = 1 meaning the start). With this interpretation, the recursive formula is a₁ = 200, aₙ₊₁ = 3aₙ, and the explicit formula is aₙ = 200·3^(n-1). Choice B is correct because it properly identifies the starting amount as a₁ = 200 and the growth factor as r = 3, giving the recursive formula a₁ = 200, aₙ₊₁ = 3aₙ and explicit formula aₙ = 200·3^(n-1). Great work! Choice A confuses arithmetic with geometric: the bacteria triples (multiplies by 3), not adds 3. The pattern is exponential growth, not linear growth. Real-world clue: if the context mentions 'percent growth,' 'doubles each,' or 'multiplies by,' think geometric. The language tells you which type!

Question 4

Is the sequence $9, 3, 1, \tfrac{1}{3}, \dots$ arithmetic, geometric, or neither?

Arithmetic
Geometric (correct answer)
Neither
Both arithmetic and geometric

Explanation: This question tests your understanding of arithmetic and geometric sequences and how to write them both recursively (each term from the previous) and explicitly (any term directly from its position). Geometric sequences have a constant ratio between consecutive terms: you multiply by the same factor each time. The explicit formula aₙ = a₁ · r^(n-1) uses the first term a₁ and common ratio r, while the recursive form a₁ = [value], aₙ₊₁ = r·aₙ shows the 'multiply to get next term' pattern. To identify this as geometric and write formulas, we check the ratio: 3/9=1/3, 1/3=1/3, (1/3)/(1)=1/3, confirming common ratio r=1/3. With a₁=9, the explicit formula is aₙ=9·(1/3)^{n-1}. The recursive version: a₁=9, aₙ₊₁=(1/3)·aₙ. This captures 'start at 9, multiply by 1/3 each time.' Choice B is correct because it properly identifies the sequence as geometric with constant ratios, not differences. Great work! Choice A confuses arithmetic with geometric: this sequence multiplies by the same amount each time, so it's geometric, not arithmetic. The pattern is multiplying by 1/3, not subtracting a constant. To identify arithmetic vs geometric: subtract consecutive terms (if differences are constant, it's arithmetic with d = that difference), or divide consecutive terms (if ratios are constant, it's geometric with r = that ratio). If neither pattern works, it's neither!

Question 5

A geometric sequence has $g_3 = 12$ and $g_6 = 96$ . What is the recursive formula for this sequence, and what is the value of $g_1$ ?

$g_n = 4 \cdot g_{n-1}$ ; $g_1 = 0.75$
$g_n = 2 \cdot g_{n-1}$ ; $g_1 = 3$ (correct answer)
$g_n = 2 \cdot g_{n-1}$ ; $g_1 = 6$
$g_n = 8 \cdot g_{n-1}$ ; $g_1 = 1.5$

Explanation: When you encounter a geometric sequence problem with two known terms, you need to find the common ratio first, then work backwards to find the first term and recursive formula. In a geometric sequence, each term equals the previous term multiplied by a constant ratio $r$ . Since $g_6 = g_3 \cdot r^3$ (because you multiply by $r$ three times to go from the 3rd to 6th term), you can find: $96 = 12 \cdot r^3$ , so $r^3 = 8$ , which means $r = 2$ . Now you can work backwards to find $g_1$ . Since $g_3 = g_1 \cdot r^2$ , you have $12 = g_1 \cdot 2^2 = 4g_1$ , so $g_1 = 3$ . The recursive formula is $g_n = 2 \cdot g_{n-1}$ , confirming answer B. Let's check why the other options fail. Choice A uses $r = 4$ , but if $g_3 = 12$ and $r = 4$ , then $g_6$ would be $12 \cdot 4^3 = 768$ , not 96. Choice C has the correct ratio but wrong first term—if $g_1 = 6$ , then $g_3 = 6 \cdot 2^2 = 24$ , not 12. Choice D uses $r = 8$ , making $g_6 = 12 \cdot 8^3 = 6,144$ , which is way too large. Study tip: Always find the common ratio first by using the relationship between the given terms, then work systematically backwards to find earlier terms. Double-check by verifying that your values produce the original given terms.

Question 6

A bacteria culture starts with 200 bacteria and doubles every 3 hours.

Which function best models the number of bacteria $B(t)$ after $t$ hours, and how many complete doubling periods will have occurred after 10 hours?

$B(t) = 200 \cdot 2^{t/3}$ ; 3 complete periods (correct answer)
$B(t) = 200 \cdot 2^{3t}$ ; 10 complete periods
$B(t) = 200 + 2^{t/3}$ ; 3 complete periods
$B(t) = 200 \cdot 3^{t/2}$ ; 5 complete periods

Explanation: This is a geometric sequence where the bacteria doubles every 3 hours. The initial amount is 200, and after every 3 hours the population is multiplied by 2. So after $t$ hours, there have been $t/3$ doubling periods, giving $B(t) = 200 \cdot 2^{t/3}$ . After 10 hours, there have been $10/3 = 3.33...$ hours worth of doubling, so 3 complete doubling periods. Choice B incorrectly uses $2^{3t}$ . Choice C uses addition instead of multiplication. Choice D uses the wrong base and exponent.

Question 7

Two sequences are given: Sequence X has the explicit formula $x_n = 15 - 2n$ and Sequence Y is defined recursively as $y_1 = 13$ and $y_n = y_{n-1} - 2$ . How do these sequences compare?

One is arithmetic and the other is geometric with different growth patterns
They have the same common difference but different starting terms
They are both arithmetic but have different common differences
They are identical sequences with the same terms in the same order (correct answer)

Explanation: When comparing sequences, you need to determine whether they're arithmetic or geometric, then examine their specific terms and patterns. Let's start with Sequence X: $x_n = 15 - 2n$ . This is an explicit formula where you can find any term directly. For the first few terms: $x_1 = 15 - 2(1) = 13$ , $x_2 = 15 - 2(2) = 11$ , $x_3 = 15 - 2(3) = 9$ . The sequence is 13, 11, 9, 7, ... with a common difference of -2, making it arithmetic. Now for Sequence Y: Starting with $y_1 = 13$ and the recursive rule $y_n = y_{n-1} - 2$ , we get: $y_1 = 13$ , $y_2 = 13 - 2 = 11$ , $y_3 = 11 - 2 = 9$ . This gives us 13, 11, 9, 7, ... which is also arithmetic with a common difference of -2. Both sequences produce exactly the same terms: 13, 11, 9, 7, 5, ... This makes answer choice D correct. Looking at the wrong answers: A is incorrect because both sequences are arithmetic, not one arithmetic and one geometric. B is wrong because they have the same starting term (13) and the same common difference (-2). C is false because while both are arithmetic, they have identical common differences of -2, not different ones. Study tip: When comparing sequences, always calculate the first few terms of each sequence explicitly. This reveals patterns and relationships that might not be obvious from the formulas alone, especially when one is given explicitly and another recursively.

Question 8

A recursive formula for a sequence is given as $a_1 = 7$ and $a_n = a_{n-1} + 4$ for $n \geq 2$ . Which explicit formula represents the same sequence and what is the value of $a_{15}$ ?

$a_n = 7 + 4n$ ; $a_{15} = 67$
$a_n = 4 + 7n$ ; $a_{15} = 109$
$a_n = 7 + 4(n-1)$ ; $a_{15} = 63$ (correct answer)
$a_n = 4 + 7(n-1)$ ; $a_{15} = 102$

Explanation: The recursive formula shows this is an arithmetic sequence with first term $a_1 = 7$ and common difference $d = 4$ . The explicit formula for an arithmetic sequence is $a_n = a_1 + d(n-1) = 7 + 4(n-1)$ . Therefore $a_{15} = 7 + 4(15-1) = 7 + 4(14) = 7 + 56 = 63$ . Choice A incorrectly uses $4n$ instead of $4(n-1)$ . Choice B switches the first term and common difference. Choice D also switches the values and uses the wrong formula structure.

Question 9

The first four terms of a sequence are 1, 4, 16, 64. A student claims this can be modeled by both $a_n = 4^{n-1}$ and the recursive formula $a_1 = 1$ , $a_n = 4a_{n-1}$ . Is the student correct, and what is $a_7$ ?

Student is incorrect; the explicit formula should be $a_n = 4n - 3$ ; $a_7 = 25$
Student is correct; both formulas are equivalent; $a_7 = 4096$
Student is incorrect; the recursive formula should be $a_n = a_{n-1} + 4^{n-2}$ ; $a_7 = 1365$
Student is correct; both formulas are equivalent; $a_7 = 16384$ (correct answer)

Explanation: Checking the sequence 1, 4, 16, 64: each term is 4 times the previous term (geometric with ratio 4). Using $a_n = 4^{n-1}$ : $a_1 = 4^0 = 1$ , $a_2 = 4^1 = 4$ , $a_3 = 4^2 = 16$ , $a_4 = 4^3 = 64$ ✓. The recursive formula $a_1 = 1$ , $a_n = 4a_{n-1}$ also gives the same sequence. So $a_7 = 4^{7-1} = 4^6 = 16384$ . Choice A incorrectly suggests an arithmetic sequence. Choice B has the wrong calculation for $a_7$ ( $4^5 = 1024$ , not $4^6$ ). Choice C proposes an incorrect recursive formula.

Question 10

Consider the sequence: 12, 8, 4, 0, -4, ... If $a_n$ represents the $n$ th term where $a_1 = 12$ , which recursive formula correctly describes this sequence, and what type of sequence is it?

$a_n = a_{n-1} - 4$ ; arithmetic sequence with common difference -4 (correct answer)
$a_n = a_{n-1} \cdot \frac{2}{3}$ ; geometric sequence with common ratio $\frac{2}{3}$
$a_n = a_{n-1} + 4$ ; arithmetic sequence with common difference 4
$a_n = a_{n-1} \div 4$ ; geometric sequence with common ratio $\frac{1}{4}$

Explanation: Looking at consecutive terms: $8 - 12 = -4$ , $4 - 8 = -4$ , $0 - 4 = -4$ , $-4 - 0 = -4$ . Since there's a constant difference of -4 between consecutive terms, this is an arithmetic sequence. The recursive formula is $a_n = a_{n-1} + (-4) = a_{n-1} - 4$ . Choice B incorrectly identifies it as geometric ( $12 \times \frac{2}{3} = 8$ , but $8 \times \frac{2}{3} = \frac{16}{3} \neq 4$ ). Choice C has the wrong sign for the common difference. Choice D also incorrectly identifies it as geometric.