College Algebra Quiz: Geometric Sequences Explicit And Recursive Forms

What this quiz covers

This quiz focuses on Geometric Sequences Explicit And Recursive Forms, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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.

Question 1

1. $a_n = 80(0.8)^n$
2. $a_n = 64(0.8)^{n-1}$
3. $a_n = 80(0.8)^{n-1}$ (correct answer)
4. $a_n = 51.2(0.8)^{n-1}$

Explanation: When you encounter a geometric sequence with a recursive relation, you need to identify the common ratio and find the first term to write the explicit formula in the standard form $a_n = a_1 \cdot r^{n-1}$. From the recursive relation $a_{n+1} = 0.8a_n$, you can see that the common ratio is $r = 0.8$. To find the first term, work backwards from the given information $a_4 = 40.96$. Since each term is obtained by multiplying the previous term by 0.8, you can find earlier terms by dividing by 0.8: $a_3 = \frac{a_4}{0.8} = \frac{40.96}{0.8} = 51.2$ $a_2 = \frac{a_3}{0.8} = \frac{51.2}{0.8} = 64$ $a_1 = \frac{a_2}{0.8} = \frac{64}{0.8} = 80$ Therefore, the explicit formula is $a_n = 80(0.8)^{n-1}$, which is choice C. Choice A uses the correct first term (80) but has the exponent as $n$ instead of $n-1$, which would make $a_1 = 80(0.8)^1 = 64$ instead of 80. Choice B incorrectly uses 64 as the coefficient, which would be $a_2$ rather than $a_1$. Choice D uses 51.2 as the coefficient, which represents $a_3$ instead of $a_1$. Remember: the standard form $a_n = a_1 \cdot r^{n-1}$ always uses the first term as the coefficient and $n-1$ as the exponent. When given information about a later term, always work backwards to find $a_1$.

Question 2

1. $a_n = 7 \cdot 2^n$
2. $a_n = 7 \cdot 2^{n-1}$ (correct answer)
3. $a_n = 14 \cdot 2^{n-1}$
4. $a_n = 7 \cdot (\sqrt{2})^{n-1}$

Explanation: When you encounter a geometric sequence problem, you need to connect the recursive definition with the explicit formula. A geometric sequence has the form $a_n = a_1 \cdot r^{n-1}$ where $a_1$ is the first term and $r$ is the common ratio. First, let's find the common ratio $k$. Since $a_1 = 7$ and $a_5 = 112$, we can use the relationship $a_5 = a_1 \cdot k^4$: $112 = 7 \cdot k^4$ $k^4 = 16$ $k = 2$ Now we can write the explicit formula: $a_n = 7 \cdot 2^{n-1}$. Let's verify: $a_5 = 7 \cdot 2^{5-1} = 7 \cdot 2^4 = 7 \cdot 16 = 112$ ✓ Looking at the wrong answers: Choice A gives $a_n = 7 \cdot 2^n$, which would make $a_5 = 7 \cdot 2^5 = 224$—this formula has an extra factor of 2 because it uses $2^n$ instead of $2^{n-1}$. Choice C gives $a_n = 14 \cdot 2^{n-1}$, which would make $a_1 = 14$, contradicting the given condition that $a_1 = 7$. Choice D gives $a_n = 7 \cdot (\sqrt{2})^{n-1}$, which would make $a_5 = 7 \cdot (\sqrt{2})^4 = 7 \cdot 4 = 28$—this uses the wrong common ratio. The answer is B. Study tip: Always remember that the standard explicit formula for geometric sequences uses $r^{n-1}$, not $r^n$, because when $n=1$, you want $r^0 = 1$ to get back to $a_1$.

Question 3

1. $-3$ (correct answer)
2. $3$
3. $-\frac{1}{3}$
4. $\frac{1}{3}$

Explanation: In a geometric sequence, $a_6 = a_3 \cdot r^{6-3} = a_3 \cdot r^3$. Substituting the given values: $-648 = 24 \cdot r^3$, so $r^3 = \frac{-648}{24} = -27$. Taking the cube root: $r = \sqrt[3]{-27} = -3$. We can verify: if $r = -3$ and $a_3 = 24$, then $a_4 = 24(-3) = -72$, $a_5 = -72(-3) = 216$, and $a_6 = 216(-3) = -648$ ✓. Choice B gives $r^3 = 27$, which would make $a_6 = 648$ (positive). Choices C and D would give much smaller values for $a_6$.

Question 4

1. $a_1 = 48$, $a_{n+1} = 4a_n$ for $n \geq 1$
2. $a_1 = 12$, $a_{n+1} = \frac{1}{4}a_n$ for $n \geq 1$
3. $a_1 = 48$, $a_{n+1} = \frac{1}{4}a_n$ for $n \geq 1$ (correct answer)
4. $a_1 = 192$, $a_{n+1} = \frac{1}{4}a_n$ for $n \geq 1$

Explanation: When converting between explicit and recursive forms of geometric sequences, you need to identify two key components: the first term and the common ratio that connects consecutive terms. From the explicit formula $a_n = 48 \cdot (\frac{1}{4})^{n-1}$, you can find the first term by substituting $n = 1$: $a_1 = 48 \cdot (\frac{1}{4})^{1-1} = 48 \cdot (\frac{1}{4})^0 = 48 \cdot 1 = 48$. The common ratio is the base of the exponential term, which is $\frac{1}{4}$. This means each term is $\frac{1}{4}$ times the previous term, so $a_{n+1} = \frac{1}{4}a_n$. Choice C correctly identifies both components: $a_1 = 48$ and $a_{n+1} = \frac{1}{4}a_n$. Choice A has the correct first term but uses $a_{n+1} = 4a_n$, which would make the sequence increase rather than decrease. This represents the reciprocal of the actual common ratio. Choice B uses the wrong first term ($a_1 = 12$ instead of 48). This might come from incorrectly calculating $48 \cdot \frac{1}{4}$, but remember that when $n = 1$, the exponent becomes zero, not one. Choice D has the correct recursive relationship but starts with $a_1 = 192$. This appears to come from multiplying 48 by 4, perhaps confusing the direction of the ratio. Study tip: Always verify your recursive formula by computing the first few terms both ways. The explicit formula $a_n = a_1 \cdot r^{n-1}$ directly shows you $a_1$ and $r$, which become the building blocks of your recursive definition.

Question 5

1. $a_{n+1} = 3a_n$
2. $a_{n+1} = -3a_n$ (correct answer)
3. $a_{n+1} = -\frac{1}{3}a_n$
4. $a_{n+1} = \frac{1}{3}a_n$

Explanation: When you encounter a geometric sequence problem with two given terms, your goal is to find the common ratio that connects consecutive terms through multiplication. In a geometric sequence, each term equals the previous term multiplied by the common ratio $r$. Since $a_2 = 15$ and $a_5 = -405$, you can use the relationship $a_5 = a_2 \cdot r^3$ (since you multiply by $r$ three times to go from the 2nd to 5th term). Substituting the known values: $-405 = 15 \cdot r^3$ Solving for $r$: $r^3 = \frac{-405}{15} = -27$ Therefore: $r = \sqrt[3]{-27} = -3$ The recursive relationship is $a_{n+1} = r \cdot a_n = -3a_n$, which is answer B. Let's check why the other options are incorrect: A) $a_{n+1} = 3a_n$: This uses $r = 3$, which would give $a_5 = 15 \cdot 3^3 = 405$, not $-405$. The sign is wrong. C) $a_{n+1} = -\frac{1}{3}a_n$: This uses $r = -\frac{1}{3}$, making the sequence decrease in magnitude rather than increase. D) $a_{n+1} = \frac{1}{3}a_n$: This uses $r = \frac{1}{3}$, which would make $a_5 = \frac{5}{9}$, far too small and positive. Study tip: Always verify your common ratio by checking that it produces the correct second given term. Calculate $a_2 \cdot r^{(n-2)}$ to find $a_n$, and pay careful attention to negative signs—they often determine the correct answer in geometric sequence problems.

Question 6

1. $a_1 = \frac{3}{2}$, $a_{n+1} = 2a_n$ for $n \geq 1$
2. $a_1 = \frac{3}{2}$, $a_{n+1} = \frac{1}{2}a_n$ for $n \geq 1$
3. $a_1 = 3$, $a_{n+1} = 2a_n$ for $n \geq 1$
4. $a_1 = 3$, $a_{n+1} = \frac{1}{2}a_n$ for $n \geq 1$ (correct answer)

Explanation: When converting between explicit and recursive formulas for geometric sequences, you need to identify two key components: the first term and the common ratio that connects consecutive terms. Let's find the first term by substituting $n = 1$ into the given formula: $a_1 = 3 \cdot (\frac{1}{2})^{1-1} = 3 \cdot (\frac{1}{2})^0 = 3 \cdot 1 = 3$. So our recursive formula must start with $a_1 = 3$. For the recursive relationship, we need the common ratio. In any geometric sequence, each term equals the previous term multiplied by the common ratio. Looking at our formula $a_n = 3 \cdot (\frac{1}{2})^{n-1}$, the base $\frac{1}{2}$ is our common ratio. This means $a_{n+1} = \frac{1}{2} \cdot a_n$. Let's verify: $a_1 = 3$, $a_2 = 3 \cdot \frac{1}{2} = \frac{3}{2}$, $a_3 = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4}$. Using our explicit formula: $a_2 = 3 \cdot (\frac{1}{2})^1 = \frac{3}{2}$ ✓. Choice A uses the wrong first term ($\frac{3}{2}$ instead of $3$) and wrong ratio ($2$ instead of $\frac{1}{2}$). Choice B has the correct ratio but wrong first term. Choice C has the correct first term but uses ratio $2$, which would make the sequence grow instead of shrink. Only choice D correctly identifies both $a_1 = 3$ and $a_{n+1} = \frac{1}{2}a_n$. Study tip: Always check your recursive formula by computing the first few terms both ways—you should get identical sequences regardless of which formula you use.

Question 7

1. $a_n = 5(-2)^{n-1}$ (correct answer)
2. $a_n = 5(-2)^n$
3. $a_n = -10 \cdot 2^{n-1}$
4. $a_n = 5 \cdot 2^{n-1}$

Explanation: From the recursive formula $a_{n+1} = -2a_n$, we can identify that the common ratio is $r = -2$. The general form of a geometric sequence is $a_n = a_1 \cdot r^{n-1}$. Substituting $a_1 = 5$ and $r = -2$, we get $a_n = 5(-2)^{n-1}$. We can verify: $a_1 = 5(-2)^0 = 5$, $a_2 = 5(-2)^1 = -10$, $a_3 = 5(-2)^2 = 20$. Choice B is incorrect because it uses $(-2)^n$ instead of $(-2)^{n-1}$, giving $a_1 = 5(-2)^1 = -10$. Choice C incorrectly separates the negative sign. Choice D ignores the negative in the common ratio.