Algebra 3 Quiz: Explicit And Recursive Formulas
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Explicit And Recursive FormulasQuestion 1 of 12

A sequence is defined by b0=6b_0 = 6 and bn=−3bn−1b_n = -3b_{n-1} for n≥1n \ge 1. Which explicit formula gives bnb_n for n≥0n \ge 0?

bn=6(−3)nb_n=6(-3)^n
bn=6(−3)n−1b_n=6(-3)^{n-1}
bn=(−18)nb_n=(-18)^n
bn=6(−3)n+1b_n=6(-3)^{n+1}
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Algebra 3 Quiz

Algebra 3 Quiz: Explicit And Recursive Formulas

Practice Explicit And Recursive Formulas in Algebra 3 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 Explicit And Recursive Formulas, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

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

A sequence is defined by b0=6b_0 = 6 and bn=−3bn−1b_n = -3b_{n-1} for n≥1n \ge 1. Which explicit formula gives bnb_n for n≥0n \ge 0?

  1. bn=6(−3)nb_n=6(-3)^n (correct answer)
  2. bn=6(−3)n−1b_n=6(-3)^{n-1}
  3. bn=(−18)nb_n=(-18)^n
  4. bn=6(−3)n+1b_n=6(-3)^{n+1}
Explanation: Whenever you see a recursive rule like b_n = -3b_{n-1}}, recognize it as a geometric sequence: the ratio is constant (-3), and the initial term is b0=6b_0=6. For a geometric sequence, the explicit formula is bn=b0rnb_n = b_0 r^n, so bn=6(−3)nb_n=6(-3)^n. That formula works: at n=0n=0, it gives 66; at n=1n=1, it gives −18-18, exactly −3⋅6-3\cdot 6. Now check the wrong options. The choice 6(-3)^{n-1}} uses an off-by-one exponent. It gives n=0n=0 as 6(−3)−1=−26(-3)^{-1}=-2, not 66, and it would only match if indices started at 11. The choice (−18)n(-18)^n multiplies the first term and ratio together as the base, giving n=0n=0 as 11, not 66, and n=2n=2 as 324324, while the real value is 5454. It confuses the first term b1=−18b_1=-18 with the initial term. The choice 6(-3)^{n+1}} shifts the exponent in the opposite direction: at n=0n=0 it gives −18-18, so it actually represents the sequence starting with b1b_1 rather than b0b_0. A quick habit: whenever you write or choose an explicit formula for a recursive sequence, test it at n=0n=0 and n=1n=1. If it gives the stated initial term and the first recursive term, it is likely correct. Here, only 6(−3)n6(-3)^n passes both checks.

Question 2

The sequence is defined by a1=8a_1 = 8 and an=an−1−12a_n = a_{n-1} - \frac{1}{2} for n≥2n \ge 2. Which explicit formula represents the same sequence?

  1. an=8−12(n−1)a_n=8-\frac{1}{2}(n-1) (correct answer)
  2. an=8−12na_n=8-\frac{1}{2}n
  3. an=8+12(n−1)a_n=8+\frac{1}{2}(n-1)
  4. an=152−12na_n=\frac{15}{2}-\frac{1}{2}n
Explanation: Whenever you see a recursive definition like an=an−1−12a_n=a_{n-1}-\frac12, recognize it as an arithmetic sequence: each term is made by adding the same common difference, here d=−12d=-\frac12. The explicit formula for an arithmetic sequence is an=a1+(n−1)da_n=a_1+(n-1)d, so substituting a1=8a_1=8 and d=−12d=-\frac12 gives an=8−12(n−1).a_n=8-\frac12(n-1). That formula correctly produces a1=8a_1=8, then a2=7.5a_2=7.5, matching the recursion. The choice 8−12n8-\frac12 n is a classic off-by-one error: it uses ndnd instead of (n−1)d(n-1)d, so at n=1n=1 it gives 7.57.5, not 88. The choice 8+12(n−1)8+\frac12(n-1) uses the wrong sign for the common difference; it describes a sequence that increases by 0.50.5, the opposite direction of the given rule. The choice 152−12n\frac{15}{2}-\frac12 n has the correct slope but the wrong starting point: at n=1n=1 it gives 77, not 88, so it represents a different arithmetic sequence entirely. A fast way to check any explicit formula for a recursive sequence is to plug in n=1n=1 and n=2n=2 and compare with the first two terms from the recursion. That simple test catches sign errors, off-by-one mistakes, and wrong starting values before you commit.

Question 3

A sequence is defined by an=5(2)n+1a_n = 5(2)^{n+1} for n≥1n \ge 1. Which recursive formula generates the same sequence?

  1. a1=10,an=2an−1a_1=10, a_n=2a_{n-1} for n≥2n \ge 2
  2. a1=20,an=2an−1a_1=20, a_n=2a_{n-1} for n≥2n \ge 2 (correct answer)
  3. a1=40,an=2an−1a_1=40, a_n=2a_{n-1} for n≥2n \ge 2
  4. a1=5,an=2an−1a_1=5, a_n=2a_{n-1} for n≥2n \ge 2
Explanation: Whenever you see a sequence given by an explicit formula like an=5(2)n+1a_n = 5(2)^{n+1}, your task is to translate it into recursive language: find the first term, then identify how each term is built from the previous one. The structure an=5⋅2n+1a_n = 5 \cdot 2^{n+1} is exponential with base 2, so the common ratio is 2. To find the starting term, plug in n=1n=1: a1=5(2)1+1=5⋅4=20a_1 = 5(2)^{1+1} = 5 \cdot 4 = 20. That means the sequence begins at 20 and every later term is twice the one before it, so the recursive form is a1=20, an=2an−1a_1 = 20,\ a_n = 2a_{n-1} for n≥2n \ge 2. The wrong choices each come from a common slip. Starting with a1=10a_1=10 would be correct if the formula were 5(2)n5(2)^n, but the extra n+1n+1 exponent doubles the first term to 20. Starting with a1=40a_1=40 gives the value of a2a_2, not a1a_1, because a2=5(2)3=40a_2 = 5(2)^3 = 40. Starting with a1=5a_1=5 ignores the power-of-2 factor entirely, treating the base coefficient as the first term. Each of those incorrect initial terms would produce a sequence shifted, doubled, or scaled incorrectly. Your quick strategy: given an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}, identify a1a_1 by substituting n=1n=1, and recognize the base of the exponent as the recursive multiplier. Then write "first term, multiply by that ratio" — and check your first term before moving on.

Question 4

An arithmetic sequence has a3=14a_3 = 14 and a7=30a_7 = 30. Which recursive formula defines the sequence for n≥1n \ge 1?

  1. a1=2,an=an−1+4a_1=2, a_n=a_{n-1}+4 for n≥2n \ge 2
  2. a1=6,an=an−1+4a_1=6, a_n=a_{n-1}+4 for n≥2n \ge 2 (correct answer)
  3. a1=6,an=an−1+16a_1=6, a_n=a_{n-1}+16 for n≥2n \ge 2
  4. a3=14,an=an−1+4a_3=14, a_n=a_{n-1}+4 for n≥4n \ge 4
Explanation: Whenever you see a question about arithmetic sequences, remember that a recursive formula needs two pieces of information: the first term a1a_1 and the common difference dd. To find dd, use the given terms: the difference between a7a_7 and a3a_3 is 30−14=1630 - 14 = 16, and there are 7−3=47-3 = 4 steps between them. So d=16/4=4d = 16/4 = 4. Now find a1a_1: since a3=a1+2da_3 = a_1 + 2d, you have 14=a1+2(4)14 = a_1 + 2(4), which gives a1=6a_1 = 6. Thus the correct recursive formula is a1=6,an=an−1+4a_1=6, a_n=a_{n-1}+4 for n≥2n \ge 2. The choice a1=2,an=an−1+4a_1=2, a_n=a_{n-1}+4 correctly uses d=4d=4 but miscalculates a1a_1 (perhaps subtracting 33 steps instead of 22). The choice a1=6,an=an−1+16a_1=6, a_n=a_{n-1}+16 correctly finds a1a_1 but uses the raw difference 1616 as the common difference instead of dividing by the number of steps. The choice a3=14,an=an−1+4a_3=14, a_n=a_{n-1}+4 for n≥4n \ge 4 is a valid rule for the tail of the sequence, but it fails the requirement of defining the sequence for n≥1n \ge 1 because it never specifies a1a_1 or a2a_2. Your study tip: always solve for dd by dividing the difference of terms by the index gap, then back-solve for a1a_1. Never use a given term like a3a_3 as your starting point unless the domain explicitly starts there.

Question 5

A sequence satisfies an+1=an−5a_{n+1}=a_n-5 for n≥1n \ge 1, and a3=12a_3=12. Which explicit formula gives ana_n for n≥1n \ge 1?

  1. an=17−5na_n=17-5n
  2. an=12−5na_n=12-5n
  3. an=22−5na_n=22-5n
  4. an=27−5na_n=27-5n (correct answer)
Explanation: Whenever you see a sequence like an+1=an−5a_{n+1}=a_n-5, recognize that this is an arithmetic sequence with common difference d=−5d=-5. Your job is to shift from the recursive form to an explicit formula using any known term. Since a3=12a_3=12, you can anchor the formula at n=3n=3: an=a3+(n−3)(−5)=12−5n+15=27−5n.a_n=a_3+(n-3)(-5)=12-5n+15=27-5n. So the correct explicit formula is an=27−5na_n=27-5n. Quick check: at n=3n=3, it gives 27−15=1227-15=12. The choice 17−5n17-5n comes from treating a3=12a_3=12 as if it were a1a_1: that would mean 12−5(n−1)=17−5n12-5(n-1)=17-5n, but it makes a3=2a_3=2, not 1212. The choice 12−5n12-5n simply attaches −5n-5n to the given term, ignoring that nn must count steps from n=3n=3; it gives −3-3 when n=3n=3. The choice 22−5n22-5n finds a1=22a_1=22 correctly but then forgets the (n−1)(n-1) adjustment: the correct form is 22−5(n−1)=27−5n22-5(n-1)=27-5n, not 22−5n22-5n. Your strategy: anchor the explicit rule at the known term, using an=ak+(n−k)da_n=a_k+(n-k)d. This prevents both off-by-one mistakes and the trap of plugging the given term into the wrong position.

Question 6

The sequence is defined by a1=12a_1=12 and an=23an−1a_n=\frac{2}{3}a_{n-1} for n≥2n \ge 2. Which explicit formula gives ana_n for n≥1n \ge 1?

  1. an=8(23)na_n=8\left(\frac{2}{3}\right)^n
  2. an=12(23)na_n=12\left(\frac{2}{3}\right)^n
  3. an=8(23)n−1a_n=8\left(\frac{2}{3}\right)^{n-1}
  4. an=12(23)n−1a_n=12\left(\frac{2}{3}\right)^{n-1} (correct answer)
Explanation: Whenever you see a recursive definition like an=23an−1a_n=\frac23 a_{n-1}, you're looking at a geometric sequence: each term is the previous term multiplied by the same common ratio, r=23r=\frac23. The explicit form is always an=a1⋅rn−1a_n=a_1\cdot r^{n-1}. Here a1=12a_1=12, so an=12(23)n−1a_n=12\left(\frac23\right)^{n-1}. Check: plugging n=1n=1 gives 12(23)0=1212\left(\frac23\right)^0=12, and n=2n=2 gives 12(23)=812\left(\frac23\right)=8, matching the recursive step. Now look at the traps. The choice 12(23)n12\left(\frac23\right)^n is close, but it uses exponent nn instead of n−1n-1. That would make a1=8a_1=8, not 1212 — a common off-by-one error. The choice 8(23)n−18\left(\frac23\right)^{n-1} incorrectly starts from the second term, a2=8a_2=8, as if it were the first term; it also gives a1=8a_1=8. Finally, the choice 8(23)n8\left(\frac23\right)^n combines both mistakes, using the second term as the starting point and shifting the exponent, so it fails even for n=1n=1. Your best strategy on explicit-formula questions is to identify a1a_1 and the common ratio rr, then write an=a1rn−1a_n=a_1r^{n-1}. Before choosing, always test your formula at n=1n=1 to confirm it returns a1a_1. That one quick check will catch most exponent and first-term errors.

Question 7

A sequence is defined by an=3⋅2n+28a_n=\frac{3\cdot 2^{n+2}}{8} for n≥1n \ge 1. Which recursive formula generates the sequence?

  1. a1=12,an=2an−1a_1=12, a_n=2a_{n-1} for n≥2n \ge 2
  2. a1=6,an=2an−1a_1=6, a_n=2a_{n-1} for n≥2n \ge 2
  3. a1=3,an=2an−1a_1=3, a_n=2a_{n-1} for n≥2n \ge 2 (correct answer)
  4. a1=3,an=12an−1a_1=3, a_n=\frac{1}{2}a_{n-1} for n≥2n \ge 2
Explanation: Whenever you see a sequence defined by an explicit formula, your first move should be to simplify it and then find the first term and the common ratio between consecutive terms. Here, notice that 8=238=2^3, so an=3⋅2n+223=3⋅2n−1.a_n=\frac{3\cdot 2^{n+2}}{2^3}=3\cdot 2^{n-1}. This is a geometric sequence: it starts at a1=3⋅20=3a_1=3\cdot 2^{0}=3, and each next term is multiplied by 22. For example, a1=3a_1=3, a2=6a_2=6, a3=12a_3=12, and so on. The recursive formula must therefore state a1=3a_1=3 and use the rule an=2an−1a_n=2a_{n-1} for n≥2n\ge2. The choice that starts with a1=12a_1=12 and doubles gives 12,24,48,…12,24,48,\dots, which has the right ratio but the wrong starting point. The choice that starts with a1=6a_1=6 and doubles gives 6,12,24,…6,12,24,\dots, again missing the true first term of 33. The choice that starts with a1=3a_1=3 but uses an=12an−1a_n=\frac12 a_{n-1} produces 3,1.5,0.75,…3,1.5,0.75,\dots, which reverses the growth into decay. Only the formula with a1=3a_1=3 and an=2an−1a_n=2a_{n-1} matches both the first term and the doubling pattern. A quick strategy: to convert an explicit geometric formula to a recursive one, simplify the expression first, then compute a1a_1 and the ratio an/an−1a_n/a_{n-1}. That ratio becomes the multiplier in the recursive rule.

Question 8

A sequence is defined by an=12−3(n+1)a_n = 12 - 3(n+1) for n≥0n \ge 0. Which recursive formula with a0a_0 as the initial value generates the same sequence?

  1. a0=9,an=an−1−3a_0=9, a_n=a_{n-1}-3 for n≥1n \ge 1 (correct answer)
  2. a0=12,an=an−1−3a_0=12, a_n=a_{n-1}-3 for n≥1n \ge 1
  3. a0=9,an=an−1+3a_0=9, a_n=a_{n-1}+3 for n≥1n \ge 1
  4. a0=6,an=an−1−3a_0=6, a_n=a_{n-1}-3 for n≥1n \ge 1
Explanation: Whenever you see a sequence defined by a linear explicit formula, recognize it as an arithmetic sequence: the number multiplied by nn is the common difference, and plugging in n=0n=0 gives the initial value. Here a_n = ́12 - ́3(n+1). Simplify: \,a_n = ́12 - ́3n - ́3 = ́9 - ́3n. So a_0 = ́9, and each step decreases by 33, giving $$ a_0=9,\quad a_n=a_{n-1}-3 \text{ for } n\ge 1. The choice with $a_0=9,\ a_n=a_{n-1}-3$ is exactly the correct recursive form. The choice with $a_0=12,\ a_n=a_{n-1}-3$ uses the right common difference but wrong starting value; $12$ would come from ignoring the extra $-3$ in the formula. The choice with $a_0=9,\ a_n=a_{n-1}+3$ has the right initial value but wrong direction; adding $3$ yields an increasing sequence, whereas computing terms like $a_1=6,\ a_2=3$ shows values decrease. The choice with $a_0=6,\ a_n=a_{n-1}-3$ has the right common difference but initial value $6$ is what you might get by plugging in $n=1$, not $n=0$. Study tip: for an arithmetic sequence, get the first term and the common difference — the constant subtracted from one term to the next — then write the recursive rule as $a_0=\text{first},\, a_n=a_{n-1}+\text{difference}$.

Question 9

The balance of an investment after nn years is modeled by Bn=500(1.02)nB_n = 500(1.02)^n for n≥0n \ge 0. Which recursive formula with B0B_0 as the initial value gives the same balances?

  1. B0=500,Bn=1.02nBn−1B_0=500, B_n=1.02^n B_{n-1} for n≥1n \ge 1
  2. B0=500,Bn=Bn−1+0.02B_0=500, B_n=B_{n-1}+0.02 for n≥1n \ge 1
  3. B1=500,Bn=1.02Bn−1B_1=500, B_n=1.02B_{n-1} for n≥2n \ge 2
  4. B0=500,Bn=1.02Bn−1B_0=500, B_n=1.02B_{n-1} for n≥1n \ge 1 (correct answer)
Explanation: Whenever you see an explicit formula like Bn=500(1.02)nB_n = 500(1.02)^n, it describes exponential growth: the initial balance is 500500, and each year the balance is multiplied by 1.021.02. So a matching recursive formula must start with B0=500B_0=500 and use the rule "multiply the previous balance by the same growth factor." That is exactly Bn=1.02Bn−1B_n=1.02B_{n-1} for n≥1n\ge 1. Why the others fail? The choice Bn=1.02nBn−1B_n=1.02^nB_{n-1} uses a changing multiplier each step, so it grows too fast: after two years it gives 500⋅1.023500 \cdot 1.02^3 instead of 500⋅1.022500 \cdot 1.02^2. The choice Bn=Bn−1+0.02B_n=B_{n-1}+0.02 treats the 2% as an addition of 0.020.02 dollars, producing a linear sequence like 500,500.02,500.04500, 500.02, 500.04, not the exponential values 510,520.2510, 520.2. The choice with B1=500B_1=500 and Bn=1.02Bn−1B_n=1.02B_{n-1} has the right multiplier but the wrong starting year: since B1=1.02B0B_1=1.02B_0, setting B1=500B_1=500 makes B0≈490.20B_0\approx 490.20, so it models a different initial investment. Only the formula with B0=500B_0=500 and multiplier 1.021.02 reproduces the explicit balances exactly. Your takeaway: to convert an explicit exponential formula Bn=a⋅rnB_n=a\cdot r^n into a recursive one, use B0=aB_0=a and Bn=rBn−1B_n=rB_{n-1}. Watch for percent growth meaning "multiply by 1+1+percent," not "add the percent."

Question 10

A sequence is defined by a1=7a_1=7 and an+1=an+3a_{n+1}=a_n+3 for n≥1n \ge 1. Which explicit formula gives ana_n for n≥1n \ge 1?

  1. an=3n+10a_n=3n+10
  2. an=3n+7a_n=3n+7
  3. an=3n+4a_n=3n+4 (correct answer)
  4. an=7n+3a_n=7n+3
Explanation: Whenever you see a sequence defined by a starting value plus a repeated addition, think arithmetic sequence. The key is identifying the first term a1a_1 and the common difference dd, then using the general form an=a1+(n−1)da_n=a_1+(n-1)d. Here a1=7a_1=7 and each step adds 33, so d=3d=3. Substitute: an=7+3(n−1)=7+3n−3=3n+4a_n=7+3(n-1)=7+3n-3=3n+4. This matches the correct choice an=3n+4a_n=3n+4. A quick check: for n=1n=1, it gives 3(1)+4=73(1)+4=7, exactly the first term. . Now the wrong formulas: an=3n+10a_n=3n+10would give a1=13a_1=13, so it seems to use a1+3(n+1)a_1+3(n+1) — an index shift too far. an=3n+7a_n=3n+7would give a1=10a_1=10, because it forgets to subtract 11 from nn; this is the common error of using a1+dna_1+dn instead of a1+d(n−1)a_1+d(n-1). an=7n+3a_n=7n+3would give a1=10a_1=10and increases by 77 each time, confusing the first term with the common difference — those roles are not interchangeable. . A stronger approach: any time you need an explicit formula, evaluate it at n=1n=1 and at n=2n=2. The correct formula must produce the givens first term and common difference. This quick check catches index shifts and reversed coefficients instantly.

Question 11

An arithmetic sequence is defined by an=2n+1a_n=2n+1 for n≥1n \ge 1. Which recursive formula using an+1a_{n+1} also defines the sequence?

  1. a1=1,an+1=an+2a_1=1, a_{n+1}=a_n+2 for n≥1n \ge 1
  2. a1=3,an+1=an−2a_1=3, a_{n+1}=a_n-2 for n≥1n \ge 1
  3. a1=3,an+1=an+2a_1=3, a_{n+1}=a_n+2 for n≥1n \ge 1 (correct answer)
  4. a1=3,an+1=2ana_1=3, a_{n+1}=2a_n for n≥1n \ge 1
Explanation: Whenever you see an explicit formula for an arithmetic sequence, your task is to extract the starting value and the common difference. The explicit form an=a1+(n−1)da_n = a_1 + (n-1)d has the same structure as an=2n+1a_n = 2n+1. Rewriting 2n+12n+1 as 3+(n−1)23 + (n-1)2 shows that a1=3a_1=3 and the common difference is d=2d=2. So the recursive definition must start at 3 and add 2 each time: a1=3, an+1=an+2a_1=3,\ a_{n+1}=a_n+2. The choice starting with a1=1a_1=1 misidentifies the first term—plugging n=1n=1 into 2n+12n+1 gives 3, not 1. The choice with an+1=an−2a_{n+1}=a_n-2 starts correctly at 3 but subtracts 2, producing 3,1,−1,…3,1,-1,\dots, a different arithmetic sequence. The choice with an+1=2ana_{n+1}=2a_n also starts at 3 but multiplies each term by 2, giving 3,6,12,…3,6,12,\dots, which is geometric, not the linear pattern 3,5,7,…3,5,7,\dots we need. A useful strategy: compute the first two terms from the explicit formula, then check which recursive rule generates those same values from the same starting point. That quickly eliminates all traps and confirms the rule.

Question 12

A geometric sequence has a4=24a_4=24 and a7=192a_7=192. Which explicit formula gives ana_n for n≥1n \ge 1?

  1. an=3(2)na_n=3(2)^n
  2. an=3(2)n−1a_n=3(2)^{n-1} (correct answer)
  3. an=6(2)n−1a_n=6(2)^{n-1}
  4. an=24(2)n−3a_n=24(2)^{n-3}
Explanation: This question tests whether you can find the common ratio and translate it into the explicit form of a geometric sequence: an=a1rn−1a_n=a_1 r^{n-1}. Since the terms are four apart? Actually a7a_7 is three steps after a4a_4, so a7=a4r3a_7=a_4 r^3. Substitute: 192=24r3192=24r^3, so r3=8r^3=8 and r=2r=2. Then use a4=24a_4=24: 24=a1(2)3=8a124=a_1(2)^3=8a_1, so a1=3a_1=3. Therefore, an=3(2)n−1a_n=3(2)^{n-1}. Check: at n=4n=4, 3(2)3=243(2)^3=24; at n=7n=7, 3(2)6=1923(2)^6=192. The choice 3(2)n3(2)^n and the choice 6(2)n−16(2)^{n-1} are actually the same sequence; both are twice the correct formula, giving a4=48a_4=48 instead of 24. They trap students who write a1rna_1 r^n instead of a1rn−1a_1 r^{n-1}. The choice 24(2)n−324(2)^{n-3} treats 24 as if it were a1a_1, but 24 is a4a_4; at n=4n=4 it produces 48, not 24. If you anchor at a4a_4, the correct form would be 24(2)n−424(2)^{n-4}, which simplifies to 3(2)n−13(2)^{n-1}. Study tip: always plug the given term numbers into any formula you choose. If both known values match, the formula is correct. And remember: for a sequence starting at n=1n=1, the exponent is n−1n-1, not nn.