AP Precalculus Flashcards: Change In Arithmetic And Geometric Sequences

What this deck covers

This deck focuses on Change In Arithmetic And Geometric Sequences, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: State the formula for the sum of the first $n$ terms of an arithmetic sequence.

Answer: $S_n = \frac{n}{2} (a_1 + a_n)$. Uses the average of first and last terms, multiplied by number of terms.

Flashcard 2: What is the first term if $a_{5} = 21$ and $d = 5$?

Answer: $1$. From $a_5 = a_1 + 4d = 21$, solve: $a_1 = 21 - 4(5) = 1$.

Flashcard 3: Find the common difference: $7, 10, 13, 16, \text{...}$

Answer: $3$. Subtract consecutive terms: $10 - 7 = 3$.

Flashcard 4: What type of sequence is $a_n = 5n$?

Answer: Arithmetic sequence. Linear form $5n$ indicates constant difference (here $d = 5$).

Flashcard 5: Find the number of terms in the geometric sequence: $1, 3, 9, \text{...}, 81$

Answer: $5$. Solve $81 = 1 \times 3^{n-1}$ to find $n = 5$.

Flashcard 6: Identify the 4th term of the sequence: $3, 9, 27, \text{...}$

Answer: $81$. Common ratio is $3$, so $a_4 = 3 \times 3^{4-1} = 3 \times 27 = 81$.

Flashcard 7: Describe the sequence $7, 21, 63, \text{...}$

Answer: Geometric sequence. Each term is multiplied by $3$ (constant ratio).

Flashcard 8: Which type of sequence is defined by $a_n = 3n + 2$?

Answer: Arithmetic sequence. Linear form $3n + 2$ indicates constant difference between consecutive terms.

Flashcard 9: Calculate the 7th term: $4, 8, 12, 16, \text{...}$

Answer: $28$. Common difference is $4$, so $a_7 = 4 + (7-1) \times 4 = 28$.

Flashcard 10: State the formula for the $n$th term of a geometric sequence.

Answer: $a_n = a_1 \times r^{n-1}$. Start with first term $a_1$, then multiply by $r$ raised to $(n-1)$ power.

Flashcard 11: Find the common ratio: $5, 15, 45, \text{...}$

Answer: $3$. Divide consecutive terms: $15 \div 5 = 3$.

Flashcard 12: If $a_1 = 2$ and $r = 4$, what is the $n$th term of the sequence?

Answer: $a_n = 2 \times 4^{n-1}$. Substitute given values into the geometric sequence formula.

Flashcard 13: State the formula for the $n$th term of an arithmetic sequence.

Answer: $a_n = a_1 + (n-1)d$. Start with first term $a_1$, then add $(n-1)$ times the common difference.

Flashcard 14: Determine the 6th term: $1, 4, 7, 10, \text{...}$

Answer: $16$. Common difference is $3$, so $a_6 = 1 + (6-1) \times 3 = 16$.

Flashcard 15: What is $a_{10}$ if $a_1 = 3$ and $d = 2$ in an arithmetic sequence?

Answer: $21$. Use formula: $a_{10} = 3 + (10-1) \times 2 = 21$.

Flashcard 16: Identify the 5th term of the sequence: $2, 5, 8, 11, \text{...}$

Answer: $14$. Common difference is $3$, so $a_5 = 2 + (5-1) \times 3 = 14$.

Flashcard 17: Find the sum of the first 4 terms: $2, 6, 18, \text{...}$

Answer: $80$. Sum the terms: $2 + 6 + 18 + 54 = 80$.

Flashcard 18: What is the common difference in an arithmetic sequence?

Answer: The constant difference between consecutive terms. Each term is found by adding this same value to the previous term.

Flashcard 19: Find the $n$th term if $a_1 = 6$ and $d = 4$.

Answer: $a_n = 6 + (n-1) \times 4$. Substitute given values into the arithmetic sequence formula.

Flashcard 20: Describe the sequence $9, 14, 19, 24, \text{...}$

Answer: Arithmetic sequence. Each term increases by $5$ (constant difference).

Flashcard 21: Identify the sequence: $5, 9, 13, 17, \text{...}$

Answer: Arithmetic sequence. Each term adds $4$ to the previous term (constant difference).

Flashcard 22: What is the common ratio in a geometric sequence?

Answer: The constant factor between consecutive terms. Each term is found by multiplying the previous term by this same value.

Flashcard 23: What is the first term if $a_{4} = 81$ and $r = 3$?

Answer: $3$. From $a_4 = a_1 \times r^3 = 81$, solve: $a_1 = 81 \div 3^3 = 3$.

Flashcard 24: If $a_1 = 5$ and $d = 3$, what is the $n$th term of the sequence?

Answer: $a_n = 5 + (n-1) \times 3$. Substitute given values into the arithmetic sequence formula.

Flashcard 25: Find the sum of the first 6 terms: $3, 6, 9, 12, \text{...}$

Answer: $63$. Sum formula: $S_6 = \frac{6}{2}(3 + 18) = 3 \times 21 = 63$.

Flashcard 26: What is the sum of the first 4 terms of the sequence: $2, 4, 8, 16, \text{...}$?

Answer: $30$. Add the terms: $2 + 4 + 8 + 16 = 30$.

Flashcard 27: What type of sequence is $a_n = 3 \times 2^n$?

Answer: Geometric sequence. Exponential form $3 \times 2^n$ indicates constant ratio (here $r = 2$).

Flashcard 28: What is the sum of the first 5 terms of the sequence: $1, 3, 5, 7, \text{...}$?

Answer: $25$. Add the terms: $1 + 3 + 5 + 7 + 9 = 25$.

Flashcard 29: Calculate the 5th term: $10, 30, 90, \text{...}$

Answer: $810$. Common ratio is $3$, so $a_5 = 10 \times 3^{5-1} = 10 \times 81 = 810$.

Flashcard 30: Determine the 3rd term: $8, 24, 72, \text{...}$

Answer: $72$. Given sequence shows $a_3 = 72$ directly from the pattern.

Flashcard 31: Which type of sequence is defined by $a_n = 4 \times 2^{n-1}$?

Answer: Geometric sequence. Exponential form $4 \times 2^{n-1}$ indicates constant ratio between consecutive terms.

Flashcard 32: State the formula for the sum of the first $n$ terms of a geometric sequence.

Answer: $S_n = a_1 \frac{1-r^n}{1-r}$, $r \neq 1$. Uses first term times the finite geometric series formula when $r \neq 1$.

Flashcard 33: Find the $n$th term if $a_1 = 7$ and $r = 2$.

Answer: $a_n = 7 \times 2^{n-1}$. Substitute given values into the geometric sequence formula.

Flashcard 34: Calculate the sum: $2, 6, 18, \text{...}, 162$

Answer: $242$. Use geometric sum formula with $a_1 = 2$, $r = 3$, and $n = 5$.

Flashcard 35: Identify the sequence: $10, 20, 40, 80, \text{...}$

Answer: Geometric sequence. Each term doubles the previous term (constant ratio of $2$).

Flashcard 36: How many terms are in the arithmetic sequence: $2, 5, 8, \text{...}, 20$?

Answer: $7$. Solve $20 = 2 + (n-1) \times 3$ to find $n = 7$.