Algebra 2 Flashcards: Arithmetic And Geometric Sequences As Functions

What this deck covers

This deck focuses on Arithmetic And Geometric Sequences As Functions, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

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: Model: A population starts at $800$ and grows by $6\%$ each year. What is $a_n$?

Answer: $a_n=800\cdot 1.06^{n-1}$. Geometric sequence: each year multiplied by growth factor $1.06$.

Flashcard 2: What is the explicit formula for the arithmetic sequence with $a_1=12$ and $d=-3$?

Answer: $a_n=12+(n-1)(-3)$. Uses the standard explicit formula with $a_1=12$ and $d=-3$.

Flashcard 3: Find the common difference $d$ if an arithmetic sequence has $a_3=10$ and $a_8=35$.

Answer: $d=5$. Using $d=\frac{a_m-a_k}{m-k}$: $d=\frac{35-10}{8-3}=\frac{25}{5}=5$.

Flashcard 4: What is the explicit formula for the geometric sequence with $a_1=2$ and $r=\frac{1}{4}$?

Answer: $a_n=2\left(\frac{1}{4}\right)^{n-1}$. Uses the standard explicit formula with $a_1=2$ and $r=\frac{1}{4}$.

Flashcard 5: Find $a_7$ for the arithmetic sequence $a_n=2+5(n-1)$.

Answer: $a_7=32$. Using $a_n=2+5(n-1)$: $a_7=2+5(6)=2+30=32$.

Flashcard 6: Model: A bouncing ball reaches $120$ cm, then $75\%$ of the previous height each bounce. What is $a_n$?

Answer: $a_n=120\cdot 0.75^{n-1}$. Geometric sequence: each bounce multiplied by factor $0.75$.

Flashcard 7: Model: A car value is \18{,}000$and depreciates$12%$yearly. What is$a_n$?

Answer: $a_n=18000\cdot 0.88^{n-1}$. Geometric sequence: each year multiplied by decay factor $0.88$.

Flashcard 8: Model: A salary starts at \40{,}000$and increases by$$1{,}500$yearly. What is$a_n$?

Answer: $a_n=40000+1500(n-1)$. Arithmetic sequence: constant yearly increase of \1{,}500$.

Flashcard 9: Translate to explicit form: $a_1=5$, $a_n=\frac{1}{2}a_{n-1}$ for $n\ge^2$.

Answer: $a_n=5\left(\frac{1}{2}\right)^{n-1}$. Converting recursive to explicit: $r=\frac{1}{2}$, so $a_n=5\left(\frac{1}{2}\right)^{n-1}$.

Flashcard 10: Use $a_n=a_k\cdot r^{n-k}$: If $a_3=16$ and $r=\frac{1}{2}$, what is $a_7$?

Answer: $a_7=1$. Using the general form: $a_7=16\cdot\left(\frac{1}{2}\right)^{7-3}=16\cdot\frac{1}{16}=1$.

Flashcard 11: What is the explicit formula for an arithmetic sequence with first term $a_1$ and common difference $d$?

Answer: $a_n=a_1+(n-1)d$. Standard explicit form adds $(n-1)$ copies of $d$ to the first term.

Flashcard 12: Find the explicit formula if an arithmetic sequence has $a_1=-1$ and $a_5=11$.

Answer: $a_n=-1+3(n-1)$. From $a_1=-1$ and $a_5=11$: $d=\frac{11-(-1)}{5-1}=3$.

Flashcard 13: Identify the missing term to keep it arithmetic: $2,\,\square,\,10,\,14$.

Answer: $6$. Arithmetic mean of $2$ and $10$ gives the middle term.

Flashcard 14: Find $a_5$ for the geometric sequence with $a_1=3$ and $r=2$.

Answer: $a_5=48$. Using $a_n=a_1\cdot r^{n-1}$: $a_5=3\cdot 2^{5-1}=3\cdot 16=48$.

Flashcard 15: What is the explicit formula for an arithmetic sequence written from term $a_k$ instead of $a_1$?

Answer: $a_n=a_k+(n-k)d$. General explicit form starting from any known term $a_k$.

Flashcard 16: Find $r$ if a geometric sequence has $a_n=10\cdot\left(\frac{3}{2}\right)^{n-1}$.

Answer: $r=\frac{3}{2}$. The base of the exponent gives the common ratio directly.

Flashcard 17: What is the explicit formula for a geometric sequence written from term $a_k$ instead of $a_1$?

Answer: $a_n=a_k\cdot r^{n-k}$. General explicit form starting from any known term $a_k$.

Flashcard 18: Find the geometric mean between $4$ and $36$ (assume positive).

Answer: $12$. Using $\sqrt{ab}$: $\sqrt{4\cdot 36}=\sqrt{144}=12$.

Flashcard 19: What is the formula for the common ratio $r$ using two terms $a_k$ and $a_m$ of a geometric sequence?

Answer: $r=\left(\frac{a_m}{a_k}\right)^{\frac{1}{m-k}}$. Ratio formula using the $(m-k)$th root of the term quotient.

Flashcard 20: Identify whether $a_n=2n+1$ defines an arithmetic or geometric sequence.

Answer: Arithmetic. Linear form $a_n=2n+1$ has constant differences between consecutive terms.

Flashcard 21: Find and correct the error: Arithmetic explicit written as $a_n=a_1+nd$.

Answer: Correct: $a_n=a_1+(n-1)d$. The exponent should be $(n-1)$, not $n$, for standard form.

Flashcard 22: Find the recursive form if a geometric sequence has $a_1=10$ and $a_2=-5$.

Answer: $a_1=10$; $a_n=-\frac{1}{2}a_{n-1}$. From consecutive terms: $r=\frac{a_2}{a_1}=\frac{-5}{10}=-\frac{1}{2}$.

Flashcard 23: Identify the arithmetic sequence: $3,7,11,15,\dots$; what is the common difference $d$?

Answer: $d=4$. Each term increases by $4$: $7-3=4$, $11-7=4$, etc.

Flashcard 24: What is the recursive definition for the geometric sequence with $a_1=9$ and $r=-2$?

Answer: $a_1=9$; $a_n=-2\,a_{n-1}$. Standard recursive form with given first term and ratio.

Flashcard 25: Translate to recursive form: $a_n=8\cdot(-3)^{n-1}$ with $n\ge^1$.

Answer: $a_1=8$; $a_n=-3\,a_{n-1}$. Converting explicit to recursive: $a_1=8$ and $r=-3$.

Flashcard 26: Find $a_1$ for a geometric sequence with $a_3=36$ and $r=3$.

Answer: $a_1=4$. Working backwards: $a_1=\frac{a_3}{r^2}=\frac{36}{3^2}=\frac{36}{9}=4$.

Flashcard 27: Translate to explicit form: $a_1=7$, $a_n=a_{n-1}-2$ for $n\ge^2$.

Answer: $a_n=7+(n-1)(-2)$. Converting recursive to explicit: $d=-2$, so $a_n=7+(n-1)(-2)$.

Flashcard 28: Identify which is geometric: $a_n=12-3n$ or $a_n=12\left(-3\right)^{n-1}$?

Answer: $a_n=12\left(-3\right)^{n-1}$. Exponential function has constant ratios; linear has constant differences.

Flashcard 29: Find $a_6$ for the arithmetic sequence with $a_1=4$ and $d=7$.

Answer: $a_6=39$. Using $a_n=a_1+(n-1)d$: $a_6=4+(6-1)(7)=4+35=39$.

Flashcard 30: Identify which is arithmetic: $a_n=5n-2$ or $a_n=5\cdot 2^{n-1}$?

Answer: $a_n=5n-2$. Linear function has constant differences; exponential has constant ratios.

Flashcard 31: Identify the geometric sequence: $5,15,45,135,\dots$; what is the common ratio $r$?

Answer: $r=3$. Each term is multiplied by $3$: $15÷5=3$, $45÷15=3$, etc.

Flashcard 32: Find the arithmetic mean between $8$ and $20$.

Answer: $14$. Using $\frac{a+b}{2}$: $\frac{8+20}{2}=\frac{28}{2}=14$.

Flashcard 33: What is the explicit formula for a geometric sequence with first term $a_1$ and common ratio $r$?

Answer: $a_n=a_1\cdot r^{n-1}$. Standard explicit form multiplies $a_1$ by $r$ raised to power $(n-1)$.

Flashcard 34: What is the arithmetic mean formula that gives the middle term between $a$ and $b$ in an arithmetic sequence?

Answer: $\frac{a+b}{2}$. Average of two consecutive terms in an arithmetic sequence.

Flashcard 35: Model: You save $50 the first week and increase by $10 weekly. What is $a_n$?

Answer: $a_n=50+10(n-1)$. Arithmetic sequence: constant weekly increase of $10.

Flashcard 36: Identify whether $a_n=7\cdot 5^{n}$ defines an arithmetic or geometric sequence.

Answer: Geometric. Exponential form $a_n=7\cdot 5^n$ has constant ratios between consecutive terms.

Flashcard 37: Find $a_6$ for the geometric sequence $a_n=\left(-2\right)\cdot 3^{n-1}$.

Answer: $a_6=-486$. Using $a_n=(-2)\cdot 3^{n-1}$: $a_6=(-2)\cdot 3^5=(-2)\cdot 243=-486$.

Flashcard 38: What is the recursive definition for the arithmetic sequence with $a_1=-5$ and $d=6$?

Answer: $a_1=-5$; $a_n=a_{n-1}+6$. Standard recursive form with given first term and difference.

Flashcard 39: What is the recursive definition for an arithmetic sequence with first term $a_1$ and difference $d$?

Answer: $a_1$ given; $a_n=a_{n-1}+d$ for $n\ge^2$. Each term equals the previous term plus the common difference.

Flashcard 40: What is the formula for the common difference $d$ using two terms $a_k$ and $a_m$ of an arithmetic sequence?

Answer: $d=\frac{a_m-a_k}{m-k}$. Difference formula: change in terms divided by change in positions.

Flashcard 41: Which option is the common ratio for the geometric sequence $-4,12,-36,108,\dots$?

Answer: $r=-3$. Each term is multiplied by $-3$: $12÷(-4)=-3$.

Flashcard 42: Find and correct the error: Geometric explicit written as $a_n=a_1\cdot r^{n}$.

Answer: Correct: $a_n=a_1\cdot r^{n-1}$. The exponent should be $(n-1)$, not $n$, for standard form.

Flashcard 43: Translate to recursive form: $a_n=3+4(n-1)$ with $n\ge^1$.

Answer: $a_1=3$; $a_n=a_{n-1}+4$. Converting explicit to recursive: $a_1=3$ and $d=4$.

Flashcard 44: Find the common ratio $r$ if a geometric sequence has $a_2=6$ and $a_5=48$.

Answer: $r=2$. Using $r=\left(\frac{a_m}{a_k}\right)^{\frac{1}{m-k}}$: $r=\left(\frac{48}{6}\right)^{\frac{1}{3}}=8^{\frac{1}{3}}=2$.

Flashcard 45: Identify the missing term to keep it geometric: $3,\,\square,\,12,\,24$.

Answer: $6$. Geometric mean of $3$ and $12$ gives the middle term.

Flashcard 46: Which option is the common difference for the arithmetic sequence $-2,1,4,7,\dots$?

Answer: $d=3$. Each term increases by the same amount: $1-(-2)=3$.

Flashcard 47: Find the recursive form if an arithmetic sequence has $a_1=9$ and $a_2=4$.

Answer: $a_1=9$; $a_n=a_{n-1}-5$. From consecutive terms: $d=a_2-a_1=4-9=-5$.

Flashcard 48: What is the geometric mean formula that gives the middle positive term between $a$ and $b$ in a geometric sequence?

Answer: $\sqrt{ab}$. Square root of the product of two consecutive positive terms.

Flashcard 49: Find $d$ if an arithmetic sequence has $a_n=18-6(n-1)$.

Answer: $d=-6$. The coefficient of $(n-1)$ gives the common difference directly.

Flashcard 50: Use $a_n=a_k+(n-k)d$: If $a_4=9$ and $d=2$, what is $a_{10}$?

Answer: $a_{10}=21$. Using the general form: $a_{10}=9+(10-4)(2)=9+12=21$.

Flashcard 51: Find the explicit formula if a geometric sequence has $a_1=6$ and $a_4=48$.

Answer: $a_n=6\cdot 2^{n-1}$. From $a_1=6$ and $a_4=48$: $r=\left(\frac{48}{6}\right)^{\frac{1}{3}}=2$.

Flashcard 52: Find $a_1$ for an arithmetic sequence with $a_4=20$ and $d=3$.

Answer: $a_1=11$. Working backwards: $a_1=a_4-3d=20-3(3)=20-9=11$.

Flashcard 53: What is the recursive definition for a geometric sequence with first term $a_1$ and ratio $r$?

Answer: $a_1$ given; $a_n=r\cdot a_{n-1}$ for $n\ge^2$. Each term equals the previous term multiplied by the common ratio.