Algebra 2 Flashcards: Deriving Applying The Geometric Series Formula

What this deck covers

This deck focuses on Deriving Applying The Geometric Series Formula, 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: What is the first term $a_1$ in terms of $a_n$, $r$, and $n$ for a geometric sequence?

Answer: $a_1=\frac{a_n}{r^{n-1}}$. Solve $a_n=a_1r^{n-1}$ for $a_1$ by dividing by $r^{n-1}$.

Flashcard 2: What is $S_n-rS_n$ for a geometric series with first term $a_1$ and ratio $r$?

Answer: $S_n-rS_n=a_1-a_1r^n$. Most terms cancel, leaving only first and last terms.

Flashcard 3: Find $S_5$ for a geometric series with $a_1=\frac{1}{2}$ and $r=4$.

Answer: $\frac{341}{2}$. Apply $S_5=\frac{1}{2}\frac{1-4^5}{1-4}$ with large ratio.

Flashcard 4: Find $S_5$ for $a_1=5$, $r= \frac{1}{2}$, and $n=5$.

Answer: $\frac{155}{16}$. Apply $S_5=5 \frac{1-( \frac{1}{2})^5}{1- \frac{1}{2}}$ with fractional ratio.

Flashcard 5: State the finite geometric series sum formula for $r \neq 1$ using $a_1$, $r$, and $n$ (alternate form).

Answer: $S_n=a_1\frac{r^n-1}{r-1}$. Equivalent form obtained by factoring out $-1$ from numerator and denominator.

Flashcard 6: Find the sum $S_4$ if the first term is $a_1=8$ and the last term is $a_4=1$ with $r=\frac{1}{2}$.

Answer: $15$. Use $S_n=\frac{a_1-a_nr}{1-r}=\frac{8-1\cdot\frac{1}{2}}{1-\frac{1}{2}}$.

Flashcard 7: Find $PV$ if $P=500$, $i=0.05$, and $n=1$ using $PV=P\frac{1-(1+i)^{-n}}{i}$.

Answer: $PV\approx476.19$. Apply present value formula with single payment scenario.

Flashcard 8: Find $S_4$ for the geometric series $6-3+\frac{3}{2}-\frac{3}{4}$.

Answer: $\frac{15}{4}$. Apply formula with $a_1=6$, $r=-\frac{1}{2}$, $n=4$.

Flashcard 9: Find $S_3$ for the geometric series $\frac{1}{3}+\frac{1}{6}+\frac{1}{12}$.

Answer: $\frac{7}{12}$. Apply formula with $a_1=\frac{1}{3}$, $r=\frac{1}{2}$, $n=3$.

Flashcard 10: Find $S_3$ for $a_1=\frac{2}{5}$, $r=\frac{5}{2}$, and $n=3$.

Answer: $\frac{39}{10}$. Apply $S_3=\frac{2}{5}\frac{1-(\frac{5}{2})^3}{1-\frac{5}{2}}$ with ratio $>1$.

Flashcard 11: Identify $r$ if $S_2=12$ and terms are $a_1=4$ and $a_2=4r$ in a geometric series.

Answer: $r=2$. From $S_2=a_1+a_2=4+4r=12$, solve $4r=8$ to get $r=2$.

Flashcard 12: State the finite geometric series sum formula for $r\ne^1$ using $a_1$, $r$, and $n$ (alternate form).

Answer: $S_n=a_1\frac{r^n-1}{r-1}$. Equivalent form obtained by factoring out $-1$ from numerator and denominator.

Flashcard 13: Identify the error: Using $S_n=a_1\frac{1-r^n}{1-r}$ when $r=1$.

Answer: Formula invalid; use $S_n=na_1$ when $r=1$. When $r=1$, the formula has division by zero; use $S_n=na_1$ instead.

Flashcard 14: Find $S_3$ for $a_1=-5$, $r=\frac{1}{2}$, and $n=3$.

Answer: $-\frac{35}{4}$. Apply $S_3=(-5)\frac{1-(\frac{1}{2})^3}{1-\frac{1}{2}}$ with negative first term.

Flashcard 15: Find $S_4$ for $a_1=1$, $r=-1$, and $n=4$.

Answer: $0$. With $r=-1$ and even $n$, alternating terms cancel completely.

Flashcard 16: What is the last term $a_n$ of a geometric sequence in terms of $a_1$, $r$, and $n$?

Answer: $a_n=a_1r^{n-1}$. Formula for the $n$-th term of a geometric sequence.

Flashcard 17: Find the common ratio $r$ for the geometric sequence $4,12,36,\dots$.

Answer: $3$. Each term is 3 times the previous: $\frac{12}{4}=3$.

Flashcard 18: Find $PV$ if $P=100$, $i=0.10$, and $n=2$ using $PV=P\frac{1-(1+i)^{-n}}{i}$.

Answer: $PV\approx173.55$. Apply present value formula with higher interest rate.

Flashcard 19: State the geometric sum formula in terms of first term $a_1$ and last term $a_n$.

Answer: $S_n=\frac{a_1-a_nr}{1-r}$. Substitute $a_n=a_1r^{n-1}$ into the standard formula.

Flashcard 20: What is $S_4$ for the geometric series $81+27+9+3$?

Answer: $120$. Apply $S_4=81\frac{1-(\frac{1}{3})^4}{1-\frac{1}{3}}$ with decreasing terms.

Flashcard 21: What is the sum $1+r+r^2+\cdots+r^{n-1}$ for $r \neq 1$?

Answer: $\frac{1-r^n}{1-r}$. Standard geometric series sum formula for powers of $r$.

Flashcard 22: Find $n$ if $a_1=2$, $r=3$, and $a_n=162$ for a geometric sequence.

Answer: $n=5$. From $a_n=a_1r^{n-1}=162$ with $a_1=2$, $r=3$, solve for $n$.

Flashcard 23: Which symbol is commonly used for the sum of the first $n$ terms of a series?

Answer: $S_n$. Standard notation for partial sums of series.

Flashcard 24: What is the common ratio $r$ in terms of consecutive terms $a_n$ and $a_{n-1}$?

Answer: $r=\frac{a_n}{a_{n-1}}$. Ratio between consecutive terms is constant in geometric sequences.

Flashcard 25: What factor is pulled out when simplifying $S_n-rS_n$ to solve for $S_n$?

Answer: $(1-r)S_n$. Factor $(1-r)$ from the left side to isolate $S_n$.

Flashcard 26: What is the sum of $n$ terms if $r=1$ and each term equals $a_1$?

Answer: $S_n=na_1$. When $r=1$, all terms equal $a_1$, so sum is $n$ times $a_1$.

Flashcard 27: What is the present value of an annuity formula for payment $P$, rate $i$, and $n$ payments?

Answer: $PV=P\frac{1-(1+i)^{-n}}{i}$. Sum of discounted payments using geometric series with $r=(1+i)^{-1}$.

Flashcard 28: What is the definition of a finite geometric series sum $S_n$ in summation notation?

Answer: $S_n=\sum_{k=1}^{n}a_1r^{k-1}$. Sum of terms $a_1r^{k-1}$ from $k=1$ to $n$ in a geometric series.

Flashcard 29: Find $S_4$ for $a_1=-2$, $r=3$, and $n=4$.

Answer: $-80$. Apply $S_4=(-2)\frac{1-3^4}{1-3}=(-2)\frac{-80}{-2}=-80$.

Flashcard 30: What is $rS_n$ if $S_n=a_1+a_1r+\cdots+a_1r^{n-1}$?

Answer: $rS_n=a_1r+a_1r^2+\cdots+a_1r^n$. Multiplying $S_n$ by $r$ shifts each term up one power of $r$.

Flashcard 31: What is the definition of a geometric sequence using first term $a_1$ and ratio $r$?

Answer: $a_n=a_1r^{n-1}$. Each term multiplies the previous by ratio $r$, starting from $a_1$.

Flashcard 32: What is $S_6$ for the geometric series $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}$?

Answer: $\frac{63}{32}$. Apply formula with $a_1=1$, $r=\frac{1}{2}$, $n=6$.

Flashcard 33: Find $S_4$ for $a_1=9$, $r=\frac{1}{3}$, and $n=4$.

Answer: $\frac{40}{3}$. Apply $S_4=9\frac{1-(\frac{1}{3})^4}{1-\frac{1}{3}}$ with small ratio.

Flashcard 34: Find $S_4$ if $a_2=6$, $r=2$, and the series is geometric.

Answer: $45$. Find $a_1$ from $a_2=a_1r=6$, then calculate $S_4$.

Flashcard 35: Find $S_3$ if the geometric sequence has $a_1=2$ and $a_3=18$.

Answer: $26$. Find $r$ from $a_3=a_1r^2=18$, then calculate $S_3$.

Flashcard 36: Find the monthly interest rate $i$ if the annual APR is $6\%$ compounded monthly.

Answer: $i=\frac{0.06}{12}=0.005$. Convert annual rate to monthly by dividing by 12 periods.

Flashcard 37: Identify the common ratio $r$ in the present value series $P+(P)(1+i)^{-1}+\cdots$.

Answer: $r=(1+i)^{-1}$. Each payment is discounted by factor $(1+i)^{-1}$ from previous.

Flashcard 38: Find $S_6$ for $a_1=3$, $r=2$, and $n=6$.

Answer: $189$. Apply $S_6=3\frac{1-2^6}{1-2}=3\frac{-63}{-1}=189$.

Flashcard 39: Find $S_3$ for $a_1=10$, $r=\frac{1}{10}$, and $n=3$.

Answer: $11.1$. Apply $S_3=10\frac{1-(0.1)^3}{1-0.1}$ with decimal ratio.

Flashcard 40: Find $S_5$ if $a_1=2$, $r=3$, and $n=5$.

Answer: $242$. Direct application of $S_5=2\frac{1-3^5}{1-3}$ gives 242.

Flashcard 41: State the finite geometric series sum formula for $r\ne^1$ using $a_1$, $r$, and $n$.

Answer: $S_n=a_1\frac{1-r^n}{1-r}$. Derived by multiplying by $r$, subtracting, and solving for $S_n$.

Flashcard 42: State the finite geometric series sum formula for $r e 1$ using $a_1$, $r$, and $n$.

Answer: $S_n=a_1\frac{1-r^n}{1-r}$. Derived by multiplying by $r$, subtracting, and solving for $S_n$.

Flashcard 43: Find $S_5$ for $a_1=5$, $r=\frac{1}{2}$, and $n=5$.

Answer: $\frac{155}{16}$. Apply $S_5=5\frac{1-(\frac{1}{2})^5}{1-\frac{1}{2}}$ with fractional ratio.

Flashcard 44: Find $a_1$ if $a_4=54$ and $r=3$ for a geometric sequence.

Answer: $2$. From $a_4=a_1r^3=54$ with $r=3$, solve $a_1\cdot27=54$.

Flashcard 45: Find $S_3$ for $a_1=7$, $r=-2$, and $n=3$.

Answer: $7$. Apply $S_3=7\frac{1-(-2)^3}{1-(-2)}$ with alternating signs.

Flashcard 46: What restriction on $r$ is required to use $S_n=a_1\frac{1-r^n}{1-r}$?

Answer: $r \neq 1$. Prevents division by zero in the denominator $(1-r)$.

Flashcard 47: What is $S_5$ for the geometric series $1+2+4+8+16$?

Answer: $31$. Powers of 2 series: $S_5=1\frac{1-2^5}{1-2}=31$.

Flashcard 48: Find $P$ if $PV=1000$, $i=0.01$, and $n=2$ using $P=PV\frac{i}{1-(1+i)^{-n}}$.

Answer: $P\approx507.51$. Substitute values into payment calculation formula.

Flashcard 49: State the geometric sum formula in terms of first term $a_1$ and last term $a_n$ (alternate form).

Answer: $S_n=\frac{a_nr-a_1}{r-1}$. Alternative form by factoring out $-1$ from numerator and denominator.

Flashcard 50: What is $S_5$ for the geometric series $2+6+18+54+162$?

Answer: $242$. Apply $S_5=2\frac{1-3^5}{1-3}$ with $a_1=2$, $r=3$, $n=5$.

Flashcard 51: State the loan payment formula for principal $PV$, monthly rate $i$, and $n$ payments.

Answer: $P=PV\frac{i}{1-(1+i)^{-n}}$. Solve the present value formula for payment $P$.

Flashcard 52: Find $S_3$ using $S_n=\frac{a_1-a_nr}{1-r}$ when $a_1=3$, $a_3=12$, and $r=2$.

Answer: $21$. Substitute known values into alternate sum formula.

Flashcard 53: Find $P$ if $PV=1000$, $i=0.05$, and $n=1$ using $P=PV\frac{i}{1-(1+i)^{-n}}$.

Answer: $P=1050$. With one payment, $P=PV(1+i)$ for simple interest.

Flashcard 54: Find $1+3+9+27+81$ by using $S_n=a_1\frac{1-r^n}{1-r}$.

Answer: $121$. Apply formula with $a_1=1$, $r=3$, $n=5$ for powers of 3.

Flashcard 55: Identify the error: Using $r=1+i$ as the ratio in the present value geometric series for loan payments.

Answer: Use $r=(1+i)^{-1}$ for present value discounting. Present value requires discount factor $r=(1+i)^{-1}$, not growth factor.

Flashcard 56: Find $PV$ if $P=100$, $i=0.01$, and $n=2$ using $PV=P\frac{1-(1+i)^{-n}}{i}$.

Answer: $PV\approx197.04$. Substitute values into present value annuity formula.

Flashcard 57: Find $S_5$ for $a_1=1$, $r=-1$, and $n=5$.

Answer: $1$. With $r=-1$ and odd $n$, alternating terms give net sum of first term.