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: 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 2: 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 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 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 5: 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 6: 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 7: 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 8: 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 9: 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 10: 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 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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 19: 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 20: 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 21: 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 22: 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 23: 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 24: 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 25: 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 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 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 28: 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 29: 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 30: 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.