Algebra Flashcards: Deriving Applying The Geometric Series Formula

Study Deriving Applying The Geometric Series Formula in Algebra with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

Algebra

Deriving Applying The Geometric Series Formula

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QUESTION
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Convert an annual interest rate RR compounded monthly to periodic rate ii. What is ii?

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ANSWER

i=R12i=\frac{R}{12}. Divide annual rate by 12 for monthly compounding.

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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.

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Flashcard 1: Convert an annual interest rate RR compounded monthly to periodic rate ii. What is ii?

Answer: i=R12i=\frac{R}{12}. Divide annual rate by 12 for monthly compounding.

Flashcard 2: Find S5S_5 for a1=3a_1=-3 and r=2r=-2.

Answer: S5=31(2)51(2)=33S_5=-3\frac{1-(-2)^5}{1-(-2)}=-33. Apply formula with a1=3a_1=-3, r=2r=-2, n=5n=5.

Flashcard 3: What is the sum of the first nn terms if r=1r=1 and first term is a1a_1?

Answer: Sn=na1S_n=na_1. When r=1r=1, all terms equal a1a_1, so sum is nn times a1a_1.

Flashcard 4: Identify the number of terms nn in the geometric series 5+15+45+1355+15+45+135.

Answer: n=4n=4. Count the terms: 5,15,45,1355, 15, 45, 135 gives n=4n=4.

Flashcard 5: Compute nn for a 1515-year loan with monthly payments.

Answer: n=180n=180. Calculate n=15×12=180n=15\times 12=180 monthly payments.

Flashcard 6: Identify the common ratio rr of the geometric sequence 7,21,63,7,21,63,\cdots.

Answer: r=3r=3. Divide consecutive terms: 217=3\frac{21}{7}=3 and 6321=3\frac{63}{21}=3.

Flashcard 7: Find the sum S3S_3 for a1=1a_1=1 and r=10r=10 using the finite sum formula.

Answer: S3=11103110=111S_3=1\frac{1-10^3}{1-10}=111. Apply geometric sum formula with a1=1a_1=1 and r=10r=10.

Flashcard 8: For a loan, what is the common ratio rr in the present value geometric series of discounted payments?

Answer: r=11+ir=\frac{1}{1+i}. Each payment is discounted by factor (1+i)1(1+i)^{-1}.

Flashcard 9: Find a1a_1 if S3=21S_3=21, r=2r=2, and n=3n=3.

Answer: a1=3a_1=3. Use S3=a11r31r=a11812=21S_3=a_1\frac{1-r^3}{1-r}=a_1\frac{1-8}{1-2}=21 to solve.

Flashcard 10: Compute S4S_4 for a1=13a_1=\frac{1}{3} and r=3r=3.

Answer: S4=1313413=403S_4=\frac{1}{3}\frac{1-3^4}{1-3}=\frac{40}{3}. Apply formula with a1=13a_1=\frac{1}{3}, r=3r=3, n=4n=4.

Flashcard 11: What is the first term a1a_1 in the discounted-payment series PMT(1+i)1+\text{PMT}(1+i)^{-1}+\cdots?

Answer: a1=PMT(1+i)1a_1=\text{PMT}(1+i)^{-1}. First payment discounted by one period.

Flashcard 12: Identify the common ratio rr for the series 3,6,12,-3,6,-12,\cdots.

Answer: r=2r=-2. Divide consecutive terms: 63=2\frac{6}{-3}=-2.

Flashcard 13: Compute PMT\text{PMT} if PV=1000PV=1000, i=0.25i=0.25, and n=1n=1.

Answer: PMT=10000.251(1.25)1=1250\text{PMT}=1000\frac{0.25}{1-(1.25)^{-1}}=1250. Solve payment formula: 1000×0.2510.8=12501000\times\frac{0.25}{1-0.8}=1250.

Flashcard 14: Compute S3S_3 for a1=6a_1=6 and r=12r=-\frac{1}{2}.

Answer: S3=61(12)31+12=92S_3=6\frac{1-(-\frac{1}{2})^3}{1+\frac{1}{2}}=\frac{9}{2}. Apply formula with a1=6a_1=6, r=12r=-\frac{1}{2}, and n=3n=3.

Flashcard 15: Find nn if a1=3a_1=3, r=2r=2, and an=96a_n=96.

Answer: n=6n=6. Use an=a1rn1a_n=a_1r^{n-1} to solve: 96=32n196=3\cdot 2^{n-1}.

Flashcard 16: Find a5a_5 for a geometric sequence with a1=2a_1=2 and r=3r=3.

Answer: a5=234=162a_5=2\cdot 3^4=162. Use an=a1rn1a_n=a_1r^{n-1} with n=5n=5, a1=2a_1=2, r=3r=3.

Flashcard 17: What is the common ratio rr in a geometric sequence in terms of consecutive terms?

Answer: r=an+1anr=\frac{a_{n+1}}{a_n}. The ratio of any term to the previous term gives the common ratio.

Flashcard 18: Find and correct the missing condition in Sn=a11rn1rS_n=a_1\frac{1-r^n}{1-r}: when is it valid?

Answer: Valid only when r1r\ne 1. Formula undefined when r=1r=1 due to division by zero.

Flashcard 19: Find the sum S3S_3 for a1=1a_1=1 and r=1r=-1.

Answer: S3=11(1)31(1)=1S_3=1\frac{1-(-1)^3}{1-(-1)}=1. Apply formula: 11(1)31(1)=122=11\frac{1-(-1)^3}{1-(-1)}=1\frac{2}{2}=1.

Flashcard 20: State the formula for the finite geometric sum Sn=a1+a1r++a1rn1S_n=a_1+a_1r+\cdots+a_1r^{n-1} for r1r\ne 1.

Answer: Sn=a11rn1rS_n=a_1\frac{1-r^n}{1-r}. Derived by multiplying by rr and subtracting to eliminate middle terms.

Flashcard 21: What is the last term ana_n in a geometric series expressed using a1a_1, rr, and nn?

Answer: an=a1rn1a_n=a_1r^{n-1}. The final term uses the same geometric sequence formula.

Flashcard 22: What is the nnth term ana_n in terms of a1a_1 and rr for a geometric sequence?

Answer: an=a1rn1a_n=a_1r^{n-1}. General term formula where position determines the power of rr.

Flashcard 23: Find S5S_5 for a1=3a_1=-3 and r=2r=-2.

Answer: S5=31(2)51(2)=33S_5=-3\frac{1-(-2)^5}{1-(-2)}=-33. Apply formula with a1=3a_1=-3, r=2r=-2, n=5n=5.

Flashcard 24: Identify the first term a1a_1 of the geometric series 7+21+63+7+21+63+\cdots.

Answer: a1=7a_1=7. The first term is the initial value in the sequence.

Flashcard 25: Find PVPV if PMT=200\text{PMT}=200, i=0.01i=0.01, and n=2n=2 for an ordinary annuity.

Answer: PV=2001(1.01)20.01PV=200\frac{1-(1.01)^{-2}}{0.01}. Apply present value formula with given payment and rate.

Flashcard 26: Find a1a_1 if a3=20a_3=20 and r=2r=2 for a geometric sequence.

Answer: a1=2022=5a_1=\frac{20}{2^2}=5. Use a3=a1r2a_3=a_1r^2 to solve: 20=a12220=a_1\cdot 2^2.

Flashcard 27: Find S2S_2 for a1=8a_1=8 and r=14r=\frac{1}{4}.

Answer: S2=81(14)2114=10S_2=8\frac{1-(\frac{1}{4})^2}{1-\frac{1}{4}}=10. Apply formula with a1=8a_1=8, r=14r=\frac{1}{4}, n=2n=2.

Flashcard 28: What is the definition of a geometric sequence using first term a1a_1 and ratio rr?

Answer: an=a1rn1a_n=a_1r^{n-1}. Each term is the first term multiplied by rr raised to the term position minus 1.

Flashcard 29: Compute S4S_4 for the geometric series 21+12142-1+\frac{1}{2}-\frac{1}{4}.

Answer: S4=54S_4=\frac{5}{4}. Geometric series with a1=2a_1=2, r=12r=-\frac{1}{2}, n=4n=4.

Flashcard 30: Compute S5S_5 for a1=1a_1=1 and r=12r=\frac{1}{2}.

Answer: S5=11(12)5112=3116S_5=1\frac{1-(\frac{1}{2})^5}{1-\frac{1}{2}}=\frac{31}{16}. Apply formula with a1=1a_1=1, r=12r=\frac{1}{2}, and n=5n=5.

Flashcard 31: What is the key multiplication step used to derive the geometric sum (multiply SnS_n by what)?

Answer: Multiply by rr to form rSnrS_n. Creates alignment to subtract and cancel telescoping middle terms.

Flashcard 32: What is the last term ana_n in a geometric series expressed using a1a_1, rr, and nn?

Answer: an=a1rn1a_n=a_1r^{n-1}. The final term uses the same geometric sequence formula.

Flashcard 33: If a loan has term tt years with monthly payments, what is the number of payments nn?

Answer: n=12tn=12t. Multiply years by 12 payments per year.

Flashcard 34: Compute the sum S4S_4 for the series 5+15+45+1355+15+45+135.

Answer: S4=513413=200S_4=5\frac{1-3^4}{1-3}=200. Apply formula with a1=5a_1=5, r=3r=3, n=4n=4.

Flashcard 35: Find rr if a1=2a_1=2, a4=54a_4=54, and the sequence is geometric.

Answer: r=3r=3. Use a4=a1r3a_4=a_1r^3 to solve: 54=2r354=2r^3, so r3=27r^3=27.

Flashcard 36: Identify the error: using Sn=a11rn1rS_n=a_1\frac{1-r^n}{1-r} when r=1r=1. What is the correct sum?

Answer: Use Sn=na1S_n=na_1 when r=1r=1. When r=1r=1, the geometric sum formula has zero denominator.

Flashcard 37: Compute the monthly rate ii for an APR of 6%6\% compounded monthly.

Answer: i=0.0612=0.005i=\frac{0.06}{12}=0.005. Divide APR by 12: 6%12=0.5%=0.005\frac{6\%}{12}=0.5\%=0.005.

Flashcard 38: Identify the geometric series that represents PVPV of nn payments PMT\text{PMT} at rate ii (ordinary annuity).

Answer: PV=PMTk=1n(1+i)kPV=\text{PMT}\sum_{k=1}^{n}(1+i)^{-k}. Sum of discounted payments using geometric series.

Flashcard 39: In the discounted-payment series, what is the nnth term ana_n?

Answer: an=PMT(1+i)na_n=\text{PMT}(1+i)^{-n}. Last payment discounted by nn periods.

Flashcard 40: Compute S4S_4 for a1=3a_1=3 and r=2r=-2.

Answer: S4=31(2)41(2)=15S_4=3\frac{1-(-2)^4}{1-(-2)}=-15. Use formula with negative ratio r=2r=-2 and n=4n=4.

Flashcard 41: Find the sum S3S_3 for a1=1a_1=1 and r=10r=10 using the finite sum formula.

Answer: S3=11103110=111S_3=1 \frac{1-10^3}{1-10}=111. Apply geometric sum formula with a1=1a_1=1 and r=10r=10.

Flashcard 42: Compute S3S_3 for the geometric series 10+5+2.510+5+2.5.

Answer: S3=17.5S_3=17.5. Direct addition: 10+5+2.5=17.510+5+2.5=17.5.

Flashcard 43: Compute S2S_2 for a1=9a_1=9 and r=23r=\frac{2}{3}.

Answer: S2=91(23)2123=15S_2=9\frac{1-(\frac{2}{3})^2}{1-\frac{2}{3}}=15. Apply formula with a1=9a_1=9, r=23r=\frac{2}{3}, and n=2n=2.

Flashcard 44: Compute S4S_4 for a1=5a_1=5, r=2r=2 using the geometric sum formula.

Answer: S4=512412=75S_4=5\frac{1-2^4}{1-2}=75. Apply the geometric sum formula with n=4n=4 terms.

Flashcard 45: Compute the common ratio rr for discounted payments when i=0.05i=0.05 per period.

Answer: r=11.05r=\frac{1}{1.05}. Common ratio is the discount factor 11+i\frac{1}{1+i}.

Flashcard 46: What condition must hold to divide by (1r)(1-r) when deriving the sum formula?

Answer: r1r\ne 1. Division by zero occurs if r=1r=1, making the formula undefined.

Flashcard 47: Identify the first term a1a_1 of the geometric series 7+21+63+7+21+63+\cdots.

Answer: a1=7a_1=7. The first term is the initial value in the sequence.

Flashcard 48: What is the present value formula for an ordinary annuity with payment PMT\text{PMT}, rate ii, and nn payments?

Answer: PV=PMT1(1+i)niPV=\text{PMT}\frac{1-(1+i)^{-n}}{i}. Geometric series formula for discounted future payments.

Flashcard 49: In the derivation, what expression results from subtracting: SnrSnS_n-rS_n?

Answer: (1r)Sn=a1a1rn(1-r)S_n=a_1-a_1r^n. Factor out (1r)(1-r) from the left side after subtraction.

Flashcard 50: Find the common ratio rr for the sequence 81,27,9,81,27,9,\cdots.

Answer: r=13r=\frac{1}{3}. Divide consecutive terms: 2781=13\frac{27}{81}=\frac{1}{3}.

Flashcard 51: What is the loan payment formula for principal PVPV, periodic rate ii, and nn payments?

Answer: PMT=PVi1(1+i)n\text{PMT}=PV\frac{i}{1-(1+i)^{-n}}. Solve the present value formula for the payment amount.

Flashcard 52: Compute S6S_6 for a1=4a_1=4 and r=12r=\frac{1}{2}.

Answer: S6=41(12)6112=638S_6=4\frac{1-(\frac{1}{2})^6}{1-\frac{1}{2}}=\frac{63}{8}. Apply formula with a1=4a_1=4, r=12r=\frac{1}{2}, and n=6n=6.

Flashcard 53: State an equivalent finite geometric sum formula for SnS_n using denominator (r1)(r-1) for r1r\ne 1.

Answer: Sn=a1rn1r1S_n=a_1\frac{r^n-1}{r-1}. Multiply numerator and denominator by 1-1 to get equivalent form.

Flashcard 54: What is the definition of a geometric sequence using first term a1a_1 and ratio rr?

Answer: an=a1rn1a_n=a_1r^{n-1}. Each term is the first term multiplied by rr raised to the term position minus 1.

Flashcard 55: Compute PVPV if PMT=100\text{PMT}=100, i=0.25i=0.25, and n=1n=1.

Answer: PV=1001(1.25)10.25=80PV=100\frac{1-(1.25)^{-1}}{0.25}=80. Apply present value formula: PMT11.2510.25=80\text{PMT}\frac{1-1.25^{-1}}{0.25}=80.

Flashcard 56: Find rr if S2=12S_2=12, a1=3a_1=3, and the series is geometric with n=2n=2.

Answer: r=3r=3. Use S2=a1(1+r)=3(1+r)=12S_2=a_1(1+r)=3(1+r)=12 to solve for rr.

Flashcard 57: Compute S3S_3 for the geometric series 10+5+2.510+5+2.5.

Answer: S3=17.5S_3=17.5. Direct addition: 10+5+2.5=17.510+5+2.5=17.5.