Arithmetic and Geometric Sequences, Exponential Functions, and Series Study Guide

Algebraic Expressions, Exponents, and Radicals

• Key procedures and simplifications for algebraic expressions involving exponents and radicals:

  • Exercise 21: Simplify $3 \times (27x)^{\frac{1}{3}}$

  • Factor the base inside the fractional exponent: $27^{\frac{1}{3}} = 3$

  • Rewrite: $3 \times 3 \times x^{\frac{1}{3}} = 9x^{\frac{1}{3}}$

  • Exercise 22: Simplify $[(5x + 3)(5x - 3)]^{\frac{1}{2}}$

  • Apply the difference of squares identity: $(5x + 3)(5x - 3) = 25x^2 - 9$

  • Express in radical form: (25x^2 - 9)^{\frac{1}{2}} = \root\sqrt{25x^2 - 9}

  • Exercise 23: Simplify $x^{\frac{1}{3}}(x^{\frac{1}{2}} - 1)(x^{\frac{1}{2}} + 1)$

  • Multiply the binomial conjugate pair: $(x^{\frac{1}{2}} - 1)(x^{\frac{1}{2}} + 1) = x - 1$

  • Distribute $x^{\frac{1}{3}}$: $x^{\frac{1}{3}}(x - 1) = x^{\frac{4}{3}} - x^{\frac{1}{3}}$

  • Exercise 24: Simplify $\frac{(x+y)^{-1}}{(x+y)^{-2}}$

  • Apply exponent quotient rule $a^m / a^n = a^{m-n}$: $(x+y)^{-1 - (-2)} = (x+y)^1 = x + y$

  • Exercise 25: Ratio of term expressions: $\frac{a_n}{a_{n+1}}$

  • Exercise 26: Simplify $\left(\frac{u^3 m^{-2}}{7 m u^{-2}}\right)^{-2}$

  • Simplify interior exponents: $\frac{u^{3 - (-2)}}{7 m^{1 - (-2)}} = \frac{u^5}{7m^3}$

  • Apply negative exponent: $\left(\frac{7m^3}{u^5}\right)^2 = \frac{49m^6}{u^{10}}$

  • Exercise 27: Simplify $\left(\frac{2}{7} x^{-3}\right)^{-2}$

  • Distribute negative exponent: $\left(\frac{7}{2}\right)^2 x^6 = \frac{49}{4} x^6$

  • Exercise 28: Simplify $\frac{3^{n+1} + 3^{n+1} + 3^{n+1}}{3^{n+1} + 3^{n+1}}$

  • Combine like terms in numerator and denominator: $\frac{3 \times 3^{n+1}}{2 \times 3^{n+1}} = \frac{3}{2}$

  • Exercise 29: Simplify $\left(\frac{81}{16}\right)^{-\frac{3}{4}} \times \left(\frac{27}{8}\right)^{-\frac{2}{3}}$

  • Evaluate first factor: $\left(\frac{81}{16}\right)^{-\frac{3}{4}} = \left(\frac{3}{2}\right)^{-3} = \left(\frac{2}{3}\right)^3 = \frac{8}{27}$

  • Evaluate second factor: $\left(\frac{27}{8}\right)^{-\frac{2}{3}} = \left(\frac{3}{2}\right)^{-2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$

  • Multiply results: $\frac{8}{27} \times \frac{4}{9} = \frac{32}{243}$

  • Exercise 30: Simplify $-\sqrt{(c - 7d)^2}$

  • Apply principal root definition: $-|c - 7d|$

  • Exercise 31: Simplify $\sqrt{1 - 6c + 9c^2}$

  • Perfect square trinomial factorization: $\sqrt{(1 - 3c)^2} = |1 - 3c|$

  • Exercise 32: Simplify $\pm\sqrt{(4e - 1)^2}$

  • Simplify radical: $\pm(4e - 1)$

  • Exercise 33: Simplify $ Kud\sqrt{(8 + \sqrt{3})(8 - \sqrt{3})}\n - Multiply under radical: \sqrt{64 - 3} = \sqrt{61}\n\n - Exercise 34: Simplify \sqrt{4a^2 - 20a + 25}\n - Perfect square trinomial factorization: \sqrt{(2a - 5)^2} = |2a - 5|\n\n - Exercise 36: Simplify \sqrt[3]{m^{12}(n - 5)^6}\n - Convert to rational exponents: m^{\frac{12}{3}} (n - 5)^{\frac{6}{3}} = m^4 (n - 5)^2\n\n - Exercise 37: Simplify \sqrt{\frac{4m - 20}{m^2 - 2m - 15}}\n - Factor numerator and denominator: \sqrt{\frac{4(m - 5)}{(m - 5)(m + 3)}} = \sqrt{\frac{4}{m + 3}} = \frac{2}{\sqrt{m + 3}}\n\n\n# Arithmetic and Geometric Sequences\n\n- Definitions and formulas:\n - Arithmetic Sequence Explicit Rule: a_n = a_1 + d(n - 1) = dn + a_0\n - Geometric Sequence Explicit Rule: a_n = a_1 r^{n - 1}\n - Common Difference: d = a_{n} - a_{n-1}\n - Common Ratio: r = \frac{a_n}{a_{n-1}}\n\n- Exercises 1–4: Arithmetic Sequences\n - Exercise 1: Sequence 6, 10, 14, 18, \dots\n - a) Common difference: d = 10 - 6 = 4\n - b) 10th term: a_{10} = 6 + 4(10 - 1) = 6 + 36 = 42\n - c) Explicit rule: a_n = 6 + 4(n - 1) = 4n + 2\n\n - Exercise 2: Sequence -4, 1, 6, 11, \dots\n - a) Common difference: d = 1 - (-4) = 5\n - b) 10th term: a_{10} = -4 + 5(10 - 1) = -4 + 45 = 41\n - c) Explicit rule: a_n = -4 + 5(n - 1) = 5n - 9\n\n - Exercise 3: Sequence -5, -2, 1, 4, \dots\n - a) Common difference: d = -2 - (-5) = 3\n - b) 10th term: a_{10} = -5 + 3(10 - 1) = -5 + 27 = 22\n - c) Explicit rule: a_n = -5 + 3(n - 1) = 3n - 8\n\n - Exercise 4: Sequence 36, 25, 14, 3, \dots\n - a) Common difference: d = 25 - 36 = -11\n - b) 10th term: a_{10} = 36 - 11(10 - 1) = 36 - 99 = -63\n - c) Explicit rule: a_n = 36 - 11(n - 1) = -11n + 47\n\n- Exercises 5–8: Geometric Sequences\n - Exercise 5: Sequence 2, 6, 18, 54, \dots\n - a) Common ratio: r = \frac{6}{2} = 3\n - b) 7th term: a_7 = 2 \times 3^{7 - 1} = 2 \times 3^6 = 2 \times 729 = 1458\n - c) Explicit rule: a_n = 2 \times 3^{n - 1}\n\n - Exercise 6: Sequence 3, 6, 12, 24, \dots\n - a) Common ratio: r = \frac{6}{3} = 2\n - b) 7th term: a_7 = 3 \times 2^{7 - 1} = 3 \times 2^6 = 3 \times 64 = 192\n - c) Explicit rule: a_n = 3 \times 2^{n - 1}\n\n - Exercise 7: Sequence 1, -2, 4, -8, 16, \dots\n - a) Common ratio: r = \frac{-2}{1} = -2\n - b) 7th term: a_7 = 1 \times (-2)^{7 - 1} = (-2)^6 = 64\n - c) Explicit rule: a_n = 1 \times (-2)^{n - 1} = (-2)^{n - 1}\n\n - Exercise 8: Sequence -2, 2, -2, 2, \dots\n - a) Common ratio: r = \frac{2}{-2} = -1\n - b) 7th term: a_7 = -2 \times (-1)^{7 - 1} = -2 \times (-1)^6 = -2\n - c) Explicit rule: a_n = -2 \times (-1)^{n - 1}\n\n- Exercise 9: Arithmetic sequence with a_3 = -8$and$a_6 = 4\n - Find common difference d$:$d = \frac{a_6 - a_3}{6 - 3} = \frac{4 - (-8)}{3} = \frac{12}{3} = 4\n - Find a_0$:$a_3 = 4(3) + a_0 = -8 \implies 12 + a_0 = -8 \implies a_0 = -20\n - Explicit rule: a_n = 4n - 20\n\n- Exercise 10: Geometric sequence with a_1 = 3$and$a_7 = 192\n - Find common ratio r$:$a_7 = a_1 r^6 \implies 192 = 3 r^6 \implies r^6 = 64 \implies r = 2\n - Find a_0$:$a_0 = \frac{a_1}{r} = \frac{3}{2} = 1.5\n - Explicit rule: a_n = 3 \times 2^{n - 1}\n\n\n# Summation Notation, Series, and Infinite Sums\n\n- Table-Based Sequence Analysis (Problems 11–12):\n - Problem 11: Growth of the bungy-gungy tree in the Amazon rain forest\n - Data: Time in weeks x = [0, 1, 2, 3, 4, 5]$, Height in cm$y = [700, 702.3, 704.6, 706.9, 709.2, 711.5]\n - Constant rate of change: 702.3 - 700 = 2.3\,cm/week\n - a) Function type: Linear\n - b) Equation: y = 2.3x + 700\n\n - Problem 12: Half-life decay of Thorium-232\n - Data: Number of half-lives x = [0, 1, 2, 3, 4, 5]$, Mass in grams$y = [16, 8, 4, 2, 1, 0.5]\n - Constant ratio: \frac{8}{16} = 0.5\n - a) Function type: Exponential\n - b) Equation: y = 16 \times (0.5)^x\n\n- Evaluation of Series using Summation Notation (Problems 13–16):\n - Problem 13: -7 - 1 + 5 + 11 + \dots + 53\n - Pattern: Arithmetic with a_1 = -7$and$d = 6\n - General term: a_n = -7 + 6(n - 1) = 6n - 13\n - Number of terms: 6n - 13 = 53 \implies 6n = 66 \implies n = 11\n - Summation notation: \sum_{n=1}^{11} (6n - 13)\n - Sum evaluation: S_{11} = \frac{11}{2}(-7 + 53) = \frac{11}{2}(46) = 253\n\n - Problem 14: 2 + 5 + 8 + 11 + \dots + 29\n - Pattern: Arithmetic with a_1 = 2$and$d = 3\n - General term: a_n = 3n - 1\n - Number of terms: 3n - 1 = 29 \implies 3n = 30 \implies n = 10\n - Summation notation: \sum_{n=1}^{10} (3n - 1)\n - Sum evaluation: S_{10} = \frac{10}{2}(2 + 29) = 5 \times 31 = 155\n\n - Problem 15: 6 - 12 + 24 - 48 + \dots\n - Pattern: Infinite geometric series with a_1 = 6$and$r = -2\n - Convergence test: |r| = |-2| = 2 \ge 1, series diverges\n - Summation notation: \sum_{n=1}^{\infty} 6 \times (-2)^{n-1}\n - Sum evaluation: DNE (Does Not Exist)\n\n - Problem 16: 80 + 60 + 45 + 33.75 + \dots\n - Pattern: Infinite geometric series with a_1 = 80$and$r = \frac{60}{80} = 0.75\n - Convergence test: |r| = 0.75 < 1, series converges\n - Summation notation: \sum_{n=1}^{\infty} 80 \times (0.75)^{n-1}\n - Infinite sum formula: S_\infty = \frac{a_1}{1 - r} = \frac{80}{1 - 0.75} = \frac{80}{0.25} = 320\n\n- Converting Sequences to Summation Notation and Exact Sums (Problems 17–18):\n - Problem 17: 2, 5, 8, \dots$for$n = 10\n - Formula for n\text{th}$term:$a_n = 3n - 1\n - 10th term: a_{10} = 3(10) - 1 = 29\n - Summation notation: \sum_{n=1}^{10} (3n - 1)\n - Exact sum: S_{10} = \frac{10}{2}(2 + 29) = 155\n\n - Problem 18: 4, -2, 1, \dots$for$n = 12\n - Formula for n\text{th}$term:$a_n = 4 \times \left(-\frac{1}{2}\right)^{n-1}\n - 12th term: a_{12} = 4 \times \left(-\frac{1}{2}\right)^{11} = -\frac{1}{512}\n - Summation notation: \sum_{n=1}^{12} 4 \times \left(-\frac{1}{2}\right)^{n-1}\n - Exact sum formula: S_{12} = \frac{4\left(1 - (-\frac{1}{2})^{12}\right)}{1 - (-\frac{1}{2})} = \frac{4\left(1 - \frac{1}{4096}\right)}{\frac{3}{2}} = \frac{8}{3} \times \frac{4095}{4096} = \frac{1365}{512}\n\n\n# Continuous Functions and AP Test Preparation\n\n- Discrete Sequences to Continuous Functions (Problems 19–20):\n - Problem 19: Sequence 2, \frac{5}{2}, \frac{25}{8}, \frac{125}{32}, \dots\n - a) Common ratio: r = \frac{5/2}{2} = \frac{5}{4} = 1.25\n - b) Function type: Exponential\n - c) Continuous equation: y = 2 \times (1.25)^{x-1}\n\n - Problem 20: Sequence 3.14, 2.9, 2.66, 2.42, \dots\n - a) Common difference: d = 2.9 - 3.14 = -0.24\n - b) Function type: Linear\n - c) Continuous equation: y = -0.24x + 3.38\n\n- Infinite Series Convergence Tests (Problems 21–24):\n - Problem 21: \sum_{n=1}^{\infty} 2 \times (1.5)^n\n - Analysis: Geometric series ratio r = 1.5 \ge 1. Diverges.\n\n - Problem 22: -3 + 3 - 3 + 3 - 3 + \dots\n - Analysis: Geometric series with r = -1$,$|r| = 1 \ge 1. Diverges.\n\n - Problem 23: \sum_{n=1}^{\infty} (5 - 4n)\n - Analysis: Arithmetic series with non-zero limit. Diverges.\n\n - Problem 24: 3 + 0.03 + 0.003 + 0.0003 + \dots\n - Analysis: Series split into constant term 3$plus geometric series$\sum_{n=1}^{\infty} 0.03 \times (0.1)^{n-1}.\n - Ratio r = 0.1 < 1, series converges.\n - Sum calculation: S = 3 + \frac{0.03}{1 - 0.1} = 3 + \frac{0.03}{0.9} = 3 + \frac{1}{30} = \frac{91}{30}\n\n- AP Test Prep Questions:\n - Question 1: The first two terms of an arithmetic sequence are 2$and$8. What is the fourth term?\n - Options: a. 20$, b.$26$, c.$64$, d.$128\n - Solution: Common difference d = 8 - 2 = 6$. Term$a_4 = 2 + 3(6) = 20\n - Correct Choice: **a. 20**\n\n - Question 2: A geometric sequence begins with 2, 6, \dots$. What is$a_5?\n - Options: a. 3$, b.$4$, c.$9$, d.$162\n - Solution: Common ratio r = \frac{6}{2} = 3$. Term$a_5 = 2 \times 3^4 = 2 \times 81 = 162\n - Correct Choice: **d. 162**\n\n\n# Exponential Inequalities, Identities, and Graphs\n\n\n\n- Solving Exponential Inequalities (Problems 23–25):\n - Problem 23: Solve 9^x < 4\n - Apply natural logarithm: \ln(9^x) < \ln(4) \implies x \ln(9) < \ln(4)\n - Isolating x$:$x < \frac{\ln(4)}{\ln(9)} = \frac{\ln(2^2)}{\ln(3^2)} = \frac{\ln(2)}{\ln(3)} \approx 0.631\n - Solution interval: (-\infty, 0.631)\n\n - Problem 24: Solve 6^{-x} > 8^{-x}\n - Express with positive exponent bases: \left(\frac{1}{6}\right)^x > \left(\frac{1}{8}\right)^x\n - Compare bases: Since \frac{1}{6} > \frac{1}{8}$,$\left(\frac{1}{6}\right)^x > \left(\frac{1}{8}\right)^x$holds for all$x > 0\n - Solution interval: (0, \infty)\n\n - Problem 25: Solve \left(\frac{1}{4}\right)^x > \left(\frac{1}{3}\right)^x\n - Compare bases: Since \frac{1}{4} < \frac{1}{3}$, the fraction$\left(\frac{1}{4}\right)^x$is larger when powers are negative ($x < 0\n - Solution interval: (-\infty, 0)\n\n- Exponential Identities Proofs (Problems 26–27):\n - Problem 26: Show that y_1 = 3^{2x+4}$and$y_2 = 9^{x+2} are identical.\n - Proof:\n      y_2 = 9^{x+2} = (3^2)^{x+2} = 3^{2(x+2)} = 3^{2x+4} = y_1\n\n - Problem 27: Show that y_1 = 2 \times 2^{3x-2}$and$y_2 = 2^{3x-1} are identical.\n - Proof:\n      y_1 = 2^1 \times 2^{3x-2} = 2^{1 + (3x - 2)} = 2^{3x-1} = y_2\n\n- Graphing Exponential Functions (Problem 28):\n - Problem 28a: Graph y = e^x\n - Point at x = 0$:$(0, 1)\n - Point at x = 1$:$(1, e) \approx (1, 2.718)\n - Horizontal Asymptote: y = 0\n\n - Problem 28b: Graph y = 2e^{x-3} + 1\n - Transformations: Shift right by 3$units, vertical stretch by factor of$2$, shift up by$1 unit\n - Key reference point at x = 3$:$y = 2e^0 + 1 = 3 \implies (3, 3)\n - Horizontal Asymptote: y = 1\n\n\n# Exponential Modeling and Word Problems\n\n\n\n- Determining Exponential Formulas from Graphs (Problems 13–14):\n - Problem 13: Graph passing through (0, 4)$and$(1, 50)\n - Form: y = a \times b^x\n - Initial value at x = 0$:$a = 4\n - Base calculation using (1, 50)$:$50 = 4 \times b^1 \implies b = 12.5\n - Formula: y = 4 \times (12.5)^x\n\n - Problem 14: Graph passing through (0, 3)$and$(4, 1.49)\n - Form: y = a \times b^x\n - Initial value at x = 0$:$a = 3\n - Base calculation using (4, 1.49)$:$1.49 = 3 \times b^4 \implies b^4 = 0.496667 \implies b = (0.496667)^{\frac{1}{4}} \approx 0.839\n - Formula: y = 3 \times (0.839)^x\n\n- Real-World Applications:\n - Problem 15: Jacksonville, Florida Population Growth\n - Initial population in 2020 (t = 0$):$736000\n - Annual growth rate: 1.49\% = 0.0149\n - a) Population equation: f(t) = 736000 \times (1.0149)^t\n - b) Predicted population in 2050 (t = 30):\n      f(30) = 736000 \times (1.0149)^{30} \approx 736000 \times 1.55836 \approx 1146953\n - c) Time to reach 1 million (f(t) = 1000000):\n      1000000 = 736000 \times (1.0149)^t \implies 1.358696 = (1.0149)^t\n      t = \frac{\ln(1.358696)}{\ln(1.0149)} \approx 20.72\,\text{years}\n      The population reaches 1 million during the year 2040 (or 2041).\n\n - Problem 16: Radioactive Substance Decay\n - Half-life: 14\,\text{days}$, Initial mass:$6.6\,\text{grams}\n - a) Amount remaining function: A(t) = 6.6 \times \left(\frac{1}{2}\right)^{\frac{t}{14}}\n - b) Time until less than 1\,\text{gram} remaining:\n      1 = 6.6 \times \left(\frac{1}{2}\right)^{\frac{t}{14}} \implies \left(\frac{1}{2}\right)^{\frac{t}{14}} = \frac{1}{6.6} \approx 0.151515\n      \frac{t}{14} \ln(0.5) = \ln(0.151515) \implies \frac{t}{14} = 2.7225 \implies t \approx 38.11\,\text{days}\n\n - Problem 17: Bacterial Growth Culture\n - Growth model: B(t) = 100 e^{0.693t}\n - a) Initial amount (t = 0):\n      B(0) = 100 e^0 = 100\,\text{bacteria}\n - b) Time to reach 200 bacteria:\n      200 = 100 e^{0.693t} \implies 2 = e^{0.693t} \implies \ln(2) = 0.693t\n      t = \frac{\ln(2)}{0.693} \approx \frac{0.693147}{0.693} \approx 1\,\text{hour}\n\n - Problem 18: Carbon-14 Decay & Half-Life Estimation\n - Decay model: C(t) = 20 e^{-0.0001216 t}\n - Initial mass (t = 0$):$C(0) = 20\,\text{grams}\n - Half-life calculation (C = 10\n      10 = 20 e^{-0.0001216 t} \implies 0.5 = e^{-0.0001216 t}\n      \ln(0.5) = -0.0001216 t \implies -0.693147 = -0.0001216 t\n      t = \frac{0.693147}{0.0001216} \approx 5700\,\text{years}$$