Pre-Calculus 30S

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Infinite Geometric Series

An infinite geometric series is a geometric series that continues on to positive or negative infinity, denoted with three periods. Geometric series: 1 + 2 + 4 + 8 Infinite geometric series: 1 + 2 + 4 + 8 + … A "sum to infinity" is denoted as S∞, and this occurs when an infinite geometric series becomes convergent. If an infinite geometric sequence is convergent, then, as its number of terms is increasing to positive infinity, the term values are converging into a constant value. Generally the convergent value is 0. This convergent value is considered the finite sum of the infinite geometric series. Infinite Geometric Series, S∞

Notice how, in order for an infinite geometric series to have a finite sum, the common ratio, r, must equal a number between -1 and 1. S∞ = finite sum of an infinite geometric series t₁ = first term r = common ratio Finding the Finite Sum of an Infinite Geometric Series 1. Make sure if the infinite geometric series is valid to have a finite sum by finding its common ratio, r. If the common ratio, r, is in between -1 and 1, then the infinite geometric series has a finite sum. 2. Find the first term, t₁, by using the initial term in the infinite geometric series. 3. With all variables found, substitute them into the infinite geometric series formula and solve for the finite sum.

PRACTICE QUESTIONS (BOOKLET) Page: 68 #3, 5

Given the following infinite geometric series: 32 + 8 + 2 + 0.5 + ... Determine S∞. 1. Make sure if the infinite geometric series is valid to have a finite sum by finding its common ratio, r. If the common ratio, r, is in between -1 and 1, then the infinite geometric series has a finite sum. Finding the common ratio, r, where t₂ = 8, t₁ = 32:

Since 0.25 is in between -1 and 1, the infinite geometric series has a finite sum. 2. Find the first term, t₁, by using the initial term in the infinite geometric series. The initial term of the infinite geometric series is 32, therefore the first term, t₁, is 32. 3. With all variables found, substitute them into the infinite geometric series formula and solve for the finite sum. Substituting values, where t₁ = 32, r = 0.25:

Therefore, the infinite geometric series has the finite sum of approximately 42.667.

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Geometric Series

A geometric series is transforming a geometric sequence into an equation so that it shows the importance of its sum. Geometric sequence: 1, 2, 4, 8, ... Geometric series: 1 + 2 + 4 + 8 + ... Just like geometric sequences, geometric series have infinite geometric series. Geometric series: 1 + 2 + 4 + 8 Infinite geometric series: 1 + 2 + 4 + 8 + ... Geometric Series, Sn

Sn = the sum of the first n terms, said as “sum of first n terms” t₁ = first term r = common ratio n = number of terms in the series Finding the Sum of a Geometric Series 1. Using the geometric series formula, set up the formula with respect to the sum needed to be found. 2. Find the common ratio, r, using the geometric series and common ratio formula, and find the first term, t₁, by using the initial term in the geometric series. 3. With all variables found, substitute them into the geometric series formula and solve for the missing sum value. Finding the First Term, t₁, in a Geometric Series 1. Using the geometric series formula, set up the formula with respect to the sum of the given term and term value. 2. Solve for t₁. Finding the Sum of a Geometric Series if n is Unknown 1. Find the common ratio, r, using the geometric series and common ratio formula, and find the first term, t₁, by using the initial term in the geometric series. 2. Set up the geometric series formula with respect the Sn. 3. Since the last term, a, is equal to tn, use the geometric sequence formula to solve for the number of terms, n. 4. With all variables found, substitute them into the geometric series formula and solve for the missing sum value.

PRACTICE QUESTIONS (BOOKLET) Page: 48 #5, 6, 9, 10 (solve for t₁)

Given the following infinite geometric series: 3 + 12 + 48 + 192 + ... Determine S12. 1. Using the geometric series formula, set up the formula with respect to the sum needed to be found. Since S12 is the sum needed to be found, we will set up the formula with respect to S12.

2. Find the common ratio, r, using the geometric series and common ratio formula, and find the first term, t₁, by using the initial term in the geometric series. Finding the common ratio, where t₂ = 12, t₁ = 3:

Finding the first term, t₁, by looking at the geometric series and seeing the initial term: t₁ = 3 3. With all variables found, substitute them into the geometric series formula and solve for the missing sum value. Substituting values, where r = 4, t₁ = 3:

Therefore, the sum of the first 12 terms is 16777215, or S12 = 16777215.

Given the following information: S14 = 16383 and r = -2 What is the first term, t₁? 1. Using the geometric series formula, set up the formula with respect to the sum of the given term and term value. The given sum is S14 = 16383, so we will set up the formula with respect to S14.

2. Solve for t₁.

Therefore, the first term, t₁, is -3, or t₁ = -3.

Given the following geometric series: -3 - 15 - 75 - ... - 46875 Determine Sn. 1. Find the common ratio, r, using the geometric series and common ratio formula, and find the first term, t₁, by using the initial term in the geometric series. Finding the common ratio, where t₂ = -15, t₁ = -3:

Finding the first term, t₁, by looking at the geometric series and seeing the initial term: t₁ = -3 2. Set up the geometric series formula with respect the Sn. Setting up the formula, where r = 5, t₁ = -3:

3. Since the last term, a, is equal to tn, use the geometric sequence formula to solve for the number of terms, n. Solving for n, where t₁ = -3, r = 5, tn = -46875: tn = t₁rn - 1 -46875 = (-3)(5)n - 1 Dividing both sides by -3: -46875 = (-3)(5)n - 1 15625 = 5n - 1 Finding the exponent to which 5a = 15625 by guessing and checking: 5a = 15625 a = 6 5⁶ = 15625 Using exponent laws to solve for the number of terms, n, where if ax + y = az + w, then x + y = z + w: 15625 = 5n - 1 5⁶ = 5n - 1 6 = n - 1 n = 7 4. With all variables found, substitute them into the geometric series formula and solve for the missing sum value. Substituting values, where n = 7, r = 5, t₁ = -3:

Therefore, the sum of the first 7 terms is -58593, or S₇ = -58593.

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Geometric Sequences

A geometric sequence is a sequence formed when each consecutive term after the first term, t₁, is multiplied by a constant, with the constant being the common ratio, r. A geometric sequence increases or decreases at a higher rate than an arithmetic sequence. An infinite geometric sequence is when the sequence continues on into positive or negative infinity, normally indicated by three periods. Geometric sequence: 1, 2, 4, 8 Infinite geometric sequence: 1, 2, 4, 8, ... Common Ratio, r

An infinite geometric sequence can either be convergent or divergent. If an infinite geometric sequence is convergent, then, as its number of terms is increasing to positive infinity, the term values are converging into a constant value. Generally the convergent value is 0. If an infinite geometric sequence's term values are not approaching a specific constant, such as zero, then it is not convergent. Instead it is divergent. If an infinite geometric sequence's term values are fluctuating from positive and negative, then it is not convergent. Instead it is divergent. Convergent infinite geometric sequence: 0.5, 0.25, 0.125, 0.0625, 0.03125, ... 0 In the sequence, as the number of terms approaches infinity, the term values are approaching the constant, 0. Divergent infinite geometric sequence: 1, 2, 4, 8, 16, 32, ... Divergent infinite geometric sequence: 1, -2, 4, -8, 16, -32, ... Geometric Sequence, tn tn = t₁rn - 1 tn = general term (mostly the "imaginary" last term of the sequence), said as "nth term" t₁ = first term r = common ratio n = number of terms in the sequence Finding a Term Value in a Sequence 1. Using the geometric sequence formula, set up the formula with respect to the term needed to be found. 2. Find the common ratio, r, using the geometric sequence and common ratio formula, and find the first term, t₁, by using the initial term in the geometric sequence. 2.1 If the common ratio, r, is not given, set up the geometric sequence formula with respect to a given term other than the first term, t₁, or the term that is needed to be solved, and solve for the common ratio, r, using exponent laws. 2.1.1. After finding two possible values of the common ratio, r, set up two cases for these two possible values and solve for the missing term value. 3. With all variables found, substitute them into the geometric sequence formula and solve for the missing term value. Finding the Number of Terms in a Sequence 1. Using the geometric sequence formula, set up the formula with respect to the term value given, imagining that that term is the “last term”, tn. If this term value is positive, then the common ratio, r, will be positive too, vice versa. 2. Using exponent laws, solve for the number of terms, n, in the geometric sequence if tn was the last term. If n = a, then tn = tₐ.

PRACTICE QUESTIONS (BOOKLET) Page: 35 #4, 5, 8

Given the following information: t₁ = 7 and t₅ = 567 Determine t₆. 1. Using the geometric sequence formula, set up the formula with respect to the term needed to be found. Since the term needed to be found is t₆, we will set up the formula with respect to t₆. Setting up the formula, where tn = t₆, n = 6: tn = t₁rn - 1 t₆ = t₁r6 - 1 t₆ = t₁r⁵ 2. Find the common ratio, r, using the geometric sequence and common ratio formula, and find the first term, t₁, by using the initial term in the geometric sequence. There is no way of finding the common ratio, r, with the given information, so we skip to step 2.1. 2.1 If the common ratio, r, is not given, set up the geometric sequence formula with respect to a given term other than the first term, t₁, or the term that is needed to be solved, and solve for the common ratio, r, using exponent laws. Since there was no common ratio, r, given, we will use a term given other than the first term, t₁, or the term that is needed to be solved, and solve for the common ratio, r, using exponent laws. In this case, a given term is t₅ = 567, so we will use this term to find the common ratio, r. Setting up the formula, where tn = t₅ = 567, n = 5, t₁ = 7: tn = t₁rn - 1 t₅ = 7r5 - 1 567 = 7r5 - 1 Solving for the common ratio, r, by subtracting 5 and 1: 567 = 7r5 - 1 567 = 7r⁴ Dividing both sides by 7: 567 = 7r⁴ 81 = r⁴ Finding the fourth root of both sides: 81 = r⁴ ⁴√81 = r r = ±3 2.1.1. After finding two possible values of the common ratio, r, set up two cases for these two possible values and solve for the missing term value. We will create two cases for the value of the common ratio, r. Case 1: t₆ when r = 3 t₆ = t₁r⁵ t₆ = (7)(3)⁵ t₆ = (7)(243) t₆ = 1701 Case 1: t₆ when r = -3 t₆ = t₁r⁵ t₆ = (7)(-3)⁵ t₆ = (7)(-243) t₆ = -1701 Therefore, the 6th term, t₆, is either 1701 or -1701, or t₆ = ±1701.

Given the following information: t₁ = 7 and r = ±3 How many terms are in the geometric sequence if the last term is 45927? 1. Using the geometric sequence formula, set up the formula with respect to the term value given, imagining that that term is the “last term”, tn. If this term value is positive, then the common ratio, r, will be positive too, vice versa. Since it already states the last term, tn, is 45927, we will use this value and the common ratio, r, to set up the formula. And since 45927 is a positive number, the common ratio, r, will be positive as well. Setting up the formula, where tn = 45927, t₁ = 7, r = 3: tn = t₁rn - 1 45927 = (7)(3)n - 1 2. Using exponent laws, solve for the number of terms, n, in the geometric sequence if tn was the last term. If n = a, then tn = tₐ. Solving for the number of terms, n, by dividing both sides by 7: 45927 = (7)(3)n - 1 45927 = (3)n - 1 6561 = 3n - 1 Finding the exponent to which 3a = 6561 by guessing and checking: 3a = 3n - 1 a = 8 3⁸ = 6561 Using exponent laws to solve for the number of terms, n, where if ax + y = az + w, then x + y = z + w: 6561 = 3n - 1 3⁸ = 3n - 1 8 = n - 1 n = 9 Therefore, if 45927 was the last term of this geometric sequence, there would be a total of 9 terms, where t₉ = 45927.

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Arithmetic Series

An arithmetic series is transforming an arithmetic sequence into an equation so that it shows the importance of its sum. Arithmetic sequence: 1, 2, 3, 4, ... Arithmetic series: 1 + 2 + 3 + 4 + ... Just like arithmetic sequences, arithmetic series have infinite arithmetic series. Arithmetic series: 1 + 2 + 3 + 4 Infinite arithmetic series: 1 + 2 + 3 + 4 + ... Arithmetic Series, Sn

Sn = the sum of the first n terms, said as "sum of first n terms" tn = general term (mostly the "imaginary" last term of the sequence), said as "nth term" t₁ = first term d = common difference n = number of terms in the series Finding the Sum of an Arithmetic Series 1. Determine the appropriate arithmetic series formula to use by considering the variables given. If the question provides information on Sn and tn, but the question asks for Sₐ, use the second formula to solve for Sₐ. If the question provides an arithmetic series with a known common difference, d, and a known first term, t₁, use the second formula to determine Sn. After determining the appropriate formula, set up this formula with respect to the sum needed to be found. 2. Find the common difference, d, using the arithmetic series and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic series. 2.1 If the common difference, d, or the first term, t₁, are unable to be found, use the first arithmetic series formula with the given tn and Sn value to solve for t₁, or use the second arithmetic series formula to solve for the common difference, d. 3. With all variables found, substitute them into the arithmetic series formula and solve for the missing sum value.

PRACTICE QUESTIONS (BOOKLET) Page: 19 #4, 6, 7, 10

Given the following infinite arithmetic series: -75 - 69 - 63 - 57 - 51 - 45 - ... Determine S₆. 1. Determine the appropriate arithmetic series formula to use by considering the variables given. If the question provides information on Sn and tn, but the question asks for Sₐ, use the second formula to solve for Sₐ. If the question provides an arithmetic series with a known common difference, d, and a known first term, t₁, use the second formula to determine Sn. After determining the appropriate formula, set up this formula with respect to the sum needed to be found. Since we are given an arithmetic series with a known common difference, d, and a known first term, t₁, we will use the second arithmetic series formula. And, since the sum needed to be found is S₆, we will set up the formula with respect to S₆. Setting up the formula, where Sn = S₆, n = 6:

2. Find the common difference, d, using the arithmetic series and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic series. Finding the common difference, d, where t₂ = -69, t₁ = -75: d = t₂ - t₁ d = -69 - (-75) d = -69 + 75 d = 6 Finding the first term, t₁, by looking at the arithmetic series and seeing the initial term: t₁ = -75 3. With all variables found, substitute them into the arithmetic series formula and solve for the missing sum value. Substituting values, where d = 6, t₁ = -75:

Therefore, the sum of the first 6 terms is -360, or S₆ = -360.

Given the following information: S15 = 93.75, d = 0.75, and t15 = 11.5 Determine S₃. 1. Determine the appropriate arithmetic series formula to use by considering the variables given. If the question provides information on Sn and tn, but the question asks for Sₐ, use the second formula to solve for Sₐ. If the question provides an arithmetic series with a known common difference, d, and a known first term, t₁, use the second formula to determine Sn. After determining the appropriate formula, set up this formula with respect to the sum needed to be found. Since S15 and t15 do not relate with S₃, we will use the second formula with respect to S₃. Setting up the formula, where Sn = S₃, n = 3, d = 0.75:

2. Find the common difference, d, using the arithmetic series and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic series. There is no way of finding the first term, t₁, with the given information, so we skip to step 2.1. 2.1 If the common difference, d, or the first term, t₁, are unable to be found, use the first arithmetic series formula with the given tn and Sn value to solve for t₁, or use the second arithmetic series formula to solve for the common difference, d. We will use the second arithmetic series formula to find the first term, t₁. Setting up the formula, where Sn = S15 = 93.75, tn = t15 = 11.5, n = 15:

Solving for the first term, t₁:

3. With all variables found, substitute them into the arithmetic series formula and solve for the missing sum value. Substituting values, where t₁ = 1:

Therefore, the sum of the first 3 terms is 5.25, or S₃ = 5.25.

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Arithmetic Sequences

An arithmetic sequence is when a sequence of real numbers have a constant distance between each consecutive term, where a consecutive term is when two numbers are beside each other and the constant is called the common difference, d. An infinite arithmetic sequence is when the sequence continues on into positive or negative infinity, normally indicated by three periods. Arithmetic sequence: 1, 2, 3, 4 Infinite arithmetic sequence: 1, 2, 3, 4, ... An arithmetic sequence can either be increasing or decreasing by a constant term, d. If the arithmetic sequence is increasing, the common difference, d, is positive. If the arithmetic sequence is decreasing, the common difference, d, is negative. Common Difference, d d = t₂ - t₁ For the following infinite arithmetic sequence: 4, 7, 10, 13, ... The common difference, d, is the following: d = t₂ - t₁ d = 7 - 4 d = 3 If we found the common difference, d, for other consecutive terms, it would still equal 3. This concludes that the sequence is arithmetic. When graphing an arithmetic sequence, it will always be in linear shape, with the points (n, tn). Arithmetic Sequence, tn tn = t₁ + d(n - 1) tn = general term (mostly the "imaginary" last term of the sequence), said as "nth term" t₁ = first term d = common difference n = number of terms in the sequence Finding a Term Value in a Sequence 1. Using the arithmetic sequence formula, set up the formula with respect to the term needed to be found. 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. 2.1. If the common difference, d, and the first term, t₁, are unable to be found, construct and simplify a system of linear equations with the two terms that are given, where the largest term goes on top of the linear system. 2.1.1. Subtract them from each other, where the largest term goes on top of the linear system, eliminating the first term, t₁, enabling you to solve for the common difference, d. 2.1.2. Use any one of the two linear equations from the system to substitute the common difference, d, in, and solve for the first term, t₁. 3. With all variables found, substitute them into the arithmetic sequence formula and solve for the missing term value. Finding the Number of Terms in a Sequence 1. Using the arithmetic sequence formula, set up the formula with respect to the term value given, imagining that that term is the "last term", tn. 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. 3. With all variables found, substitute them into the arithmetic sequence formula and factor the common difference, d, and the expression (n - 1), and solve for the number of terms, n, in the arithmetic sequence if tn was the last term. If n = a, then tn = tₐ.

PRACTICE QUESTIONS (BOOKLET) Page: 8 #4, 6, 10, 13, 15

Given the following infinite arithmetic sequence: -3, 2, 7, 12, ... Determine t20. 1. Using the arithmetic sequence formula, set up the formula with respect to the term needed to be found. Since the term needed to be found is t20, we will set up the formula with respect to t20. Setting up the formula, where tn = t20, n = 20: tn = t₁ + d(n - 1) t20 = t₁ + d(20 - 1) t20 = t₁ + d(19) 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. Finding the common difference, d, where t₂ = 2, t₁ = -3: d = t₂ - t₁ d = 2 - (-3) d = 2 + 3 d = 5 Finding the first term, t₁, by looking at the arithmetic sequence and seeing the initial term: t₁ = -3 3. With all variables found, substitute them into the arithmetic sequence formula and solve for the missing term value. Substituting values, where d = 5, t₁ = -3: t20 = t₁ + d(19) t20 = (-3) + (5)(19) Multiplying 5 and 19: t20 = (-3) + (5)(19) t20 = (-3) + 95 Adding -3 and 95: t20 = (-3) + 95 t20 = 92 Therefore, the 20th term, t20, is 92, or t20 = 92.

Given the following information: t₃ = 81 and t12 = 27 Determine t23. 1. Using the arithmetic sequence formula, set up the formula with respect to the term needed to be found. Since the term needed to be found is t23, we will set up the formula with respect to t23. Setting up the formula, where tn = t23, n = 23: tn = t₁ + d(n - 1) t23 = t₁ + d(23 - 1) t23 = t₁ + d(22) 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. There is no way of finding the common difference, d, or the first term, t₁, with the given information, so we skip to step 2.1. 2.1. If the common difference, d, and the first term, t₁, are unable to be found, construct and simplify a system of linear equations with the two terms that are given. Setting up the formula, where tn = t₃ = 81, n = 81: tn = t₁ + d(n - 1) t₃ = t₁ + d(3 - 1) 81 = t₁ + d(2) 81 = t₁ + 2d Setting up the formula, where tn = t12 = 27, n = 27: tn = t₁ + d(n - 1) t12 = t₁ + d(12 - 1) 27 = t₁ + d(11) 27 = t₁ + 11d 2.1.1. Subtract them from each other, where the largest term goes on top of the linear system, eliminating the first term, t₁, enabling you to solve for the common difference, d. The largest term is t12, so its linear equation will go on top. 27 = t₁ + 11d 81 = t₁ + 2d Subtracting the linear system: 27 = t₁ + 11d 81 = t₁ + 2d -54 = 0 + 9d -54 = 9d Solving for the common difference, d, by dividing both sides by 9: -54 = 9d -6 = d d = -6 2.1.2. Use any one of the two linear equations from the system to substitute the common difference, d, in, and solve for the first term, t₁. Using the first linear equation, t12's, where d = -6, and solving for t₁: 27 = t₁ + 11d 27 = t₁ + 11(-6) Multiplying 11 and -6: 27 = t₁ + 11(-6) 27 = t₁ + (-66) Adding 66 to both sides: 27 = t₁ + (-66) 93 = t₁ t₁ = 93 3. With all variables found, substitute them into the arithmetic sequence formula and solve for the missing term value. Substituting values, where d = -6, t₁ = 93: t23 = t₁ + d(22) t23 = 93 + (-6)(22) Multiplying -6 and 22: t23 = 93 + (-6)(22) t23 = 93 + -132 Adding 93 and -132: t23 = 93 + -132 t23 = -39 Therefore, the 23rd term, t23, is -39, or t23 = -39.

Given the following infinite arithmetic sequence: -3, 2, 7, 12, ... What term in the infinite arithmetic sequence has the value of 212? 1. Using the arithmetic sequence formula, set up the formula with respect to the term value given, imagining that that term is the "last term", tn. Let us imagine that 212 is the value of the last term, tn, therefore making tn = 212. Setting up the formula, where tn = 212: tn = t₁ + d(n - 1) 212 = t₁ + d(n - 1) 2. Find the common difference, d, using the arithmetic sequence and common difference formula, and find the first term, t₁, by using the initial term in the arithmetic sequence. Finding the common difference, d, where t₂ = 2, t₁ = -3: d = t₂ - t₁ d = 2 - (-3) d = 2 + 3 d = 5 Finding the first term, t₁, by looking at the arithmetic sequence and seeing the initial term: t₁ = -3 3. With all variables found, substitute them into the arithmetic sequence formula and factor the common difference, d, and the expression (n - 1), and solve for n. If n = a, then tn = tₐ. Substituting values, where d = 5, t₁ = -3: 212 = t₁ + d(n - 1) 212 = (-3) + (5)(n - 1) Factoring 5 and (n - 1): 212 = (-3) + (5)(n - 1) 212 = (-3) + 5n - 5 Adding -3 and -5: 212 = (-3) + 5n - 5 212 = 5n - 8 Adding 8 to both sides: 212 = 5n - 8 220 = 5n Dividing both sides by 5: 220 = 5n 44 = n n = 44 Therefore, there are 44 terms in the arithmetic sequence if the last term was 212. This means, 212 is the 44th term, and so t44 = 212.

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Functions and Algebra properties of a function preliminaries Sequences and Series arithmetic sequences arithmetic series geometric sequences geometric series infinite geometric series Absolute Value Functions the absolute value of a real number graphing absolute value functions (linear and quadratic) solving absolute value equations Radical Functions simplifying radical expressions adding and subtracting radical expressions multiplying and dividing radical expressions solving radical equations Polynomial Functions factoring properties of quadratic functions graphing quadratic functions solving quadratic equations by factoring solving quadratic equations by graphing solving quadratic equations using the quadratic formula linear inequalities quadratic inequalities systems of equations Rational Functions simplifying rational expressions multiplying and dividing rational expressions adding and subtracting rational expressions solving rational expressions graphing reciprocal functions (linear and quadratic) Trigonometry angles the sine law the cosine law ambiguous cases