QA #1 Factors and Multipless

Questions and Answers Sheet 1 Factors and Multiples Question #1 ∂ What is ∂𝑥𝑖 (𝑥𝑖 !) where 𝑥𝑖 is a discrete variable? Do you consider (𝑥𝑖 !) = (𝑥𝑖 )(𝑥𝑖 − 1) ⋯ 1 and do product rule on each term, or something else? Answer: Step 1 The derivative of a function of a discrete variable doesn't really make sense in the typical calculus setting. However, there is a continuous variant of the factorial function called the 𝛾 function, for which you can take derivatives and evaluate the derivative at integer values. In particular, since 𝑛! = 𝛾 (𝑛 + 1) , there is a nice formula for γ′ at integer values: 1 γ′ (𝑛 + 1) = 𝑛! (−γ + ∑𝑛𝑘=1 ) 𝑘 where 𝛾 is the Euler-Mascheroni constant. Step 2 As has been mentioned, the Gamma function \gamma(x) is the way to go. Integration by parts yields ∞ 𝛾 (𝑥) = ∫0 𝑒 −𝑡 𝑡 𝑥−1 𝑑𝑡 ∞ = (𝑥 − 1) ∫0 𝑒 −𝑡 𝑡 𝑥−2 𝑑𝑡 = (𝑥 − 1)𝛾 (𝑥 − 1) Taking the derivative of the logarithm of 𝛾 (𝑥 ) gives γ′ (𝑥) γ(𝑥) = 1 𝑥−1 + γ′ (𝑥−1) γ(𝑥−1) Because 𝛾 (𝑥 ) is log-connvex and lim γ′ (𝑥) 𝑥⇒∞ γ(𝑥) − log(𝑥 ) = 0 we get that γ′ (𝑥) γ(𝑥) 1 1 = −𝛾 + ∑∞ 𝑘=1 (𝑘 − 𝑘+𝑥−1) For integer 𝑛, ! = 𝛾 (𝑛 + 1), so the derivative is 𝑛 𝛾 (𝑛 + 1) = 𝛾 (𝑛 + 1) (−γ + ∑∞ 𝑘=1 𝑘(𝑘+𝑛)) = 𝑛! (−γ + 𝐻𝑛 ) where 𝐻𝑛 is the 𝑛𝑡ℎ Harmonic Number (with the convention that 𝐻0 = 0) Question #2 How do you evaluate 7𝐶2 ? Answer: Explanation: 𝑛! 𝑛𝐶𝑟 = ((𝑛), (𝑟)) = 𝑟!(𝑛−𝑟)! 7! 7! 5040 5040 𝑛𝐶𝑟 = ((7), (2)) = 2!(7−2)! = 2!⋅5! = 2⋅120 = 2040 = 21 Question #3 How do you factor 𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥 − 2 ? Answer: (𝑥 − 2)(𝑥 + 𝑖 )(𝑥 − 𝑖 ) Explanation: factor the terms by grouping = 𝑥 2 (𝑥 − 2 ) + 1 (𝑥 − 2 ) take out the common factor (𝑥 − 2) = (𝑥 − 2)(𝑥 2 + 1) 𝑥 2 + 1 can be factored using difference of squares 𝑎2 − 𝑏2 = (𝑎 − 𝑏)(𝑎 + 𝑏) with 𝑎 = 𝑥 and 𝑏 = 𝑖 ⇒ (𝑖 = √−1) = (𝑥 − 2)(𝑥 + 𝑖 )(𝑥 − 𝑖 ) ⇒ in factored form Question #4 How do you factor 20𝑥 5 − 5 Answer: 20𝑥 5 − 5 = 5(4𝑥 2 − 1) = 5(2𝑥 − 1)(2𝑥 + 1) Explanation: So we have 20𝑥 2 − 5 = 5(4𝑥 2 − 1) ; dividing thru by 5 5(4𝑥 2 − 1) And 4𝑥 2 − 1 is the difference between 2 squares, 2𝑥 and 1 (4𝑥 2 − 1) = (2𝑥 − 1)(2𝑥 + 1) Putting it all together: 20𝑥 2 − 5 = 5(2𝑥 − 1)(2𝑥 + 1) 1 So there are roots at ± 2 Question #5 What is the next number in the sequence 1,2,6,24,120, ⋯ ? Answer: These are the first 5 factorials (starting at 1!) 1! = 1 2! = 2 × 1 = 2 3! = 3 × 2 × 1 = 6 4! = 4 × 3 × 2 × 1 = 24 5! = 5 × 4 × 3 × 2 × 1 = 120 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 Question #6 Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. 𝑓 (𝑥, 𝑦) = 3𝑥 + 𝑦; 𝑥 2 + 𝑦 2 = 10 Answer: Step 1 We first find our gradients of f and g ∇𝑓 =< 3,1 > ∇𝑔 =< 2𝑥, 2𝑦 > Step 2 We then set our gradients equal to each other using a LaGrange multiplier. We solve for x 3 = 𝜆2𝑥 3 2𝑥 = 1 = 𝜆2𝑦 1 2𝑦 3 2𝑥 = 1 = 2𝑦 6y=2x 3y=x Step 3 We plug 3y in for x in our constraint function. This determines what y will be equal too. We find that 𝑦 = ±1 . We can plug this back into our LaGrange equation.This yields the points (-3,-1) and (3,1). The signs must be the same due to our LaGrange equation. (3𝑦)^2 + 𝑦^2 = 10 10𝑦 2 = 10 𝑦 = ±1 @y=-1 : 3(-1)=-3=x @y=1 : 3(1)=3=x Step 4 We plug these into our f(x,y) to find that: (3,1) is a maximum (-3,-1) is a minimum f(-3,-1)=3(-3)-1=-10 f(3,1)=3(3)+1=10 result Max: 10 at (3,1) Min: -10 at (-3,-1) Question #7 Factor 𝑥 3 − 10𝑥 2 + 29𝑥 Answer: Factor the following: 𝑥 3 − 10𝑥 2 + 29𝑥 Hint: Factor common terms out of 𝑥 3 − 10𝑥 2 + 29𝑥. Factor x out of 𝑥 3 − 10𝑥 2 + 29𝑥: Answer: 𝑥(𝑥 2 − 10𝑥 + 29) Question #8 Of the multiples of 2 between 1 and 99, what fraction are also multiples of 3? 16 99 16 𝐵) 49 1 𝐶) 3 33 𝐷) 49 49 99 𝐴) Answer: Step 1 the multiples of 2 between 1 and 99 are the even numbers . Therefore, there are 49 even numbers from 1 to 99 The numbers that are multiple of 2 are 2,4,6,8,10,12,...49 out of those even numbers, the numbers that are also the multiples of 3 are the numbers that are divisible by 3. therefore from the set 2,4,6,8,10,12,...49 every third number is multiple of 3. therefore, there are 16 such numbers Step 2 therefore, the required fraction is given by 1649 the answer is 1649 Question #9 Tim always eats strawberries in multiples of three. Todd eats them in multiples of four and Rick in multiples of six. Gavin is not fussy about it but likes cream to go with them. Uncle Sam wants each of the four kids to eat the same number of strawberries. What is the minimum number of strawberries that he has to give to each kid? Select one answer: 𝐴)24 𝐵)6 𝐶)3 𝐷)1 𝐸)12 Answer: Step 1 Tim eats multiples of three Todd eats multiples of four Rick eats multiples of six. Gavin is not bussy. Thus we have numbers to work with 3,4,6. Step 2 Therefore, since uncle Sam want to give same number of strawberries. Then the number = 2𝑐𝑚(3,4,6) = 12 Therefore, uncle Sam should give 12 strawberries to each of kid. Option E is correct. Question #10 Exercise 1: Counting binary strings. Count the number of binary strings of length 10 subject to each of the following restrictions. There is only one binary string of length ten with no 1's: 00000000000. There are 210 binary strings of length ten. Therefore the number of binary strings of length ten with at least one 1 is 210 − 1 . (b) The string has at least one 1 and at least one 0. (c) The string contains exactly five 1's or it begins with a 0. Exercise 2: Counting integer multiples. (b) How many integers in the range 1 through 140 are integer multiples of 2, 5, or 7? Answer: Step 1 The string has at least one 1 and at least one 0. This removes the possibility of all one and all zeros Thus number of binary strings is 210 − 2 Step 2 The string contains exactly five 1's or it begins with a 0. First digit is fixed as zero In remaining 9 places Lets choose 5 places and place 1s there Number of ways are 𝐶59 126 Step 3 How many integers in the range 1 through 140 are integer multiples of 2, 5, or 7 There are 70 multiples of 2 23 multiples of 3 left we have 9 multiples of 5 that have not yet been included. Total is 102 Question #11 1 = 𝑎𝑥 + 𝑏𝑦 , for all integers x and y 1 = 𝑎𝑥 + 𝑏𝑦 , for some integers x and y 1 = 𝑎𝑥 + 𝑏𝑦 , for unique integers x and y All the above Answer: Step 1 Two integers are relatively ' when there are no common factors other than 1. This means that no other integer could divide both numbers. Two integers a,b are called relatively ' to each other if 𝑔𝑐𝑑 (𝑎, 𝑏) = 1 . Step 2 For any positive integers a and b, there exist integers x andy such that ax + by = gcd(a, b) . Furthermore, as x and y vary over all integers a 𝑥 + 𝑏𝑦 attains all multiples and only multiples of gcd(a, b). Here 𝑔𝑐𝑑(𝑎, 𝑏) = 1 So there exist integers x and y such that 𝑎𝑥 + 𝑏𝑦 = 1 So the correct choice is b. 1 = 𝑎𝑥 + 𝑏𝑦 , for some integers x and y. Question #12 Consider the function 𝑓 (𝑥 ) = 4 sin(π2(𝑥 − 3)) + 8. State the amplitude A, period P, and midline. State the phase shift and vertical translation. In the full period [0, P], state the maximum and minimum y-values and their corresponding x-values. Enter the exact answers. Amplitude: 𝐴 = Period: 𝑃 = Midline: 𝑦 = The phase shift is _____ The vertical translation is_______ Hints for the maximum and minimum values of f(x): The maximum value of 𝑦 = sin(𝑥 ) is 𝑦 = 1 and the corresponding x values are 𝑥 = 𝜋2 and multiples of 2𝜋 less than and more than this x value. You may want to solve 𝜋2(𝑥 − 3) = 𝜋2 . The minimum value of 𝑦 = sin(𝑥 ) is 𝑦 = −1 and the corresponding x values are 𝑥 = 3 𝜋 2 and multiples of 2𝜋 less than and more than this x value. You may want to solve 𝜋2(𝑥 − 3) = 3𝜋2 . If you get a value for x that is less than 0, you could add multiples of P to get into the next cycles. If you get a value for x that is more than P, you could subtract multiples of P to get into the previous cycles. For x in the interval [0, P], the maximum y-value and corresponding x-value is at: 𝑋= 𝑌= For x in the interval [0, P], the minimum y-value and corresponding x-value is at: 𝑋= 𝑌= Answer: Step 1 we have 𝑓 (𝑥 ) = 4 sin(2π(𝑥 − 3)) + 8 Amplitude 𝐴 = cofficient of sin(2π(𝑥 − 3)) = 4 period 𝑃 = 2 𝜋 cofficient of (𝑥 − 3) = 2𝜋2𝜋 = 1 maximum value and minimum value −1 ≤ sin(2π(𝑥 − 3)) ≤ 1 − 4 ≤ 4 sin(2π(𝑥 − 3)) ≤ 4 − 4 + 8 ≤ 4 sin(2π(𝑥 − 3)) + 8 ≤ 4 + 84 ≤ 4 sin(2π(𝑥 − 3)) + 8 ≤ 12 maximum value is 12 and minimum value is 4. Step 2 in interval [0,1] according to graph of function 𝑓 (𝑥) = 4 sin(2π(𝑥 − 3)) + 8 maximum value 𝑥 = 0.25 and minimum value at 𝑥 = 0.75 Question #13 Consider the function 𝑓 (𝑥 ) = 2 sin(π2(𝑥 − 3)) + 4 . State the amplitude A, period P, and midline. State the phase shift and vertical translation. In the full period [0, P], state the maximum and minimum y-values and their corresponding x-values. Hints for the maximum and minimum values of f(x): The maximum value of 𝑦 = sin(𝑥 ) is 𝑦 = 1 and the corresponding x values are 𝑥 = 𝜋2 and multiples of 2\pi less than and more than this x value. You may want to solve 𝜋2(𝑥 − 3) = 𝜋2 . The minimum value of 𝑦 = sin(𝑥 ) is y=-1 and the corresponding x values are 𝑥 = 3𝜋2 and multiples of 2𝜋 less than and more than this x value. You may want to solve 𝜋2(𝑥 − 3) = 3𝜋2 . If you get a value for x that is less than 0, you could add multiples of P to get into the next cycles. If you get a value for x that is more than P, you could subtract multiples of P to get into the previous cycles. For x in the interval [0, P], the maximum y-value and corresponding x-value is at: 𝑥= 𝑦= For x in the interval [0, P], the minimum y-value and corresponding x-value is at: 𝑥= 𝑦= Answer: Step 1 Consider a given function π 𝑓(𝑥 ) = 2 sin [ 2 (𝑥 − 3) + 4] . State the amplitude A, period P and mid-line. State the phase shift and vertical translation. State the minimum and maximum y-values and their corrosponding x-values in the interval [0,P]. Step 2 a) Consider a given function π 𝑓(𝑥 ) = 2 sin [ 2 (𝑥 − 3)] + 4 The amplitude of the function f is 2. Hence, 𝐴 = 2 . The period of the function f is 2π π 2 ( ) = 4 . Hence, 𝑇 = 4 . The mid-line of the trigonometric function is a point where the function attain its maximum values. Now, π 2 (𝑥 − 3 ) = (2𝑛−1)π 2 𝑥 − 3 = 2𝑛 − 1 ⇒ 𝑥 = 2𝑛 + 2 for all 𝑛 ∈ 𝑍 Hence, the mid-line of the function f is 𝑥 = 2𝑛 + 2 for all 𝑛 ∈ 𝑍 Step 3 b) Consider a given function π 𝑓(𝑥 ) = 2 sin [ 2 (𝑥 − 3)] + 4 The function f is shifted right side by 3 units. Hence, its phase shift is 3. The function f is shifted vertically upward by 4 units. Hence, its vertically shift is 4 units. Step 4 c) The range of the trigonometric function is [−1,1] . It implies that −1 ≤ sin(𝑥 ) ≤ 1 . Hence, π −1 ≤ sin ( 2 (𝑥 − 3)) ≤ 1 π −2 ≤ 2 sin ( 2 (𝑥 − 3)) ≤ 2 π −2 + 4 ≤ 2 sin ( 2 (𝑥 − 3)) + 4 ≤ 2 + 4 2 ≤ 𝑓 (𝑥 ) ≤ 6 Hence, minimum value of the function f is 2 and the maximum value of the function f is 6. It implies that 𝑓𝑚𝑖𝑛 = 2 and 𝑓𝑚𝑎𝑥 = 6 . Step 5 Further, Maximum attain at 𝑥 = 2𝑛 + 2 for all 𝑛 ∈ 𝑍 . Hence, the point of maxima is 𝑥 = 0,2,4 . π sin ( 2 (𝑥 − 3)) = 0 π 2 (𝑥 − 3 ) = 𝑛 𝑥 − 3 = 2𝑛 ⇒ 𝑥 = 2𝑛 + 3 for all 𝑛 ∈ 𝑍 Hence, the point of minima is 1,3. Question #14 Fifty raffle tickets are numbered 1 through 50, and one of them is drawn at random. What is the probability that the number is a multiple of 5 or 7? Consider the following ''solution'': Since 10 tickets bear numbers that are multiples of 5 and 7 tickets bear numbers that are multiples of 7, we concluded the required probability is as follows. 10 7 17 + = 50 50 50 Is this the correct answer? (If so, enter yes. If not, enter the correct answer.) Answer: Step 1 Fifty raffle tickets are numbered 1 through 50, and one of them is drawn at random. What is the probability that the number is a multiple of 5 or 7? Consider the following ''solution'': Since 10 tickets bear numbers that are multiples of 5 and 7 tickets bear numbers that are multiples of 7, we concluded the required probability is as follows. Step 2 Let A be the event that the number is multiple of 5 B be the event that the number is multiple of 7 Find 𝑃(𝐴 ∪ 𝐵) Therefore, 𝑃 (𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃 (𝐵) − 𝑃(𝐴 ∩ 𝐵) Number of tickets that are multiplies of both 5 and 7 are 50 5×7 50 = 35 = 1 1 𝑃(𝐴 ∩ 𝐵) = 50 𝑃 (𝐴 ∪ 𝐵 ) = 17 10 50 + 7 50 − 1 50 1 = 50 − 50 16 = 50 The probability that the number of tickets are multiplies of 5 or 7 is 16 50 Question #15 π Consider the function 𝑓 (𝑥 ) = 3 sin ( 3 (𝑥 − 4)) + 5 . State the amplitude A, period P, and midline. State the phase shift and vertical translation. In the full period [0, P], state the maximum and minimum y-values and their corresponding x-values. Enter the exact answers. Amplitude: 𝐴 = Period: 𝑃 = Midline: 𝑦 = The phase shift is: a. up 4 units b. 4 units to the right c. 4 units to the left d. 5 units to the left The vertical translation is: a. up 4 units b. down 5 units c. up 5 units d. down 4 units Hints for the maximum and minimum values of f(x): The maximum value of 𝑦 = sin(𝑥 ) is 𝑦 = 1 and the corresponding x values are 𝑥 = 2𝜋 less than and more than this x value. You may want to solve π 3 (𝑥 − 4) = π 2 π 2 . The minimum value of 𝑦 = sin(𝑥 ) is 𝑦 = −1 and the corresponding x values are 𝑥 = of 2𝜋 less than and more than this x value. You may want to solve π 3 and multiples of (𝑥 − 4) = 3π 2 3π 2 and multiples . If you get a value for x that is less than 0, you could add multiples of P to get into the next cycles. If you get a value for x that is more than P, you could subtract multiples of P to get into the previous cycles. For x in the interval [0, P], the maximum y-value and corresponding x-value is at: x= y= For x in the interval [0, P], the minimum y-value and corresponding x-value is at: 𝑥= 𝑦= Answer: Step 1 Consider the provided question, π Given function, 𝑓 (𝑥 ) = 3 sin ( 3 (𝑥 − 4)) + 5 First find Amplitude, Period and Midline. π Amplitude=coefficient of sin ( 3 (𝑥 − 4)) = 3 2π Period = 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑜𝑓 𝑥 = 2π π 3 =6 Midline: (y=vertical displacement): 𝑦 = 5 Step 2 Draw the graph of the given function, Step 3 From the above diagram it is clear that, phase shift is 4 units to the right. So, the correct answer is option (b). And the vertical translation is 5 units up (2 + 8−2 2 = 5) So, the correct answer is option (c). Step 4 Now, in the interval [0, 𝑃] = [0,6] the maximum value of y corresponding to x is, From the above graph in the interval [0, 6], The maximum value of 𝑦 = 8 corresponding to 𝑥 = 5.5 So, 𝑥 = 5.5 𝑦 = 8 Step 5 In the interval [0,P]=[0,6] the minimum value of y corresponding to x is, From the above graph in the interval [0, 6], The minimum value of 𝑦 = 2 corresponding to 𝑥 = 2.5 So, 𝑥 = 2.5 𝑦 = 2 Question #16 The Universal Set, U, consists of the natural numbers from 20 to 60 incluive a. Define or describe in words the following three (3) sets: factors of 64, ' numbers, and multiples of 3. b. List the elements in each of your sets: 𝐴 = { 𝐵 = { 𝐶 = { c. Determine the probability of each of the following: 𝐼. 𝑃(𝐶 ) 𝐼𝐼. 𝑃(𝐴 ∪ 𝐵) 𝐼𝐼𝐼. 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶 ) 𝐼𝑉. 𝑃 (𝐵\𝐶 ) 𝑉. 𝑃((𝐴\𝐵) ∩ 𝐶) Answer: Step 1: Given, The universal set U, consist of the natural numbers from 20 to 60 inclusive. We have to answer the following... Step 2: Explanation Solution (a). If we have 3 sets, 𝑆1 = factors of 64 i.e. 1,2,4,8,16,32,64 𝑆2 = ' numbers 𝑆3 = multiple of 3 Now describe the sets in words, 𝑆1 = {𝑥 ∶ 𝑥 is a factor of 64 and 20 ≤ 𝑥 ≤ 60} 𝑆2 = {𝑥 ∶ 𝑥 is ' and 20 ≤ 𝑥 ≤ 60} 𝑆3 = {𝑥 ∶ 𝑥 is multiple of 3 and 20 ≤ 𝑥 ≤ 60} Solution (b). List of the elements 𝑆1 = 32 𝑆2 = 23,29,31,37,41,43,47,53,59 𝑆3 = 21,24,27,30, … ,57,60 Solution (c). Determine the probability Since we have from the b part, 𝑆1 ∪ 𝑆2 = 32,23,29,31,37,41,43,47,53,59 𝑆1 ∩ 𝑆2 ∩ 𝑆3 = 14 7 𝐼. 𝑃(𝑆3 ) = 40 = 20 10 1 𝐼𝐼. 𝑃(𝑆1 ∪ 𝑆2 ) = 40 = 4 𝐼𝐼𝐼. 𝑃(𝑆1 ∩ 𝑆2 ∩ 𝑆3 ) = 0 40 =0 Question #17 Values for the function f(x) are shown in the table. x 0 1 2 3 4 5 f ( x) 2 4 8 16 32 64 Which statement proves that f(x) is an exponential function? A) All of the values of f(x) are odd numbers. B) All of the values of f(x) are multiples of 2. C) The function f(x) grows by equal factors over equal intervals. D) The function f(x) grows by equal differences over equal intervals. Answer: Step 1 Given a table of values of f(x) at different points. Step 2 an exponential function is of the type 𝑓(𝑥 ) = 𝑎 𝑥 where a is constant and x is variable. by examining the table the given function is 𝑓 (𝑥 ) = 2𝑥+1 . Hence option B is correct. f(x) grows by equal factor over equal interval. Question #18 Let 𝑈 = { 1, 2, 3, … ,2400 } Let S be the subset of the numbers in U that are multiples of 3, and let T be the subset of U that are multiples of 7. Since 2400 ÷ 3 = 800 , it follows that 𝑛(𝑆) = 𝑛({3 ⋅ 1,3 ⋅ 2, ⋯ ,3 ⋅ 800}) = 800 . (a) Find n(T) using a method similar to the one that showed that 𝑛(𝑆) = 800 . (b) Find 𝑛(𝑆 ∩ 𝑇) . (c) Label the number of elements in each region of a two-loop Venn diagram with the universe 𝑈 and subsets S and T. Questions:Find 𝑛(𝑇) ? Find 𝑛(𝑆𝑛𝑇) Answer: Step 1 Solution: 𝑛(∪) = 2400 ∪= {1,2,3, … . . ,2400} 2400 ÷ 3 = 800 𝑛(𝑠) = 𝑛{3 ⋅ 1,3 ⋅ 2, … .3 ⋅ 800} = 800 𝑛(𝑡) = 𝑛(7 ⋅ 1,7 ⋅ 2,7 ⋅ 3, … . ,7 ⋅ 342) = 342 2394 Step 2 𝑛(𝑡) = 342 𝑠 ∩ Multiple of 7 and 3 = 21 𝑛(𝑠 ∩ 𝑡) = 𝑛(21 ⋅ 1,21 ⋅ 2, … . .21 ⋅ 114) = 114 2394 𝑛(𝑠 ∩ 𝑡) = 114 Question #19 Find the twenty-eighth positive multiple of 4 and the sum of the first twenty-eight positive multiples of 4. 𝑎28 = 𝑆28 = Answer: Step 1 To calculate the twenty-eighth positive multiple of 4 and the sum of the first twenty-eight positive multiples of 4. Step 2 Formula used for nth term and sum of n-terms of an AP series are given below: 𝑎𝑛 = 𝑎 + ( 𝑛 − 1 ) 𝑑 𝑛 𝑆𝑛 = 2 (2𝑎 + (𝑛 − 1)𝑑 ) Where, 𝑎 = first term 𝑑 = common difference 𝑛 = number of terms Step 3 Now, Multiple of 4 is: 4, 8, 12, 16…… 𝑎 = 4 𝑑 = 8 − 4 = 12 − 8 = 4 𝑛 = 28 As the common difference is same. Therefore, the series is in AP. Step 4 Therefore, 𝑎𝑛 = 𝑎 + ( 𝑛 − 1 ) 𝑑 𝑎28 = 4 + (28 − 1)4 = 112 𝑛 𝑆𝑛 = 2 (2𝑎 + (𝑛 − 1)𝑑 ) 𝑆28 = 28 2 (2 ⋅ 4 + (28 − 1)4) = 1624 Question #20 𝑈 = the set of natural numbers between 10 and 20 𝐴 = Even numbers 𝐵 = Multiples of 3 where the number is less than 18. 𝐶 = composite numbers Find (𝐴′ 𝑈𝐶 ) − 𝐵 Answer: Step 1 𝑈 = the set of natural numbers between 10 and 20. 𝑈 = {11,12,13,14,15,16,17,18,19} 𝐴 = even numbers ⇒ {2,4,6, … 18} 𝐵 = multiples of three where the number 15 less than 18. 𝐵 = {3,6,9,12,15} 𝐶 = composite numbers = {4,6,8,9,10,12,14,15,16,18, … } Step 2 Find (𝐴′ 𝑈𝐶 ) − 𝐵 Hence, 𝐴′ = {1,3,5,7,9,11,13, … } Now, (𝐴′ 𝑈𝐶 ) = {1,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19, … } and (𝐴′ 𝑈𝐶 ) − 𝐵 = {1,4,5,7,8,10,11,13,14,16,17,19}