MATH 2 part 1

(x – 5 )2 = ± 49 Using the square root method 4 16 Add 5 to both sides.x– 5 =± 7 2 44 Solve for x x=5±7 44x = 5 + 7 and x = 5 - 744 44x = 12 x=- 2 4 4x=3 x=- 1 2The solutions are 3 and – 1 . 2Note that this equation can also be solved by factoring: 2x² - 5x – 3 = 0 ( 2x + 1 )( x – 3 ) = 0 2x + 1 = 0 or x – 3 = 0 2x = -1 x=3 x=-1 2Check: 2x² - 5x – 3 = 0 If x = - 1 2 2(- 1 )² - 5(- 1 ) – 3 = 0 22 2( 1 ) + 5 –3 = 0 42 1 + 5 – 6 =0 222 6 – 6 =0 22 0=0 9

If x = 3 2x² - 5x – 3 = 0 2(3)² - 5(3) – 3 = 0 2(9) + 15 – 3 = 0 18 + 15 – 3 = 0 3–3=0 0=0The solutions check.Example 4:Use completing the square to solve equation: -x² -2x + 3 = 0Solution: In this equation, note that the coefficient of x² is –1. This means that you just cannot apply completing the square right away. You need to make the coefficient of x² equal to 1. To do this, multiply both sides of the equation by –1. Thus, -1(-x² - 2x + 3) = 0(-1) x² + 2x – 3 = 0 x ² + 2x + [ 1 (2)]2 = 3 + [ 1 (2)]2 22 x² + 2x + 1 = 3 + 1 Solve for x (x + 1)² = 4 x+1=± 4 x = -1 ± 2 x = -1 + 2 or x = -1 –2 x=1 or x = -3 The solutions are 1 and -3.Note that the equation can be solved by factoring. -x² -2x + 3 = 0 -(x² + 2x – 3) = 0 x² + 2x – 3 = 0 (x + 3)(x – 1) = 0 x + 3 = 0 or x–1=0 x = -3 or x=1 10

Check:: -x² -2x + 3 = 0 If x = -3 -(-3)² - 2(-3) + 3 = 0 -(9) + 6 +3 = 0 -9 + 9 = 0 0=0 If x = 1 -x² -2x + 3 = 0 -(1)² - 2(1) + 3 = 0 -1 – 2 +3 = 0 -3 + 3 = 0 0=0The solutions check.Example 5:Use completing the square to solve equation: -2x2 - 4x +2 = 0.Solution: This equation cannot be solved by factoring. -2x2 -4x +2 = 0 Divide both sides by -2 x2 +2x -1 = 0 Add 1 to both sides x2 + 2x = 0 Add[ 1 (2)]2 to both sidesx2 + 2x+[ 1 (2)]2 = 1 +[ 1 (2)]2 2 22 to complete the square x2 + 2x +1 = 1+ 1 Factor and Simplify (x + 1 )2 = 2 Solve by the square root x + 1=± 2 method x = -1 ± 2 Solve for x x = -1+ 2 and x = -1 - 2 The solutions are -1+ 2 and -1+ 2 . 11

Check: -2x2 - 4x + 2 = 0 If x = -1+ 2-2(-1+ 2 )² - 4(-1+ 2 ) + 2 = 0-2[1 - 2 2 + ( 2 )²] + 4 - 4 2 + 2 = 0-2 + 4 2 - 2(2) + 4 - 4 2 + 2 = 0 -2 - 4 + 4 + 2 = 0 -6 + 6 = 0 0=0If x = -1 - 2 -2x2 - 4x + 2 = 0-2(-1 - 2 )² - 4(-1 - 2 ) + 2 = 0-2[1 + 2 2 + ( 2 )²] + 4 + 4 2 + 2 = 0-2 - 4 2 - 2(2) + 4 + 4 2 + 2 = 0 -2 – 4 + 4 + 2 = 0 -6 + 6 = 0 0=0Both solutions check. From the preceding examples, you must be able to identify the steps insolving quadratic equations by completing the square method.To solve an equation by completing the square, follow these steps: 1. Make sure that coefficient of x2 is 1. If it is not, make it 1 by dividing both sides of the equation by the coefficient of x2. 2. Isolate the constant (numerical term) by transposing it on the right-hand side of the equation. 3. Complete the square: a) Identify the coefficient of x. b) Take half the coefficient of x. c) Square half the coefficient of x d) Add the numbers in c to both sides of the equation 4. Factor the left side and simplify the right sides of the equation. 5. Solve the resulting equation by square root method.6. Check the solution. 12

Try this outSolve the following by completing the square. For Nos. 1 – 15, see Example 1 1. x2 + 8x + 6 = 0 2. x2 – 4x – 3 = 0 3. x2 – 10x = 15 4. u2 – 5u – 2 = 0 5. x2 – 6x = 19 6. a2 – 4a – 5 = 0 7. x2 + 10x + 20 = 0 8. x² -2x – 5 = 0 9. a² - 8a – 20 = 0 10. z² + 5 = 7z 11. x² + 8x – 9 = 0 12. x² + 6x – 7 = 0 13. x² + 6x + 8 = 0 14. x² + 5x – 6 = 0 15. x² = 4x + 3 For Nos. 16 – 20, see Examples 2 and 3 16. 2x2 = a – 4x 17. 3x2 = 12 – 6x 18. 3x² = 2 – 5x 19. 2x² = 5x + 3 20. 4x² = 2 – 7x For Nos. 21 – 25, see Examples 4 and 5 21. –x² - 12x = 6 22. -2z² + 12z – 4 = 0 23. 2x(x + 3) = 8 24. x(x + 3) -1/2 = -2 25. 20x² - 11x – 3 = 0 13

Lesson 3 The Quadratic Formula In this lesson, you will see how completing the square will be used toderive the Quadratic Formula. This method works easily in solving any quadraticequation. To do this, let us begin with solving a Quadratic Equation of the form ax2 +bx + c = 0 by completing the square.Example 1 Solve 2x2 + 7x + 4 = 0 by completing the square.Solution: 2x2 + 7x + 4 = 0 Divide both sides of the eq’n by 2 x2 + 7 x + 2 = 0 2 x2 + 7 x = - 2 Add -2 to both sides of the eq’n 2 Add the square of 1 of 7 .x2 + 7 x + [ 1 ( 7 )]2 = -2 + [ 1 ( 7 )]2 22 2 22 22x2 + 7 x + 49 = -2 + 49 2 16 16 (x + 7 )2 = - 32 + 49 Factor the left-hand side and 4 16 16 simplify the right –hand side by finding the LCD, which is 16. (x + 7 )2 = 17 4 16 x + 7 = ± 17 By the square root method 4 16 x + 7 = ± 17 Solve for x. 44 x = - 7 ± 17 44 x = − 7 + 17 and x = − 7 − 17 44 14

Now, study the following derivation of the quadratic formula by the methodof completing the square. The steps is just like what has been done in Example 1so that you can see that the process is the same. The only difference is that youwill be working with letters instead of numbers.ax2 +bx + c = 0x2 + b x + c = 0 Divide both sides by a. aax2 + b x = - c Subtract c from both sides. aa ax2 + b x + [ 1 ( b )]2 = - c + [ 1 ( b )]2 Complete the square.a 2a a 2ax2 + b x + b2 = - c + b2 Add the square of 1 of b . 2aa 4a2 a 4a2(x + b )2 = – − 4ac + b2 Factor the left-hand side 2a 4a2 and simplify the right-hand side by finding to LCD which is 4a.x+ b = ± b2 − 4ac Solve the equation by 2a 4a2 the square ro(o5t) methodx+ b = ± b2 − 4ac Solve for x. 2a 2ax = - b ± b2 − 4ac 2a 2a or x = − b ± b2 − 4ac 2aPresto! What had been derived is the Quadratic Formula.The Quadratic Formula: x = − b ± b2 − 4ac 2a 15

The quadratic formula tells you that if you have a quadratic equation in standardform ax2 + bx + c =0, then all you have to do is substitute the values of a, b and cinto the formula to get the solutions.Example 2Solve by using the quadratic formula x2 - 3x - 5 =0Solution: The equation is already in standard formula. To use the formula you must first identify a, b, and c. a=1 b = -3 c = -5Substituting these values into the formula, you should get x = − b ± b2 − 4ac 2a − (−3) ± (−3)2 − 4(1)(−5) = 2(1) = 3 ± 9 + 20 2 = 3 ± 29 2The solution are 3 + 29 and 3 − 29 22Check: If x = 3 + 29 x2 - 3x - 5 = 0 2 ( 3 + 29 )² - 3( 3 + 29 ) - 5 = 0 22 9 + 6 29 + ( 29)2 - 9 + 3 29 – 5 = 0 429 + 6 29 + ( 29)2 2(9 + 3 29) - –5=0 44 16

If x = 3 − 29 9 + 6 29 + 29 −18 − 6 29 - 5 = 0 2 4 20 - 5 = 0 4 5-5=0 0=0 x2 - 3x - 5 = 0 ( 3 − 29 )² - 3( 3 − 29 ) - 5 = 0 22 9 − 6 29 + ( 29)2 - 9 − 3 29 – 5 = 0 429 − 6 29 + ( 29)2 2(9 − 3 29) - –5=0 44 9 − 6 29 + 29 −18 + 6 29 - 5 = 0 4 20 - 5 = 0 4 5-5=0 0=0 You may find that checking is a little bit tedious but its fun doing since youget to apply your skills in simple arithmetic. Notice that the quadratic formula is a lot easier to use than completing thesquare. This is because the formula contains all the steps necessary if you wereusing completing the square.Here are some things to watch out when using the Quadratic Formula:1. If b is positive, then the – b that appear in the formula will benegative. If b is negative, then the value of – b in the formula will bepositive.2. The quantity 2a in the formula is the denominator of the entireexpression -b ± b2 − 4ac . 17

Example 3:Solve by using the quadratic formula 3x2 – 3x =5Solution: Begin by writing the equation into the standard form 3x2 +11x = 4 3x2 + 11x - 4 = 0Substitute the value of a, b and c into the Quadratic Formula a=3 b = 11 c = -4 x = − b ± b2 − 4ac 2aThen we have, x = −11 ± (11) − 4(3)(−4) 2(3) = −11 ± 121 − (−48) 6 = −11 ± 169 6 x = −11 ± 13 6 x = −11 + 13 and x = −11 −13 66 x= 2 = 1 = − 24 = -4 63 6The solution is 1 and -4. 3The checking is left for you. 18

Example 4Solve by the quadratic formula: 3 – 2t = t2Solution: Arrange the equation in standard form and identify a, b, and c. t2 + 2t - 3 = 0 a=1 b=2 c = -3Substitute in the quadratic formula and solve for t: t = − b ± b2 − 4ac 2a − (2) ± (−2)2 − 4(1)(−3) t= 2(1) t = − 2 ± 4 + 12 2 = − 2 ± 16 2 t= −2±4 2 t = − 2 + 4 and t = − 2 − 4 22 = 2 = −6 22 t=1 t = -3The solution are 1 and -3.This equation can be solved by factoring. Try solving and check..Example 5Solve for x : 2x2 – 5x + 7 = x(2x - 3) 19

Solution: Be careful! This does not mean that just because there is an x2 in the equation that the equation 2x2 – 5x + 7 = x(2x - 3) must be a quadratic equation. Putting the equation in standard form, then 2x2 – 5x + 7 = x(2x- 3) 2x2 – 5x + 7 = 2x2 - 3x - 2x2 + 3x -2x2 + 3x by addition -2x + 7 = 0 -2x = -7 x=7 2 As you can see, it’s not a quadratic equation at all. Example 5 clearlyshows, the method of solving an equation depends on the type of equation youare dealing with. Look carefully at the equation before deciding the method toapply.Example 6Solve by the quadratic formula: x2 + 3x+ 4 =0Solution: The equation is already in standard form, so that a=1 b=3 c=4 x = − b ± b2 − 4ac Not real solution 2a x = − 3 ± (3)2 − 4(1)(4) 2(1) = − 3 ± 9 −16 2 x = −3± −7 2 As soon as you see that the answer involves the square root of a negativenumber, you can stop and say that the equation has no real solution. 20

Try this outA. Write each equation in standard form, if necessary. Then, determine thevalues of a, b and c. Do not solve the equation. 1. x² + 4x + 3 = 0 2. x² - x – 4 = 0 3. 3x² - 2x + 7 =0 4. 4x² + 7x – 3 = 0 5. 4y² = 2y – 1 6. 2x = 3x² + 4 7. x(3x – 5) = 2 8. y(5y + 10) = 8Solve the equations using the Quadratic Formula.B. 1. x2 + 3x - 5 =0 2. x2 + 5x - 2 =0 3. y2 + 4y - 6 =0 4. y2 + 2y - 5 =0 5. u2 – 2u + 3 =0 6. u2 - 3u + 3 =0 7. t2 - 7t = 6 8. t2 + 6 = 6tC. 1. 5x2 - x = 2 2. 7x2 - 3 = x 3. 3x2 + 5x + 2 =0 4. 2x2 - 3x - 1 =0 5. t2 - 3t + 4 = 2t2 + 4t - 3 6. (5w + 2) (w - 1) = 3w + 1 7. (x + 2)2 = x(x + 4)D. Find out what is wrong with the following solution: x² - 3x – 1 = 0 x = − b ± b2 − 4ac 2a x = −3± 9−4 2 = −3± 5 2 21

Let’s Summarize1. Most types of quadratic equations can be solved by completing the square.2. In completing the square, it is important to remember that a. the coefficient of x2 should be 1; if not, divide both sides of the equation by the coefficient of x2. b. the term to be added is the square of one-half the coefficient of x in the equation.3. The quadratic formula is x = − b ± b2 − 4ac 2a4. Practically, any quadratic equation can be solved by the quadratic formula. What have you learnedA. Solve by completing the square: 1. x² - 4x – 2 = 0 2. x² - 2x – 4 = 0 3. x² + 2x = 0 4. 6(x² - 1) = 5x 5. (x + 2)2 – 4x = 8B. Solve using the quadratic formula: 1. x² + 6x + 9 = 0 2. x² -4x = 0 3. x² - 7 = 0 4. x² + 8x + 12 = 0 5. 2x² + x = 5 22

Answer KeyHow much do you knowA. 1. 4 ± 31 2. 1 ± 8 or 1 ± 2 2 3. − 3 ± 13 2 4. 0, - 2 3 5. 4 ± 11 5B. 1. 4, -3 2. 0, 4 3. no real solution 4. 1 ± 13 6 5. -1 ± 2Try this outLesson 1 A. 1. 4, (x + 2)2 2. 9, (x + 3)2 3. 25, (x - 5)2 4. 16, (x - 4)2 5. 36, (x + 6)2 B. 1. 1 , (x + 1 )2 42 2. 9 , (a - 3 )2 42 23

3. 169 , (b - 13 )2 424. 1 , (b + 1 )2 935. 25 , (c - 5 )2 16 4C. 1. 1, (x + 1)22. 1, (x - 1)23. 1 , (x - 1 )2 36 64. 9 , (x - 3 )2 64 85. 1 1 )2 , (x - 5 25Lesson 2 B. 1. – 4 ± 10 2. 2 ± 7 3. 5 ± 40 or 5 ± 2 10 4. 5 ± 33 2 5. 3 ± 28 or 3 ± 2 7 6. 5, -1 7. -5 ± 5 8. 1 ± 6 9. 10, -2 10. 7 ± 29 2 11. -9, 1 12. -7, 1 13. -2, -4 14. 1, -6 24

15. 2 ± 716. -1± 317. -4, 118. -2, 1 319. 3, - 1 220. -2, 1 421. -6 ± 3022. 3 ± 723. -4, 124. − 3 ± 3 225. 11 ± 181 20Lesson 3A. a=1 b=4 c=3 1. a=1 b = -1 c = -4 2. a=3 b = -2 c=7 3. a=4 b=7 c = -3 4. a=4 b = -2 c=1 5. a=3 b = -2 c=4 6. a=3 b = -5 c = -2 7. a=5 b = 10 c = -8 8. 25B. 1. − 3 ± 29 2 2. − 5 ± 33 2

3. − 2 ± 10 4. -1 ± 6 5. no real solution 6. no real solution 7. 7 ± 61 2 8. 3 ± 3C. 1. 1 ± 41 10 2. 1 ± 85 14 3. no real solution 4. 3 ± 17 4 5. − 7 ± 77 2 6. 6 ± 56 or 3 ± 14 10 5 7. no real solution D. The solution should be: x² - 3x – 1 = 0 x = − b ± b2 − 4ac 2a x = 3± 9+4 2 = 3 ± 13 2 26

What have you learned A. 1. 2 ± 6 2. 1 ± 5 3. -2, 0 4. 3 , - 2 23 5. ±2 B. 1. -3 2. 4, 0 3. ± 7 4. -2, -6 5. −1 ± 41 4 27

Module 2 Searching for Patterns in Sequences, Arithmetic , Geometric and others What this module is about This module is about finding the common difference and the nth term of anarithmetic sequence. As you go over the exercises, you will develop skills in solving thecommon difference and finding the nth term given the first few terms of an arithmeticsequence; finding the first term and common difference or a specified nth term given twoterms of arithmetic sequence; and solve problems involving arithmetic means andharmonic sequence . You will also recall some concepts on sequence. What you are expected to learnThis module is designed for you to: 1. Recall sequence and differentiate finite and infinite sequence 2. Find the common difference and nth term given the first few terms of an arithmetic sequence 3. Find the first term and common difference or a specified nth term given two terms of an arithmetic sequence 4. Solve problems involving arithmetic means and harmonic sequence.How much do you know1. Find three terms of the sequence tn = 2 + 3na. 6, 8, 10 c. 5, 7, 9b. 8, 12, 16 d. 5, 8, 112. A sequence where each succeeding term is obtained by adding a fixed number is called a/an __________________.3. The fixed number between any two succeeding terms is called ________. 1

4. Give the first four terms of the arithmetic sequence for which the first term is 9 and the common difference is 7?a. 9, 12, 16, 21 c. 9, 12, 15 , 18b. 9, 16, 23, 30 d. 9, 13, 17, 215. Give the arithmetic sequence whose 5th term is 16 and whose 7th term is 24?a. 8, 12, 16, 20, … c. 7, 11, 15, 19, …b. 8, 11, 14, 17, … d. 9, 14, 19, 24, …6. Give the common difference of # 5 _____.7. Insert two arithmetic means between 5 and 20.8. Find the 10th term of the arithmetic sequence 9, 18, 27, 36, …9. Find the 5th term of the harmonic sequence _1_ , _1_ , _1_ , … 24610. Find the arithmetic mean of 2 and 16.What you will do Lesson 1 Recalling SequenceSequence is a set of numbers arranged in a pattern.Examples:1. For the sequence denoted by tn = __n__, find the first six terms. n+1t1 = __1__ = _1_ t4 = __4__ = _4_ 1+1 2 4+1 5t2 = __2__ = _2_ t5 = __5__ = _5_ 2+1 3 5+1 6 2

t3 = __3__ = _3_ t6 = __6__ = _6_ 3+1 4 6+1 7The first six terms of a sequence are _1_ , _2_ , _3_ , _4_ , _5_ , and _6_ . 2 34 5 6 72. Find the first five terms of the sequence tn = 5n + 2n.t1 = 5(1) + 21 = 5 + 2 = 7 t4 = 5(4) + 24 = 20 + 16 = 36t2 = 5(2) + 22 = 10 + 4 = 14 t5 = 5(5) + 25 = 25 + 32 = 57t3 = 5(3) + 23 = 15 + 8 = 23The first five terms of the sequence tn = 5n + 2n are 7, 14, 23, 36 and 57.Try this outA. Find the first three terms of the sequence defined by each equation.1. tn = n + 5 6. tn = 50 – 5n2. tn = (-3)n 7. tn = (2)n+13. tn = 2n + 5 8. tn = (n – 4)n4. tn = n(n + 10) 9. tn = 1/2(n – 6)5. tn = 7n + 3 10. tn = (-2) n + (n – 1) nB. Find the seventh terms in each sequence.1. 2, 4, 6, 8, … 6. 0, 1, 1, 2, 3, …2. 100, 250, 400, … 7. 1, 3, 6, 10, …3. 3, 10, 17, … 8. 1, 2, 4, 16, 32, …4. 1, 4, 9, 16, … 9. -1, 2, -3, 4, …5. 7, 17, 27, 37, … 10. 300, 200, 100, 0, … Lesson 2 Arithmetic Sequences Mr. Lorenzo’s daughter is graduating from elementary school thisyear. To prepare his daughter’s college education he is thinking of eithertaking an education plan for her, or save a fixed amount in a bank. Hehas already saved PhP8,000 for this purpose. He decided to savePhP500 monthly in a bank starting this year. 3

Think about it.1. How much will he have saved after one year?2. Let’s include the already saved PhP8,000 to his savings after a year, what is the total savings?Is this your answer? 1. PhP500 x 12 months = Ph6,000 2. Php 8,000 + PhP6,000 = Php 14,000Think about it. 3. How much will he have saved after two years? three years? four years? 4. How much money will he have saved when his daughter enrolls in college? 5. Mr. Lorenzo estimates that his daughter will need P30,000 for her four years in college. Will he have saved this amount by the time his daughter reaches the last year in college?Let’s summarize using a table.Year Savings1 8,0002 8,000 + 6,000 = 14,0003 8,000 + 6,000 + 6,000 = 20,0004 8,000 + 6,000 + 6,000 + 6,000 = 26,0005 8,000 + 6,000 + 6,000 + 6,000 + 6,000 =32,000 As shown on the equation, PhP6,000 is added to PhP8,000 savings on the firstyear, then Php6,000 is added to the savings in a given year to find the savings for thesucceeding year. This procedure of using previous values to generate new values iscalled iteration. If this procedure is followed, you will need to get the amount of savingseach year, 1st, 2nd, 3rd, and 4th before you can answer question #5. There is a better way! This is seen in the pattern on the right side of the table.Observed that pattern.Year Savings 1 2 8,000 = 8,000 + 0(6,000) 3 4 8,000 + 6,000 = 8,000 + 1(6,000) 5 8,000 + 6,000 + 6,000 = 8,000 + 2(6,000) 8,000 + 6,000 + 6,000 + 6,000 = 8,000 + 3(6,000) 8,000 + 6,000 + 6,000 + 6,000 + 6,000 = 8,000 + 4(6,000)Can you answer the following questions based on the given table? 1. What is the relationship between the number of times 6,000 is added to 8,000 and the year corresponding to that amount of savings? 2. How would you now answer question #5 without finding the amount of savings yearly? 4

How about if Mr. Lorenzo’s daughter decided to take a 5-year course? Can youanswer it without the use of the table? We can simplify this procedure if we will use a formula. Look at the relationshipbetween time and savings as shown in the table below.Year 1 2 3 4 5 6Savings 8,000 14,000 20,000 26,000 32,000 38,000 The numbers 8,000, 14,000, 20,000, 26,000, 32,000, 38,000 form a sequenceof six terms. Each number may be denoted as tn, where n= 1,2,3,…6.The numbers 8,000 corresponds to t1 14,000 corresponds to t2 20,000 corresponds to t3 26,000 corresponds to t4 32,000 corresponds to t5 38,000 corresponds to t6and so on to tn, It can be deduced that for the nth term tn(read t sub n), tn = 8,000 + (n – 1)(6,000) The value of n (year) which you substitute to the equation to solve tn (savings) isthe independent variable or x and the resulting tn is the dependent variable (y). The equation tn = 8,000 + (n – 1)(6,000) is a sequence where n = 1.2.3.4.5.6comprise the domain and the numbers are the range.Domain is the set of all the first elements of a relation/equation.Range is the set of all the second elements of a relation/equation.Domain = { 1, 2, 3, 4, 5, 6 }Range = { 8 000 , 14 000 , 20 000 , 26 000 , 32 000 , 38 000 }Let’s solve the problem using the equation tn = 8,000 + (n – 1)(6,000) with n = 6.tn = 8,000 + (6 – 1)(6,000)tn = 8,000 + 5(6,000)tn = 8,000 + 30,000 = 38,000Consider these sequence. Find the next four terms of each.1. 5, 25, 45, 65, … 3. 1, 9, 17, 25, … 5

2. 0, 9, 18, 27, … 4. -9, -4, 1, 6, …In each sequence, how will you get the next terms?Give the common difference of the following sequence. 1. 5, 25, 45, 65, … 25 – 5 = 20; 45 – 25 = 20; 65 – 45 = 20 so, we add a fixed value of 20 for every succeeding term thus, the common difference is 202. 0, 9, 18, 27, … 9 – 0 = 9; 18 – 9 = 9; 27 – 18 = 9 the common difference is 93. 1, 9, 17, 25, … 9 – 1 = 8; 17 – 9 = 8; 25 – 18 = 8 the common difference is 84. -9, -4, 1, 6, … -4 – (-9) = -4 + 9 = 5; 1 – (-4) = 1 + 4 = 5; 6 – 1 = 5 the common difference is 5Now, you are ready to find the next terms of the given sequence.Check if your answers are correct. 1. 5, 25, 45, 65, 85, 105, 125, 145 2. 0, 9, 18, 27, 36, 45, 54, 63 3. 1, 9, 17, 25, 33, 41, 49, 57 4. -9, -4, 1, 6, 11, 16, 21, 26How many correct answer/s did you get?How did you get your answers?1. 5, 25, 45, 65, 85, 105, 125, 145 the common difference is 20 65 + 20 = 85; 85 + 20 = 105; 105 + 20 = 125; 125 + 20 = 1452. 0, 9, 18, 27, 36, 45, 54, 63 the common difference is 9 27 + 9 = 36; 36 + 9 = 45; 45 + 9 = 54; 54 + 9 = 633. 1, 9, 17, 25, 33, 41, 49, 57 the common difference is 8 25 + 8 = 33; 33 + 8 = 41; 41 + 8 = 49; 49 + 8 = 57 6

4. -9, -4, 1, 6, 11, 16, 21, 26 the common difference is 5 6 +5 = 11; 11 + 5 = 16; 16 + 5 = 21; 21 + 5 = 26A sequence where each succeeding term is obtained by adding a fixednumber is called an arithmetic sequence. This fixed number is thecommon difference d between any two succeeding terms. The sequence that begins 1, 4, 7, 10, 13, 16, . . . is an arithmetic sequence sincethe difference between consecutive terms is always 3. The sequence that begins 8, 6, 4, 2, 0, -2, -4, . . . is an arithmetic sequence sincethe difference between consecutive terms is always -2. In order to identify if a pattern is an arithmetic sequence you must examineconsecutive terms. If all consecutive terms have a common difference you canconclude that the sequence is arithmetic.Consider the sequence: 4, 11, 18, 25, 32, . . .Since: 11 - 4 = 7 18 - 11 = 7 25 - 18 = 7 32 - 25 = 7the sequence is arithmetic. We can continue to find subsequent terms by adding 7.Therefore, the sequence continues: 39, 46, 53, etc.The terms (t) of the sequence can then be expressed this way:t1 = t1 + 0dt2 = t1 + dt3 = t2 + d = ( t1 + d) + d = t1 + 2dt4 = t3 + d = ( t1 + 2d) + d = t1 + 3dt5 = t4 + d = ( t1 + 3d) + d = t1 + 4d...tn = tn - 1 + d = [(t1 + (n – 2)d) + d = t1 + (n –1)d 7

where tn is the last term t1 is the first term n is the number of terms and d is the common difference Any arithmetic sequence is defined by the equation given as tn = t1 + (n –1)d The sequence 2, 5, 8, 11, 14 is an arithmetic sequence with the commondifference of 3. The defining equation of the sequence is y = 3x – 1 with the domain {1,2, 3, 4, 5}. Since the domain consists of positive integers 1 up to 5 we call this finitesequence. If all the positive integers comprise the domain, then the sequence is infinite. Theinfinite sequence 7, 2, -3, -8, -13, … is an arithmetic sequence with the commondifference -5 and the defining equation is y = 12 – 5x. An arithmetic sequence is any sequence for which the defining equation is linear.Linear equation is defined as y = mx + b. In equation tn = t1 + (n –1)d, what is y? m?x? and b?y is tn m is d x is (n – 1) and b is t1tn = t1 + (n –1)d y = b + (x – 1)m or y = m(x – 1) + b Can you now see the relationship between the domains and ranges of the linearequation and arithmetic sequence?Examples:1. Give the first five terms of the arithmetic sequence for which the first term is 9 and the common difference is 7?a. By iteration and Since t1 = 9, then t2 = 9 + 7 = 16 t3 = 16 + 7 = 23 t4 = 23 + 7 = 30 t5 = 30 + 7 = 37b. By using the equation tn = t1 + (n – 1)d 8

Since t1 = 9, n = 5, d = 7 t5 = 9 + (5 – 1)7 t5 = 9 + (4)7 = 9 + 28 = 372. Find the 100th term of #1 using the equation. t1 = 9, n = 100, d = 7 t5 = 9 + (100 – 1)7 t5 = 9 + (99)7 = 9 + 693 = 7023. In the arithmetic sequence 1, 8, 15, 22, …, which term equals 519? tn = 519, t1 = 1, n = ?, d = 7 519 = 1 + (n – 1)7 519 = 1 + 7n - 7 519 = 7n – 6 519 + 6 = 7n 525 = n 7 n = 75 519 is the 75th term of the arithmetic sequence.4. Solve the following problem. Mr. Simon bought a house at the beginning of 1995 for PhP 150,000.00. If it increased PhP 10,000.00 in value each year, how much was it worth at the end of 2005? In1995 the amount of the house bought by Mr. Simon was PhP 150,000. In thefollowing year, 1996, PhP10,000 was added to the original amount, thus having the newvalue of PhP160,000.00. Let’s place the data in the table to solve the problem.Year Value of the House1995 150,00001996 150,0000 + 10,000 = 160,0001997 160,0000 + 10,000 = 170,0001998 170,0000 + 10,000 = 180,0001999 180,0000 + 10,000 = 190,0002000 190,0000 + 10,000 = 200,0002001 200,0000 + 10,000 = 210,0002002 210,0000 + 10,000 = 220,0002003 220,0000 + 10,000 = 230,0002004 230,0000 + 10,000 = 240,0002005 240,0000 + 10,000 = 250,000 9

This problem illustrate an arithmetic sequence.Year Value of the House 1st 150,0000 2nd 150,0000 + 10,000 = 160,000 3rd 150,0000 + 2(10,000) = 170,000 4th 150,0000 + 3(10,000) = 180,000 5th 150,0000 + 4(10,000) = 190,000 6th 150,0000 + 5(10,000) = 200,000 7th 150,0000 + 6(10,000) = 210,000 8th 150,0000 + 7(10,000) = 220,000 9th 150,0000 + 8(10,000) = 230,00010th 150,0000 + 9(10,000) = 240,00011th 150,0000 + 10(10,000) = 250,000The house worth Php250,000 at the end of 2005.Let us use tn = t1 + (n – 1)d to solve the problem.t11 = 150,000 + (11 – 1)10,000 = 150,000 + 100,000 = 250,0005. Solve this problem. A restaurant has square tables which seat four people. When two tables areplaced together, six people can be seated (see the diagram) . If 20 square tables are placed together to form one long table, how many peoplecan be seated? If 1000 square tables are placed together to form one very long table, how manypeople can be seated? One approach to solving the problem is to make a table in order to see if there isa pattern that relates the number of tables to the number of people that can be seated. 10

Number of Diagram Number of Tables Seats 1 4 2 6 3 8 4 10 The pattern that is emerging is clearly an arithmetic sequence. The numbers inthe sequence begin 4, 6, 8, 10, ... . To find the number of people that can sit at 20 tables, use the formula: tn = t1 +(n – 1)d The first element: t1 = 4. The common difference: d = 2. The term: n = 20.t20 = 4 + (20 – 1)2 The 20th term = 4 + (19 X 2) = 4 + 38 = 42 Therefore, 42 people could sit at 20 tables. To find the number of people that can sit at 1000 tables, use the formula. The first element: t1 = 4. The common difference = d = 2. The term: n =1000. 11

t1000 = t1 + (1000 – 1)2The 1000th term = 4 + (999 X 2) = 4 + 1098 = 2002Therefore, 2002 people could sit at 1000 tables.Try this outA. Give the common difference of the following sequence.1. 2, 4, 6, … d = _____2. 13, 16, 19, 22, … d = _____3. 99, 88, 77, 66, … d = _____4. _1_, _1_, _3_, _1_, … d = _____ 8 48 25. 99, 87, 75, 63, … d = _____6. -8, -3, 2, 7, … d = _____7. 91, 84, 77, 70,… d = _____ ____ d = _____8. √3 , 2√3 , 3√3 , 4√3 , …9. 25, 34, 43, 52, … d = _____10. 10, 4, -2, -8, … d = _____B. Find the term indicated in each of the following arithmetic sequences.1. 2, 4, 6, … 15th term2. 13, 16, 19, 22, … 25th term3. 99, 88, 77, 66, … 18th term4. _1_, _1_, _3_, _1_, … 20th term 8 48 25. 99, 87, 75, 63, … 12th term6. -8, -3, 2, 7, … 23rd term 12

7. 91, 84, 77, 70,… 17th term ____ 14th term8. √3 , 2√3 , 3√3 , 4√3 , …9. 25, 34, 43, 52, … 10th term10. 10, 4, -2, -8, … 22nd termC. Find the first three terms defined by the following equations.1. 8 + _3_ a2. 3b + _b + 1_ 2b3. _5_ + 8 3c4. 2b + _4 + b_ b5. b2 + __2__ b+2D. Solve the following problems. 1. In the arithmetic sequence -3, 0, 3, 6, …, which term is equal to 138? 2. In the arithmetic sequence 15, 21, 27, 33, …, which term equals to 177? 3. The force of gravity causes a body to fall 16.1 decimeters during the first second, 48.3 the next second, 80.5 the third, and so on. How far will the body fall in 10 seconds? 13

Lesson 3Finding the 1st Term and Common Difference Given Two terms of Arithmetic SequenceExamples:1. If the 6th term of an arithmetic sequence is 27 and the 12th term is 48, find the first term.We have two equations,Equation 1 t6 = 27 27 = t1 + (6 – 1)d 27 = t1 + 5dEquation 2 t12 = 48 48 = t1 + (12 – 1)d 48 = t1 + 11dSubtract Equation 1 from Equation 2. 48 = t1 + 11d 27 = t1 + 5d_ 21 = 6d 21 = 6d 66 7=d 2It can also be solved by subtracting Equation 2 from Equation 1 27 = t1 + 5d 48 = t1 + 11d -21 = - 6d -21 = - 6d -6 - 6 7=d 2Substitute the value of d in the first equation we have, 27 = t1 + 5 ( 7 ) 2 27 = t1 + 35 2 14

54 – 35 = t1 2 19 or 9 1_ = t1 22You can also use the 2nd equation to find the first term. 48 = t1 + 11( 7 ) 2 48 = t1 + 77 96 – 77 = 2 2 t1 19 or 9 1_ = t1 222. Find the arithmetic sequence of 6 terms if the first term is _2_ and the lastis 7 _1_ . 3 3You need to find d first to solve the problem. _22_ = _2_ + (6 – 1)d 33 _22_ = _2_ + 5d 33 _22_ – _2_ = 5d 33 _20_ = 5d 3 _20_ = d 15 d = _4_ 3t1 = _2_ t2 = _2_ + _4_ = _6_ = 2 3 333t3 = 2 + _4_ = _6 + 4_ = _10_ t4 = _10_ + _4_ = _14_ 333 33 3 15

t5 = _14_ + 4_ = _18_ or 6 t6 = 6 + _4_ = _18 + 4 _ = _22_ 33 3 333Therefore the required sequence is _2_ , 2, _10_ , _14_ , 6, _22_ 3 33 33. What is the arithmetic sequence whose 23rd term is -107 and whose 55th term is -267? t23 = -107 and t55 = -267Since tn = t1 + (n – 1)dthen t23 = t1 + (23 – 1)d -107= t1 + 22dand t55 = t1 + (55 – 1)d -267= t1 + 54dYou take -107= t1 + 22d as Equation 1, and take -267= t1 + 54d as Equation 2.Equations 1 and 2 are two linear equations of two unknowns.-107 = t1 + 22d -107 = t1 + 22d– (-267 = t1 + 54d) + 267 = - t1 – 54d 160 = -32d 160 = d -32 d = -5Find t1 using equation 1. t23 = t1 + (23 – 1)-5 -107 = t1 + (22)-5 -107 = t1 + -110-107 + 110 = t1 t1 = 3You can also use equation 2 to find t1 t55 = t1 + (55 – 1)-5 -267 = t1 + (54)-5 -267 = t1 + -270-267 + 270 = t1 t1 = 3The arithmetic sequence is 3, -2, -7, -12, … 16

Try this out 1. Form an arithmetic sequence with 1st term 3 and 7th term 15. 2. Find the arithmetic sequence whose 34th term is -39 and 50th term is -61. 3. Find the arithmetic sequence whose 10th term is _31_ and 20th term is _71. 44 4. How many numbers less than 400 but greater than 10 are divisible by 7? 5. Find the 29th to the 35th terms of the resulting sequence in #4. 6. What are the first five terms of an arithmetic sequence whose 9th term is 16 and 40th term is 47? 7. The 18th and 52nd terms of an arithmetic sequence are 3 and 173 respectively. Find the 25th term. 8. Find the value of m so that 8m + 4, 6m – 2, and 2m – 7 will form an arithmetic sequence. Lesson 4 Solving Problems Involving Arithmetic Means In an arithmetic sequence, the term(s) between any two terms is (are)called arithmetic mean(s) between two terms.Examples:1. Find the arithmetic means between 2 and 8. Given two terms plus two terms means there are four terms in all. Assume thatt1 = 2 and t4 = 8. Let’s have the diagram of the sequence. 2 , __, __, 8 t1 , t2 , t3 , t4 tn = t1 + (n – 1)d 8 = 2 + (4 – 1)d 8 = 2 + 3d8 – 2 = 3d 6 = 3d d=2 17

Hence, t2 = 2 + (2 – 1)2 t2 = 2 + 2 = 4 t3 = 2 + (3 – 1)2 t3 = 2 + 4 = 6The numbers 4 and 6 are the two arithmetic means between 2 and 82. Find the five arithmetic means between 5 and 47. Given two terms plus five terms means there are seven terms in all. Assume thatt1 = 5 and t7 = 47. Let’s have the diagram of the sequence. 5 , __, __, __, __, __, 47 t1 , t2 , t3 , t4 , t5 , t6 , t7 tn = t1 + (n – 1)d 47 = 5 + (7 – 1)d 47 = 5 + 6d47 – 5 = 6d 42 = 6d d=7Hence, t2 = 5 + (2 – 1)7; t2 = 5 + 7 = 12 t3 = 5 + (3 – 1)7; t3 = 5 + 14 = 19 t4 = 5 + (4 – 1)7; t4 = 5 + 21 = 26 t5 = 5 + (5 – 1)7; t5 = 5 + 28 = 33 t6 = 5 + (6 – 1)7; t6 = 5 + 35 = 40The numbers 12, 19, 26, 33, and 40 are the five arithmetic means between 5 and 47.3. Insert six arithmetic means between 2 and 16. Also prove that their sum is 6 times the arithmetic mean between 2 and 16. 2 , __, __, __, __, __, __, 16 t1 , t2 , t3 , t4 , t5 , t6 , t7 , t8 Let t1 , t2 , t3 , t4 , t5 , t6 be the six arithmetic means between 2 and 16. Then, bydefinition, 2, t1, t2, ..., t6, 16 are in arithmetic progression. Let d be the common difference. Here 16 is the 8th term. tn = t1 + (n – 1)d t8 = 2 + (8 – 1)d 16 = 2 + (8 – 1)d 16 = 2 +7d16 – 2 = 7d 18

14_ = d 7 d=2Hence, the six arithmetic means are: 4, 6, 8, 10, 12, 14.Now the sum of these means4 + 6 + 8 + 10+ 12 + 14 = 54.Find the arithmetic mean between 2 and 16.Let d be the common difference. Here 16 is the 3rd term. tn = t1 + (n – 1)d t3 = 2 + (3 – 1)d 16 = 2 + (3 – 1)d 16 = 2 +2d16 – 2 = 2d 14_ = d 2 d=7Hence the arithmetic mean between 2 and 16 is 9.The sum of the arithmetic means 54 is 6 times the arithmetic mean between 2and 16, which is 9. 54 = 6(9)Try this outSolve what is asked:1. Insert four arithmetic means between -1 and 14.2. Insert five arithmetic means between 14 and 86.3. Insert three arithmetic means between -18 and 4.4. Insert four arithmetic means between 12 and -35. Insert one arithmetic mean between 24 and 68. Such a number is called the arithmetic mean of the two numbers.6. Find the arithmetic mean of 7 and -15.7. Find the four arithmetic means between 7 and -15.8. Find the arithmetic mean of _3_ and _5_. 539. Insert 5 arithmetic means between -2 and 10. Show that their sum is 5 times the arithmetic mean between -2 and 10.10. Insert 10 arithmetic means between -5 and 17 and prove that their sum is 10 times the arithmetic mean between -5 and 17. 19

Lesson 5 Solving Problems Involving Harmonic Sequence Harmonic Sequence is a sequence of numbers whose reciprocals form anarithmetic sequence.Examples:1. Insert two harmonic means between 1 and 1. 48Get the reciprocal of _1_ 4 4 and _1_ 8 84, ___, ___, 8t1 , t2 , t3 , t4Use tn = t1 + (n – 1)d to find d. 8 = 4 + (4 – 1) d 8 – 4 = 3d 4 = 3d _4_ = d 3add d to t1: 4 + _4_ = 12 + 4 = _16_ then get the reciprocal: 333t2 = _3_ 16Solve for t3 :16 + _4_ = _20_ then get the reciprocal:33 3t3 = _3_ 20To verify : The reciprocals of the terms form an arithmetic sequence. 20

The harmonic sequence is: The arithmetic sequence:_1_ , _3_ , _3_ , _1_ 4 , _16_ , _20_ , 8 4 16 20 8 33The common difference is 4. 32. Find the 15th term of the harmonic sequence -1, 1, _1_ , _1_ , … 35Get the reciprocal and the corresponding arithmetic sequence is -1, 1, 3, 5, …Find d using the first and the fourth terms t4 = t1 + 3d 5 = -1 + (4 – 1)d5 + 1 = 3d 6 = 3d _6_ = d 3 d=2Find the 15th term using t1 = -1, d = 2 and n = 15t15 = -1 + ( 15 – 1)2 = -1 + (14)2t15 = 27Try this outA. Find the indicated term of the following harmonic sequences.1. _1_ , _1_ , _1_ , _1_ , … 9th term 2 5 8 112. _2_ , _1_ , _2_ , _1_ , … 10th term 3 2 533. _1_ , _1_ , _1_ , _1_ , … 12th term 4 9 14 194. _1_ , _1_ , _1_ , _1_ , … 15th term 5 7 9 115. _- 4_ , -4 , _4_ , … 8th term 53 21

B. Find the harmonic mean of the following. 6. 1 and 1 48 7. _1_ and _1_ 83 8. _9_ and _4_ 88 9. Insert three harmonic means between _7_ and _1_ . 6 18 10. What is the resulting 8-term harmonic sequence if the related arithmetic sequence has for its 8th term 74 and common difference 7? Let’s SummarizeSequence is a set of numbers arranged in a patternDomain is the set of all the first elements of a relation/equation.Range is the set of all the second elements of a relation/equation.Arithmetic Sequence is a sequence where each succeeding term is obtained byadding a fixed numberCommon Difference (d) is a fixed number added to a preceding term to obtain thesucceeding term.Any arithmetic sequence is defined by the equation given as tn = t1 + (n –1)dThere are two types of sequence: finite and infinite sequenceFinite Sequence is an arithmetic sequence with finite elements in the domain.Infinite Sequence is an arithmetic sequence with infinite elements in the domain.An arithmetic sequence is any sequence for which the defining equation is linear.In an arithmetic sequence, the term(s) between any two terms is (are) called ArithmeticMean(s) between two terms. 22

Harmonic Sequence is a sequence of numbers whose reciprocals form an arithmeticsequence.What have you learned1. For the sequence denoted by tn = __2n__, find the first five terms. 5+na. _2_ , _ 3_ , _5 _ , _ 6_ , _ 7_ c. _2 _ , _ 4_ , _6 _ , _ 4_ , _ 5_7 8 9 10 11 6 78 9 10b. _3_ , _ 4_ , _5 _ , _ 6_ , _ 7_ d. _2 , _ 4_ , _6 _ , _ 4_ , _ 5_ 7 8 9 10 11 8 9 9 10 112. Find the first three terms of the sequence tn = 3 + 3n.a. 6, 12, 30, 84 c. 6, 12, 18, 24b. 6, 9, 12, 15 d. 6, 9, 18, 363. Give the first four terms of the arithmetic sequence for which the first term is5 and the common difference is 4?a. 5, 9, 12, 16 c. 5, 9, 12, 15b. 5, 9, 13, 16 d. 5, 9, 13, 174. Give the arithmetic sequence whose 7th term is 23 and whose 12thterm is 38?a. 5, 9, 13, 14, 18, … c. 5, 7, 9, 11, 13, …b. 5, 8, 11, 14, 17, … d. 5, 8, 12, 17, 23, …5. The common difference of # 4 is _____.6. Find the three arithmetic means between 9 and 33.a. 15, 21, 27 c. 14, 20, 26b. 12, 17, 22 d. 13, 20, 277. Find the 25th term of the arithmetic sequence 2, 5, 8, 11, …8. Insert four arithmetic means between 12 and 47.9. Find the 10th term of the harmonic sequence _1_ , _1_ , _1_ , _1_ , … 2 5 8 1110. Find the arithmetic mean of _2_ and _3_. 32 23

Key to correctionHow much do you know 6. 4 1. d 7. 10 and 15 2. arithmetic sequence 8. 90 3. common difference 9. 1 4. b 10 5. a 10. 9Try this out 6. 45, 40, 35Lesson 1 7. 4, 8, 16 8. -3, 4, -1A. 9. -5/2 , -2, -3/2 1. 6, 7, 8 10. -2, 5, 0 2. -3,. 9, -27 3. 7, 9, 11 6. 8 4. 11, 24, 39 7. 28 5. 10, 17, 24 8. 128 9. -7B. 10. -300 1. 14 2. 1,000 6. 5 3. 45 7. -7 4. 49 8. √ 3 5. 67 9. 9 10. -6Lesson 2A. 1. 2 2. 3 3. -11 4. 1/8 5. -12 24

B. 6. 102 1. 30 2. 85 7. -21 3. -88 8. 14 √3 4. 5/2 or 2 1/2 9. 106 5. -33 10. -116C. 1. 11, 19/2, 9 2. 4, 27/4, 29/3 3. 29/3, 53/6, 77/9 4. 7, 7, 25/3 or 8 1/3 5. 1 2/3, 4 ½, 9 2/5D. 1. 138 = -3 + (n –1)3 = -3 + 3n – 3 n = 144 = 48 48th term 3 2. 177 = 15 + (n –1)6 = 15 + 6n – 6 n = 168 = 28 28th term 6 3. t t0 = 16.1 + (10 – 1)32.2 = 16.1 + (9)32.2 = 16.1 + 289.8 = 305..9 decimetersLesson 3 1. 3, 5, 7, 9,… 2. -39 = t1 + (34 – 1)d = t1 + 33d Equation 1 -61 = t1 + (50 – 1)d = t1 + 49d Equation 2 Subtract Equation 1 from 2 - 22 = 16d d = -11_ t1 = 51 8 8 Find t1 by substituting d to any of Equation 1 or 2. The arithmetic sequence is _51_, 5, _29_, _9 , … 8 84 3. 31 = t1 + 9d Equation 1 4 71 = t1 + 19d Equation 2 4 25

d = 1, and t1 = -_5_ 4The arithmetic sequence is _-5_, _ -1_ , _3_, _7_ , … 4 4 444. d = 7, t1 = 14, and t1 = 399, find n = ? n = 565. t29 = 14 + (29 – 1)7 = 210 The 29th term to 35th term are 210, 217, 224, 231, 238, 245, 2526. d = 1, t1 = 8, and the first five terms are 8, 9, 10, 11 and 127. d = 5, t1 = -82, and the 25th term = 388. Find the common difference of the 2nd and 1st terms(6m – 2) – ( 8m – 4) = -2m – 6(2m – 7) – ( 6m – 2) = -4m – 5To find m equate -2m – 6 and -4m – 5 -2m – 6 = -4m – 5 m=½Lesson 41. 2, 5, 8, 112. 26, 38, 50, 62, 743. -_25_, -7, _-3_ 224. 12, 9, 6, 3, 0, -35. 24, 46, 686. 7, -4, -157. 7, -13 , -9, -31, -53, -15 5 55 58. _-17_ 159. d = 2 The 5 arithmetic means between -2 and 10 are 0, 2, 4, 6, 8 and the sum is 20 d=6 The arithmetic mean between -2 and 10 is 4 5(4) = 2010.d = 2 The 10 arithmetic means between -5 and 17 are -3, -1, 1, 3, 5, 7, 9, 11, 13 and15 and the sum is 60 d = 11 The arithmetic mean between -5 and 17 is 6 10(6) = 60 26

Lesson 5A. 1. t9 = _1_ 26 2. t10 = _1_ 63. t12 = _1_ 594. t15 = _1_ 33 5. t8 = _4_ 23B. 6. _1_ 67. _2_ 118. _9_ 139. _7_ , _7_, _7_ 36 66 9610. 25, 32, 39, 46, 53, 60, 67, 74Harmonic Mean _1_ , _1_ , _1_ , _1_ , _1_ , _1_ , _1_ , _1_ 25 32 39 46 53 60 67 7474 = t1 + (7)7 = 25What have you learned1. c 6. a2. a 7. 743. d 8. 19, 26, 33, 404. b 9. 295. 3 10. 13 12 27

Module 2 Inverse Variation What this module is about This module deals with the relation of two quantities where one valueincreases as the other value decreases and vice versa. Inverse variationconcepts can also be used to solve problems in other fields of mathematics. What you are expected to learn1. identify relationship between quantities that are inverse variation.2. find the constant of variation and the equation of relation.3. apply the concept of proportionality.How much do you knowA. Examine the tables of values and complete the sentence using the word increases or decreases.1. x 12 16 20 24 y 3456y _____________ as x ___________2. m -3 -2 4 12 n -4 -6 -13 -14n _____________ as m __________3. L 1.0 0.4 0.2 0.1 M 0.32 0.80 1.60 3.20M ____________ as L ___________4. l 18 12 9w 2 34w____________ as l ____________

5. d 7 10 15 18 20 c 7Π 10Π 15Π 18Π 20Π c ______________ as d ______________ B. Determine if the following situations show inverse variation or not. 1. The number of hours to finish a job to the number of men working. 2. The intensity of light from a bulb to the square of the distance from the bulb. 3. Air pressure to its altitude. 4. Family expenses as to the number of children. 5. The area of quadrilateral to the length of its side. What you will do Lesson 1 Inverse Variation A car is traveling a distance of 10 km at a speed of 10 km/hr, and it willtake one hour to finish the trip. At 30 km/hr, it will take 1 hr to finish the trip and 3so on. This determines a set of pairs of numbers, all having the same product:(10,1), (20, 1 ), (30, 1 ), (40, 1 ) and so on. 234In table form the relation between the speed r and the time of traveling t isshown: r 10 20 30 40 t 1 111 234Note that as the first number gets larger, the second number gets smaller. 2

Inverse variation occurs whenever a situation produces pairs of numbers,whose product is constant. Here, the time varies inversely as the speed: rt = 10 (a constant) The statement, “ y is inversely proportional to x ,” translates to y = k , xwhere k is the proportionality constant or constant of variation.Examples: 1. Find the equation and solve for k: y varies inversely as x and y = 6 when x = 18 . Solution: The relation y varies inversely as x translates to y = k . Substitute x the values to find k: y=k x 6= k 18 k = (6)(18) k = 108 The equation of variation: y = 108 x2. Find the constant of variation and the equation of the variation: y varies inversely as x , given that y = 24 and x = 0.3. Solution: The relation y varies inversely as x translates to y = k . Substitute x the values to find k: y=k x 24 = k 0.3 k = (24)(0.3) k = 7.2 3

The constant of variation is 7.2 and the equation of variation is y = 7.2 x3. y varies inversely as x. If y = 2 when x = 4 , find the value of x when y = 1. Solution: The relation y varies inversely as x translates to y = k . Substitute x the values to find k: y=k x 2= k 4 k = (2)(4) k =8 The constant of variation is 8, the equation of variation is y = 8 x To find x when y = 1: y=8 x 1= 8 x x = (8)(1) x=84. If y varies inversely as x and y = 27 when x = 3, find y when x = 9. Solution: The relation y varies inversely as x translates to y = k . x Substitute the values to find k: y=k x 27 = k 3 4