ADDITIONAL MATHEMATICS FORM 5 MODULE 2

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PROGRAM DIDIK CEMERLANG AKADEMIK SPM ADDITIONAL MATHEMATICS FORM 5 MODULE 2 PROGRESSIONS (Geometric Progressio) ORGANISED BY: JABATAN PELAJARAN NEGERI PULAU PINANG

CHAPTER 2 : GEOMETRIC PROGRESSIONS Cotets Page 2. CONCEPT MAP (GEOMETRIC PROGRESSIONS) 2 2.2 IDENTIFY CHARACTERISTICS OF GEOMETRIC PROGRESSIONS 2.3 DETERMINE WHETHER A GIVEN SEQUENCE IS A GEOMETRIC PROGRESSION 2.4 DETERMINE BY USING FORMULA 3 4 6 a) specific terms i geometric progressios b) The umber of terms i geometric progressios 2.5 FIND a) The sum of the first terms of of geometric Progressios b) The sum of a specific umber of cosecutive terms of geometric Progressios c) The value of,give the sum of the first terms of geometric Progressios 2.6 FIND a) The sum to ifity of geometric progressios b) The first term or commo ratio, give the sum to ifiity 2.7 SOLVE PROBLEMS INVOLING GEOMETRIC PROGRESSIONS 6 4 SPM Questios 5 Assessmet test Aswers 20 8 2

CHAPTER 2 : GEOMETRIC PROGRESSION 2. CONCEPT MAP PROGRESSION ARITHMETIC PROGRESSION GEOMETRIC PROGRESSION THE th TERM T T = ar -.a = first term.r = Commo ratio. = umber of terms SUM OF THE FIRST TERMS S = S = a, r > ( r ) r OR a( r r ), r < THE r th TERM T r T r = S r - S r SUM OF INFINITY S = a r, - < r < 3

PROGRESSIONS ( Geometric Progressio) GEOMETRIC PROGRESSION th The term : T = or T = The sum of the first term: S = or S = a = r = l = = Fill i the blak 2.2 Idetify characteristics of geometric progressio: EXERCISE : Complete each of the sequece below whe give a (fist term ) ad r (commo ratio). A.r the first four terms of geometric progressio, Example: a) -3 2-3, (-3)(2) = - 6, (-3)(2) 2 = -2, (-3)(2) 3 = -24.b) 3-2.c) 4 3.d) -6-2.e) 2 3.f) 4 y y 3 3 4

2.3 Determie whether a give sequece is a geometric progressio EXERCISE 2 : 2.3. Determie whether a give sequece below is a geometric progressio. Example :.r = commo ratio a) -8, 4, -2,.r = 4 8 = 2, 2 = 4 2 (true) b) 5,, 7, 23, c) 6, -8, 4,.. d) -20, -50, -30, -35,.. e) 3 x, 9 2 x, 27 4 x f) a 5, a 4 b, a 3 b 2 g),,,. 4 2 36 h),,, 27 9 3 5

EXERCISE 3 : 2.3.2 Give that the first three terms of a geometric progressio are below. Fid the value of x Geometric progressio Example : a).x, x + 4, 2x + 2, Value x x + 4 2x + = x x + 42 (x + 4) 2 = x(2x + 2).x 2 + 8x + 6 = 2x 2 + 2x 0 = x 2 6x 9 0 = (x 8)(x + 2).x 8 = 0 @ x + 2 = 0 Hece x = 8 @ x = -2 b) x, x + 2, x + 3 c) x + 3, 5x - 3, 7x + 3 e) x 6, x, 2x + 6 6

2.4 Determie by usig formula: EXERCISE 4: T = ar - 2.4. specific terms i arithmetic progressios Example :. Fid the 7 th term of the geometric progressio. - 8, 4, -2,.. Solutio: a = - 8 r = T 7 = (-8)( = 8 2 )7-4 8 = 2 2. Fid the 8 th term of the geometric progressio. 6, -8, 4, 3. For the geometric progressio 4 2,,,..,fid the 9 th term. 9 3 4. Fid the 3 th term of the geometric progressio 50, 40, 32. 5. Fid the 0 th term of geometric progressio a 5, a 4 b, a 3 b 2 6. Give that geometric progressio 5.6,.4, 0.35, Fid the 0 th term. 7

2.4.2 Fid the umber of terms of the arithmetic progressio EXERCISE 5: Example : a) 64, - 32, 6,.- 8 b) 2, -4, 8, 52 T = - 8 a = 64, r = - 2 64 = - 2 8 = 2 64 8 = 6 2 2 2 = 9 2 2. - = 9. = 0 3. c) 405, - 35, 45, - 27 5 4 2 8.d),,..6, 9 3 8

2.5 Fid a) The sum of the first terms of geometric progressios b) The sum of a specific umber of cosecutive terms of geometric progressios c) The value of, give the sum of the first terms of geometric progressios. S = S = a( r ), r > r OR a( r ), r < r 2.5. Fid the sum of the first terms of geometric progressios EXERCISE 6: geometric progressios Fid the sum of the first term Example :.a) 2, - 4, 8,.a = 2 r = 4 2 = - 2 7 2( ( 2) ) S 7 = ( 2) = 86 b) 5, 0, 20.. S 8 c) 2, -6, 3 S 9 d) 3 x, 9 2 x, 27 4 x,. S 6 9

2.5.2 The sum of a specific umber of cosecutive terms of geometric progressios EXERCISE 7: geometric progressios Example.a) 4, 2,, 64.a = 4 r = 4 2 = 2 Fid the sum of the first term T = 64 4 = 2 64 = 2 4 64 = 2 2 2 2. = 8. = 9 S 9 = 9 [ ( / 2) ] ( / 2) 4 6 63 = 7 64 3, 3, 9,. 287. c) 24, 2, 6,. 4 3 0

2.5.3 The value of, give the sum of the first terms of geometric progressios. EXERCISE 8 : Example :.a) The first ad 4 th tems of a geometric progressio are 2 Fid the value of r Solutio : T 4 = 27 28 ad 27 28. 27 2 (r 3 ) = 28 (r 3 ) = 27 28 2 (r 3 ) = 27 64, Hece r = 3 4 b) The first ad 6 th tems of a geometric progressio are 2 2 ad 607 2. Fid the value of r c) The commo ratio ad 5 th tems of a geometric progressio are 2 3 Fid the value of a ad 7 22 27.

2.4 Fid : a) the sum to ifiity of geometric progressios b) the first term or commo ratio, give the sum to ifiity of geometric progressios. a S = r S = sum to ifiity a = first term r = commo ratio EXERCISE 9: Fid the sum to ifiity of a give geomertric progressio below:. 24, 3.6, 0.54,. Example: 2 2 6, 2,,,... 3 9 a = 6 2 r = = 6 3 a S = r 6 = - - 3 9 = 2 2. 8, -27,9,.. 3.,,,..... 2 4 8 EXERCISE 0: 2

. The sum to ifiity of a geometric progressio is 200. Give that the first term is 52. Fid the commo ratio. 2. Give that the commo ratio of geometric progressio is 25. The sum of the first terms,where is large eough such that r = 0 is 75. Fid the first term. Example: The sum to ifiity of a geometric progressio is 8. Give that the first term is 2. Fid a) the commo ratio b) the third term Solutio: a) S = 8 a = 8 r 2 = 8 r 3 r= 4 3. The sum to ifiity of a geometric progressio is 600 ad the commo ratio is 0.4. Fid a) the first term b) the miimum umber of terms such that the sum of terms to be more the 599. b) T = ar 3 =2 4 9 = 8 3 3

4. Express each of the recurrig decimal below as a fractio i its simplest form. Example: 0.3 0.3 = 0.3 + 0.03 + 0.003 +.. 0.3 = 0. 0.3 = 0.9 = 3 Example: 4.020202 4.020202... = 4 + 0.02 + 0.0002 + 0.000002 +... 0.02 = 4 + -0.0 0.02 = 4 + 0.99 2 =4 99 a) 0.444. c).222.. b) 0.232323 d) 5.070707... 4

2.5 Solve problems ivolvig geometric progressios: EXERCISE : Example: A garderer has a task of cuttig the grass of a law with a area of 000 2 m. 2 O the first day, he cut a area of 6m. O each successive day, he cuts a area. times the area that he cut the previous day the task is completed. Fid a) the area that is cut o the 0 th day. b) The umber of the days eeded to complete the task. Osma is allowed to sped a allocatio of RM millio where the maximum withdrawal each day must ot exceed twice the amout withdraw the day before. If Osma withdraws RM200 o the first day, determie after how may days the amout of moey allocated will all be used up. Solutio: a) a=6, r=. T = ar T 0 = 6(.)9 = 37.73 m 2 b) S 000 6. - 000.- 60(. ) 000. 6.25. > 7.25 log. log 7.25 log. log 7.25 log7.25 > log. >20.79 The smallest iteger value of is 2 hece,the umber of the days eeded to complete the task is 2 days. 5

SPM QUESTIONS:. 2003 (Paper : No.8) I a geometric progressio, the first term is 64 ad the fourth term is 27. Caculate (a) the commo ratio (b) the sum to ifiity of the geometric progressio. [4 marks] 2. 2004(Paper : No.9) Give a geometric progressio 4 y,2,, p,...,express p i terms of y. y 3. 2004(Paper : No.2 ) Express the recurrig decimal 0.969696 as a fractio i its simplest form. [4 marks] 6

4. 2004(Paper2: Sectio A: No.6) Diagram 2 shows the arragemet of the first three of a ifiite series of similar triagles. The first triagle has a base of x cm ad a height of y cm. The measuremets of the base ad height of each subsequet triagle are half of the measuremets of its previous oe. y cm xcm Diagram 2 (a) Show that the areas of the triagles form a geometric progressio ad state the commo ratio. [3 marks] (b) Give that x= 80 cm ad y= 40 cm, 2 (i) determie which triagle has a area of 6 cm, 4 2 (ii) fid the sum to ifiity of the areas, i cm, of the triagles. [5 marks] 5. 2005 (Paper : No.0) 7

The first three terms of a sequece are 2, x, 8 Fid the positive value of x so that the sequece is (a) a arithmetic progressio (b) a geometric progressio [4 marks] 6. 2005 (Paper : No. 2) The sum of the first terms of the geometric progressio 8,24,72,.is 8744. Fid (a) the commo ratio of the progressio (b) the value of [4 marks] ASSESSMENT: 8

. The first three terms of a geometric progressio are 2x + 3, x ad x 2 with a commo ratio r, where - < r <. Fid (a) the value of x (b) the sum of the first terms,where is large eough such that r 0 2. I the progressio 5, 0, 20, 40,. Fid the least umber of terms required such that their sum exceeds 000. 3. The third term ad the sixth term of a geometric progressio are 27 ad 8 respectively. Fid the secod term. 4. I a geometric progressio, the sum of the first five terms is 3. 8 9

Give that the commo ratio is 2. Fid (a) the first term (b) the sum of all the terms from the fourth to the sixth term. 5. The third term of a geometric progressio exceeds the secod term by 6 while the fourth term exceeds the third term by 2. Fid the sum of the first 5 terms. ANSWERS: EXERCISE : EXERCISE 2: 20

b) 3, 6, -2, 24 c) 4, 2, 36, 08 d) -6, 2, 24, 48 e) 2, 6, 8, 54 2 3 4 y 4y 6y 64y f),,, 3 9 27 8 EXERCISE 3: b) x = -4 c) x = 3 d) x = -2 @ x = 8 EXERCISE 5: b) = 9 c) = 8 d) = 7 EXERCISE 7: c) = 8, S 8 = 3280 d) = 9, S 9 = -022 a) true b) false c) true d) false e) true f) true g) false h) true EXERCISE 4:. T 7 = - 8 2. T 8 = - 2 3. T 9 = 26344 2304 4. T 3 = 32 9 b 5. T 0 = 4 a 6. T 0 = 0.00002 EXERCISE 6: b) 275 c) 8 64 d) 665 729 EXERCISE 8: b) r = 3 c) a = 3 e) = 6, S 6 = 47 4 EXERCISE 9:. 28.24 3 2. 60 4 3. EXERCISE 0: 2. r=0.74 3. a=72 4. a) 360 b) 7 4 a) 9 5. 23 b) 99 4 c) 33 7 d) 5 99 2

EXERCISE : 3 days SPM QUESTION: 3. a) r b) S 256 4 8 2. p 2 y 3. 32 33 4. a) r 4 b)i. =5 ii. 233 3 5. a) x=5 b) x=4 6. a) r=3 b) =7 ASSESSMENT:. a) x=3 b) 27 2 2. 8 3. 40 2 4. a) -2 b) 5. 40 3 7 6 22

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