11.3 Geometric Sequences and Series - ClassZone

1 Page 1 of 2666 Chapter 11 Sequences and SeriesGeometric Sequencesand SeriesUSINGGEOMETRICSEQUENCES ANDSERIESIn a the ratio of any term to the previous term is constant. This constant ratio is called the and is denoted by Geometric SequencesDecide whether each sequence is , 2, 6, 24, 120, .. , 27, 9, 3, 1, ..SOLUTIONTo decide whether a sequence is Geometric , find the ratios of consecutive aa21 = 21 = 2 aa32 = 62 = 3 aa43 = 264 = 4 aa54 = 12240 = 5 The ratios are different, so the sequence is not aa21 = 2871 = 13 aa32 = 297 = 13 aa43 = 93 = 13 aa54 = 13 The ratios are the same, so the sequence is a Rule for the nth TermWrite a rule for the nth term of the sequence 8, 12, 18, 27, .. Then find sequence is Geometric with first term a1= 8 and common ratio r= 182 = 32.

2 So, a rule for the nth term is:an= a1rn 1 Write general 8 }32} n 1 Substitute for a1and 8th term is a8= 8 32 8 1= 211687 .EXAMPLE 2 EXAMPLE 1common ratiogeometric sequence ,GOAL1 Write rules forgeometric Sequences andfind sums of geometricsequences and Series tomodel real-lifequantities,such as monthly bills forcellular telephone service in Example 6. To solve real-lifeproblems, such as finding the number of tennis matches played in Exs. 70 and you should learn itGOAL2 GOAL1 What you should nth term of a Geometric sequence with first term a1and common ratio risgiven by:an= a1rn 1 RULE FOR A Geometric SEQUENCEREALLIFEREALLIFEPage 1 of Sequences and Series667 Finding the nth Term Given a Term and the Common RatioOne term of a Geometric sequence is a3= 5.

3 The common ratio is r= a rule for the nth the by finding the first term as a1rn 1 Write general a1r3 1 Substitute 3 for = a1(2)2 Substitute for a3and = a1 Solve for , a rule for the nth term is:an= a1rn 1 Write general (2)n 1 Substitute for a1and graph is shown at the right. Notice that thepoints lie on an exponential curve. This is true foranygeometric sequence with r> the nth Term Given Two TermsTwo terms of a Geometric sequence are a2= 45 and a5= 1215. Find a rule for the nth system of equations using an=a1rn 1and substituting 2 for n(Equation 1) and then 5 for n(Equation 2).a2= a1r2 1 45 = a1rEquation 1a5= a1r5 1 1215 = a1r4 Equation 2 Solvethe system. 4r5 = a1 Solve Equation 1 for a1.

4 1215 = 4r5 (r4)Substitute for a1 in Equation 2. 1215 = 45r3 Simplify. 27 = r3 Divide each side by 45. 3 = rTake the cube root of each = a1( 3)Substitute for rin Equation 1. 15 = a1 Solve for a rule for a1rn 1 Write general 15( 3)n 1 Substitute for a1and r. A rule for the nth term is an= 15( 3)n 4 EXAMPLE 3120nan35760100140 Page 1 of 2668 Chapter 11 Sequences and SeriesThe expression formed by adding the terms of a Geometric sequence is called aAs with an arithmetic Series , the sum of the first nterms of ageometric Series is denoted by Sn. You can develop a rule for Snas a1+ a1r+ a1r2+ a1r3+ ..+ a1rn 1 rSn=0 a1r a1r2 a1r3 .. a1rn 1 a1rnSn(1 r) = a1 o0o o0o0 o0o0 .. o0o0 0 a1rnTherefore, Sn(1 r) = a1(1 rn).

5 If r 1, you can divide both sides of thisequation by 1 rto obtain the following rule for a SumConsider the Geometric Series 1 + 5 + 25 + 125 + 625 + .. the sum of the first 10 nsuch that Sn= begin, notice that a1= 1 and r= 5. Therefore:S10= a1 11 rr10 Write rule for 1 11 5510 Substitute for a1and 2,441,406 Simplify. The sum of the first 10 terms is 2,441, =SnWrite general = 3906 Substitute for a1, r, and 5n= 15,624 Multiply each side by 4. 5n= 15,625 Subtract 1 from each 15,625 Divide each side by loglo1g5,5625 = 6 Solve for n. So, Sn= 3906 when n= 5n 41 5n 1 51 rn 1 rEXAMPLE 5geometric sum Snof the first nterms of a Geometric Series with common ratio r 1 is:Sn= a1 11 rrn THE SUM OF A FINITE Geometric SERIESHOMEWORK HELPV isit our Web extra Back For help with logarithmicequations, see p.

6 1 of Sequences and Series669 GEOMETRICSEQUENCES ANDSERIES INREALLIFEW riting a Geometric SequenceCELLULARTELEPHONESIn 1990 the average monthly bill for cellular telephoneservice in the United States was $ From 1990 through 1997, the averagemonthly bill decreased by about per year. Source: Statistical Abstract of the United a rule for the average monthly cellular telephone bill an(in dollars) interms of the year. Let n = 1 represent was the average monthly cellular telephone bill in 1993? did the average monthly cellular telephone bill fall to $50? the average monthly bill decreased by the same percent each year, the average monthly bills from year to year form a Geometric sequence . Use a1= and r = 1 = A rule for the average monthly bill is:an= ( )n 1993, n= 4.

7 So, the average monthly bill was a4= ( )3 $ want to find nsuch that an= ( )n 1= 50 Write equation using rule for an.( )n 1 each side by 1 for n 1. n 6 Solve for average monthly cellular telephone bill reached $50 in 1995 (when n= 6).Finding the Sum of a Geometric SeriesUse the model for the average monthly cellular telephone bill in Example 6. Onaverage, what did a person pay for cellular telephone service during 1990 1997?SOLUTIONB ecause the model an= ( )n 1gives the average monthlybill, the model bn= 12( )( )n 1= ( )n 1gives the average annualbill. Using a1= and r= , you can estimate a person s total cost for cellulartelephone service during the 8 year period 1990 1997 to be:S8= a1 11 rr8 Write rule for 11 ( )8 Substitute for a1and r.

8 5790 Simplify. A person paid about $5790 for cellular telephone service during 1990 7 EXAMPLE 6 GOAL2 CELLULARTELEPHONESIn 1990 there were about 5 million cellular phonesubscribers. By 1997 thenumber had grown to over55 ONAPPLICATIONSREALLIFEREALLIFEC ellular PhonesDATA UPDATEV isit our Web INTERNETSTUDENTHELPPage 1 of 2670 Chapter 11 Sequences and this statement: The constant ratio in a Geometric sequence is called the ?ratio and is denoted by ?. makes a sequence Geometric ? the rule for the sum of the first nterms of a Geometric the common ratio of the Geometric , 12, 36, 108, 324, .. , 6, 36, 216, 1296, .. , 6, 18, 54, 162, .. , 14, 28, 56, 128, .. , 32, 16, 8, 4, .. , 5, 52 , 54 , 58 , ..Write the next term and find a rule for the nth term of the Geometric , 3, 9, 27.

9 , 8, 32, 128, .. , 6, 36, 216, .. , 75, 15, 3, ..14. 12 , 14 , 18 , 116 , ..15. 28, 14, 7, 72 , 74 , ..Write a rule for the nth term of the Geometric , a1= 2, a1= 3, a1= 14 , a3= 5, a4= 15 28, a5= the sum of the first 8 terms of the Geometric Series 1 + 8 + 64 + 512 +.. the model from Example 6 to find the averagemonthly bill for cellular telephone service in 1997. CLASSIFYINGSEQUENCESD ecide whether the sequence is arithmetic, Geometric , or neither. Explain your , 24, 96, 384, .. , 3, 7, 13, .. , 13, 22, 31, .. , 1, 5, 9, ..28. 11, 7, 3, 1, ..29. 12 , 32 , 92 , 227 , ..30. 13 , 23 , 1, 43 , ..31. 34 , 18 , 116 , 332 , ..32. 35 , 245 , 1525 , 6625 , ..FINDINGCOMMONRATIOSFind the common ratio of the Geometric , 4, 16, 64.

10 , 6, 12, 24, ..35. 3, 6, 12, 24, .. , 40, 320, 2560, .. , 68, 34, 17, ..38. 14 , 18 , 116 , 312 , ..WRITINGTERMSW rite a rule for the nth term of the Geometric find , 4, 16, 64, .. , 10, 20, 40, .. , 14, 98, 686, .. , 30, 150, 750, .. , 53 , 59 , 257 , .. , 43 , 89 , 1267 , ..PRACTICEANDAPPLICATIONSGUIDEDPRACTICEV ocabulary Check Concept Check Skill Check STUDENTHELPHOMEWORK HELPE xample 1:Exs. 24 32 Example 2:Exs. 33 44 Example 3:Exs. 45 49,54 59 Example 4:Exs. 50 53 Example 5:Exs. 60 69 Examples 6, 7:Exs. 70 79 Extra Practiceto help you masterskills is on p. 1 of Sequences and Series671 WRITINGRULESW rite a rule for the nth term of the Geometric , a1= 13 , a1= , a3= 18 , a1= 8, a1= 12 , a4= 10, a6= 20, a4= 30, a5= 3750 GRAPHINGSEQUENCESG raph the Geometric 4(2)n 3(5)n 2(3)n 8(3)n 5 12 n 4 32 n 1 FINDINGSUMSFor part (a), find the sum of the first nterms of the geometricseries.