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Worksheet 3 6 Arithmetic and Geometric Progressions

Worksheet3:6 ArithmeticandGeometricProgressionsSectio n1 ArithmeticProgressionAnarithmeticprogres sionis a listof numberswherethedi erencebetweensuccessive numbersis constant. Thetermsin an arithmeticprogressionareusuallydenotedas u1; u2; theinitialtermin theprogression,u2is thesecondterm,andso on;unis thenth anarithmeticprogressionis2;4;6;8;10;12;1 4; : : :Sincethedi erencebetweensuccessive termsis constant, we haveu3 u2=u2 u1andin generalun+1 un=u2 u1We willdenotethedi erenceu2 u1asd, which is a : Giventhat3,7and11arethe rstthreetermsin anarithmeticprogression,whatisd?

A geometric series is a geometric progression with plus signs between the terms instead of commas. So an example of a geometric series is 1+ 1 10 + 1 100 + 1 1000 + We can take the sum of the rst n terms of a geometric series and this is denoted by Sn: Sn = a(1 rn) 1 r Example 5 : Given the rst two terms of a geometric progression as 2 and 4, what

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Transcription of Worksheet 3 6 Arithmetic and Geometric Progressions

1 Worksheet3:6 ArithmeticandGeometricProgressionsSectio n1 ArithmeticProgressionAnarithmeticprogres sionis a listof numberswherethedi erencebetweensuccessive numbersis constant. Thetermsin an arithmeticprogressionareusuallydenotedas u1; u2; theinitialtermin theprogression,u2is thesecondterm,andso on;unis thenth anarithmeticprogressionis2;4;6;8;10;12;1 4; : : :Sincethedi erencebetweensuccessive termsis constant, we haveu3 u2=u2 u1andin generalun+1 un=u2 u1We willdenotethedi erenceu2 u1asd, which is a : Giventhat3,7and11arethe rstthreetermsin anarithmeticprogression,whatisd?

2 7 3 = 11 7 = 4 Thend= 4. Thatis, thecommondi erencebetweenthetermsis we know the rsttermin anarithmeticprogression, andthedi erencebetweenterms,thenwe canworkoutthenth term, canworkoutwhatany termwillbe. Theformulawhichtellsus whatthenth termin anarithmeticprogressionisun=a+ (n 1) dwhereais the : If the rst3 termsin anarithmeticprogressionare3,7,11thenwhat is the10thterm?The rsttermisa= 3, andthecommondi erenceisd= + (n 1)du10=3 + (10 1)4=3 + 9 4=391 Example3: If the rst3 termsin anarithmeticprogressionare8,5,2thenwhati sthe16thterm?In thisprogressiona= 8 andd= + (n 1)du16=8 + (10 1) ( 3)= 37 Example4: Giventhat2x;5 and6 xarethe rstthreetermsin anarithmeticprogression, whatisd?

3 5 2x=(6 x) 5x=4 Sincex= 4, thetermsare8, 5, 2 andthedi erenceis 3. Thenexttermin thearithmeticprogressionwillbe can ndthesumof the rstnterms,which we willdenotebySn, usinganotherformula:Sn=n2[2a+ (n 1)d]Example5: If the rst3 termsin anarithmeticprogressionare3,7,11thenwhat is thesumof the rst10 terms?Notethata= 3,d= 4 andn= (2 3 + (10 1) 4)=5(6+ 36)=210 Alternatively, butmoretediously, we addthe rst10 termstogether:S10= 3 + 7 + 11 + 15 + 19 + 23 + 27 + 31 + 35 + 39 = 210 Thismethod wouldhave drawbacks if we hadto add100termstogether!Example6: If the rst3 termsin anarithmeticprogressionare8,5,2thenwhati sthesumof the rst16 terms?

4 S16=162(2 8 + (16 1) ( 3))=8(16 45)= 2322 Exercises:1. For each of thefollowingarithmeticprogressions, ndthevaluesofa,d, andtheunindicated.(a)1, 4, 7,: : :, (u10)(b) 8, 6, 4,: : :, (u12)(c)8, 4, 0,: : :, (u20)(d) 20, 15, 10,: : :, (u6)(e)40,30,20,: : :, (u18)(f) 6, 8, 10,: : :, (u12)(g)2, 212, 3,: : :, (u19)(h)6, 534, 512,: : :, (u10)(i) 7, 612, 6,: : :, (u14)(j)0, 5, 10,: : :, (u15)2. For each of thefollowingarithmeticprogressions, ndthevaluesofa,d, andtheSnindicated.(a)1, 3, 5,: : :, (S8)(b)2, 5, 8,: : :, (S10)(c)10,7, 4,: : :, (S20)(d)6, 612, 7,: : :, (S8)(e) 8, 7, 6,: : :, (S14)(f) 2, 0, 2,: : :, (S5)(g) 20, 16, 12,: : :, (S4)(h)40,35,30,: : :, (S11)(i)12,1012, 9,: : :, (S9)(j) 8, 5, 2,: : :, (S20)Section2 GeometricProgressionsA geometricprogressionis a listof termsas in anarithmeticprogressionbutin thiscasetheratioof successive termsis a constant.

5 In otherwords,each termis a constant 'swritethetermsin a geometricprogressionasu1; u2; u3; u4andso a geometricprogressionis10;100;1000;10000; : : :Sincetheratioof successive termsis constant, we haveu3u2=u2u1andun+1un=u2u1 Theratioof successive termsis usuallydenotedbyrandthe rsttermagainis : Findrforthegeometricprogressionwhose rstthreetermsare2, 4, 2 Thenr= : Findrforthegeometricprogressionwhose rstthreetermsare5,12, 5 =120 12=110 Thenr= we know the rsttermin a geometricprogressionandtheratiobetweensu ccessive terms,thenwe canworkoutthevalueof any termin thegeometricprogression.

6 Thenthtermisgivenbyun=arn 1 Again,ais the rsttermandris thatarn 16= (ar)n : Given the rsttwo termsin a geometricprogressionas 2 and4, whatis the10thterm?a= 2r=42= 2 Thenu10= 2 29= : Given the rsttwo termsin a geometricprogressionas 5 and12, whatis the7thterm?a= 5r=110 Thenu7=5 (110)7 1=51000000=0:0000054A geometricseriesis a a geometricseriesis1 +110+1100+11000+ We cantake thesumof the rstntermsof a geometricseriesandthisis denotedbySn:Sn=a(1 rn)1 rExample5: Given the rsttwo termsof a geometricprogressionas 2 and4, whatis thesumof the rst10 terms?We know thata= 2 andr= 2.

7 ThenS10=2(1 210)1 2=2046 Example6: Given the rsttwo termsof a geometricprogressionas 5 and12, whatis thesumof the rst7 terms?We know thata= 5 andr=110. ThenS7=5(1 1107)1 110=51 1107910=5:555555In certaincases,thesumof thetermsin a geometricprogressionhasa limit(notethatthisissummingtogetheranin nitenumber of terms).A serieslike thishasa limitpartlybecauseeach successive termwe areaddingis smallerandsmaller(butthisfactin itselfis notenoughto say thatthelimitingsumexists).Whenthesumof a geometricserieshasa limitwe saythatS1existsandwe can ndthelimitof moreinformationonlimits, thatris greaterthan 1 butlessthan1, <1.

8 If thisis thecase,thenwe canusetheformulaforSnabove andletngrow arbitrarilybigso thatrnbecomesas closeas we like to ris thelimitof thegeometricprogressionso longas 1< r < : Thegeometricprogressionwhose rsttwo termsare2 and4 does nothave aS1becauser= 26< : For thegeometricprogressionwhose rsttwo termsare5 and12, ndS1. Notethatr=110sojrj<1, so r=51 110=559 Sothesumof 5 +12+120+1200+: : :is 559 Example9: Considera geometricprogressionwhose rstthreetermsare12, 6and3. Noticethatr= 12. (1 rn)1 r=12(1 ( 12)8)1 ( 12) 7:967S1=a1 r=121 ( 12)=123=2=8 Exercises:1.

9 Findthetermindicatedforeach of thegeometricprogressions.(a)1, 3, 9,: : :, (u9)(b)4, 8, 16,: : :, (u10)(c)18, 6, 2,: : :, (u12)(d)1000,100,10,: : :, (u7)(e)32, 8, 2,: : :, (u14)(f) 0:005, 0:05, 0:5,: : :, (u10)(g) 6, 12, 24,: : :, (u6)(h)1:4, 0:7, 0:35,: : :, (u5)(i)68, 34,17,: : :, (u9)(j)8, 2,12,: : :, (u11)62. Findthesumindicatedforeach of thefollowinggeometricseries(a)6 + 9 + 13:5 + (S10)(b)18 9 + 4:5 + (S12)(c)6 + 3 +32+ (S10)(d)6000+ 600+ 60 + (S20)(e)80 20 + 5 + (S9) For each of thefollowingprogressions,determinewhethe rit is Arithmetic , Geometric ,orneither:(a)5, 9, 13,17,: : :(b)1, 2, 4, 8,: : :(c)1, 1, 2, 3, 5, 8, 13,21,: : :(d)81, 9, 3,13,: : :(e)512,474,436,398,: : :2.

10 Findthesixthandtwentiethterms,andthesumo f the rst10termsof each of thefollowingsequences:(a) 15, 9, 3,: : :(b)log7, log14,log28,: : :(c)116,18,14,: : :(d) , , ,: : :(e)64, 32,16,: : :3.(a)Thethirdandeighthtermsof anAPare470and380respectively. Findthe rsttermandthecommondi : writeexpressionsforu3andu8andsolvesimult aneously.(b)Findthesumto 5 termsof thegeometricprogressionwhose rsttermis 54andfourthtermis 2.(c)Findthesecondtermof a geometricprogressionwhosethirdtermis94an dsixthtermis 1681.(d)Findthesumtontermsof anarithmeticprogressionwhosefourthand fthtermsare13 (a)A university lecturerhasanannualsalaryof $40, thisincreasesby 2%eachyear,how much willshehave grossedin totalafter10 years?


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