chaptefl: ar;thmetic and geometric series

8/14/2019 ChapteFl: Ar;Thmetic and Geometric Series

1/17

ChapteFl: Ar;thmetic and Geometric SeriesAt the cnd oftbe lesson, students should be able to :

a) recognize that a sequence is an arithmctic progression (AP) when ihere is acornmon dill'erence between consccutive terms:b) use the lonnula lbr the ,th term of an AP;c) use the formula for the sum to , tems of an arilhmetic series;(J) use lhc lhnnuld lor rl'c sunr to n n,ltural numbers. i.e. S. - 41-l):e) recognjze that a sequence is a geoDetric progression (GP) wh; there is acommon r-atio between consecutive lerms:0 use thc lbnnula for thc ,?th term of a GP;g) use thc lonnula for thc sum to 7] tenns of a geometric series;h) undeNland that r" > 0 as r --) co rvhen lrl < 1 and use it to dcduce the sum toinlinity o1_ a geometric series, and the condilion fbr the sum to infinity to exist;i) detenninc thc finitc sum ofa scries made up olarithmetic and geometric

seri.,s, lor examplc, lin.1 ) ( a,, + r,, ) where r,, and v, are the ,?th terms ofarithmetic and geometric series respectively;j) solve practical problems involving aritlunetio and geonretric serics

Delinition:A ,-tql, is a listing ofiems in a defined order with a rule lor obtaining thetcnn\A sequence can be finite or infinite. Here are some examples olsequences:Spot the pattem in thc iollowing scquences: lt { ir' iiq ill t JtncJ.(u) 2,6,10,14,ts ( :tl!l1. q'rcnc< )Difference of +4 from one term to the next(b) 11, 6, l, -4, ... ( /n.rinii *{q.n.. )Difference of 5 from one rem to me next

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un-u4 t t (usltntl. Arithmetic Prosressi1.1 Delinitior of Arithmetic Proeression

OF r'ttrq\An qrr+I^dic 'r'- (A.P.) is a sequence of numben in which each telm is obtainedAom the precedmg one by the addition ofa constant number called the commotu differeice-Examples of arithmetic progression :

{ u4') ("J "l(") 12,8, 4, 0, 4, -8, -12 ) first term: 12 , common difference = -4(b) 3+,4+,6,7+,... )firstte.-: 11 ,.o--on difference: lf1y'ole .' For an A.P, let ,l, denote the ,th term.

u1= a , uz= a+d . u3= a+2.1 , u+= al3dIn general, th itth term ofan A.P is Lt,=a+(n 1)d, where a: fimttel.rn

d = common difference

ffi: Find the 1Oth term and the nth term for the sequence 7, 10, 13, ... .sa!!1: ,, :4+ ltr-DJUt": I + lr"-')3

' 3!rtllr:t+(h-r)i= q.l ,h*

@: Find the three numbers in an arithmetic progression whose sum is 24 and whose productis48o U r *. L r-J- l,rJ.rgsln'il-J) rl r (t+J) =r'{

3l : rttx.,

fult- J=r-+6.f .roJ: -1. + 1. 6(

(r- J (ll (]+J) : qr.( t -ll(1) {t +J) : qr"r! -J' -- {.

Jr : q, J =rr


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1.2 Sum ofthe first r, terms of an Arithmetic SeriesWhen all lhe terms ofan A.P. are added, we have an arithnetic series.Consider the surl of the fi rct hundred positive integers : 1+2 +3+ 4 +5+ 6 + ... + 100Witing the arithmetic sedes in normal and in reverse order, we have

S :1 +2 +3+4+5 +.. +100 (normal order) ..----- (DS : 100+99+98+97+96+...+ I (reverse order) -- ..--(iD

Adding [(i) + (ii)] gives :25=101 + 101 +101 +..........+10125= 101 x 100Hence, S = 5050The above process is known as finding the sum from first principies.In general,let S, denote the sum ofthe first /, telms ofan alithmetic series whose last term is u,. That is,

0.{(d.rJl +... r( tr, rJl i ltt^-Jl lltt ----- ( I )\\,* \"Wiring S,, in reverse order, we gets, = lr + tt{r-.rl i ( 1,, -rJ, + ... (q+,i) +a ---- (2)

\ \(I)+(2) gives: 25,=(a+u,)+(a+u.,)+(a+rl,)+.. +(a+u,)+(a+u,) =n(a+u,)q + (r-r,J

tlra + tn-rt'l]ince &n =rthtem:a+(, l)d, we have s,=ila+u +(n r)afz'Sr " {ir-!

iience. fo, un A.P. . sum of the first ntermsis S,-'(a+r, l s,=ilz'+1" rlaf

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4/17

ffi: The sum of the first eighteen tems of an arithmetic series is -45 and thl- eighteenth termis also 45- Find the common difference and the sum ofthe first hundred tems.Soln:

s'' -{[:" '1'-r1-'lsrs = -qs = 5tb+rrJl ---(1)ut.- -qt . 4+ nJ --------------- ''\Solving equations (l) and (2) simultaneousln a=40,d=-5

51* : g[ !q4 r(qq ]l{)l: 1. i lie@: Find the sum ofthe positive integers which are less than 500 and arc not multiples of 11.

Soln : sr stRequired sum - (t0,n '8 +r^ ink.,;or k.sr 1l-" sC - (sln * hulHpur $ ll lcs +1"

sr =1+2+3 + 1+.-.+ 4s9 = v\.1-r qa&) : trlh5.str -tt+22+33+44+...+495-1l(t+2+3+4+...+45)= l\(I+ {\(r)) I : lu fs

Required sum: S| -S =124750-11385=113365)t ly'nle: Ifsum ofthe first/? terms equals ft ald sum of the 499.4.j94gq equals [, then

S,=l? and Sn*^=h+k Ut is WRONG to *:rite S. =/cl5n =h k//-.--'.''^*..\\ittt! ir',. Uh li+l An+r ll,HrnL='.

-\"_'t-J'J c4.4h k.Ftn {.tm

Srn+n =h+l


5/17

ffi: The sum of the fiIst l0 terms of an A.P. is 145 and the sum of the next 6 tems is 231.Find (i) the ll'r lcrm,and > .ro(ii) the least nurnber ofterms required for the sum to exceed 2000.-+'--r-U' u' r. 0rr 4r t16

Sra +L\t +ut2 +\3 +ua +\5 +\6 =145 +231-S,. = ]!t:a +t s1t= 145 + 2l I

.+ 8l2a +l5dJ=3'/6:>2a+l5d =47 -----(1)to.tr0 = 145 = ::1[2d + r)d] = 1452-,>2a+9d =29 ---------- (2)

Solving equations ( 1) and (2) simultaneously, a = l,d = 3u3r =.J+30rl = 1+30(3) =91

1Lt5Soln:(D 5r" = IttStrs, = li{6 t )31

(ii) Let n be the least numbem ofterms required for S, > 2000Sr )) r.."3a' -r, -,1* > -(r' - l(. rX ^ +?r. r) t-

Thercforc the least number ofterms required is 37.h) 3 ( .]tr

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6/17

. r:, , ,: - +Do It Yourself( Tutorial,l. Ql ) r,J" q: .B J-iThe sunl ofthe first 20 terms ofan A.P. is 45, and the sum ofthe first 40 tems is 290. Find thelirst term and the common differcnce. Find the number of terms in the A.P. which are less than

Geometric Prosressio2.1 Definition of Geometric Pro{ressionCue"o rhe nc\r number in lhe lolluu ing sequ;nccs:

1(I'' r"- )J) rr-. $.t,',

100. (Answer: 205) i I s .. , ,t .Do ll Yourself( Tutorial 4.O2{a) ) ' ot t'u surrl of rhe i,'"lru rerm(ro[ an A P. is 50. ard ,n, .r. J*. ".t, zo ,",-;1. so r;"4 ,rl"sum ofrhe first I00 terms ofthe progression. (Answei: 750) -5i , ir - jr, .) -- ..;r, - in -r---. -

.1!1) 2, 4, 8,tz2) 5, "2,

l) l,' l,

16, 3\8 165' 25' 125'

1, 1, l,?\{rs

For all the above 3 sequences, each successive lelm can be bbtained by multiplying a constant(common ratio) to the preceding term.A number sequence that has this property is known as a Geometric Progression.

-l

A geometric progression is of the fon'n : o, or, or2 , orf, ...That is, n,=a , uz=ar , u3= at2For any geometric progrcssion, the rrth term i, Si,r"o UV for-1 ]where a denotes thg lirst term and r denotes the common ratio between successive tems oftheprogression.

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7/17

whictr of tlre iollowing are exarnples of geometric progresslons? f : #; : / o^t;t c/(a) 3,6, 12,24,48, ... 6'p -! oth'- )'4i1" )-fu) Jz, z, zJi, q, ... FP E .d. '"- l4{' Ji(c) 1. 1.2.3.5.8. 13. ...

" -r2 tl t4(,1) l. - . -. ,, Rl. Or ! ..i,'q tdfi. -1/-\tl )) tl //


8/17

Fg.l: Tkee consecutive terms of a geometric progression are 3', 3'tl and 81. Find the value ofr. If 81 is the fifth term of the geometric progression, find the seventh term.Soln:

r?-I - -rs -t?-a=l

1=rr,llr = .31 -,3r l' -5Lb =ttar \ .(14l3lu :.fl 4=l=81=d(34)=81=4=l

2,2 Sum ofthe first n terms of a Geometric SeriesWhen all the terms ofa geometric prcgression are added, we have a geometric series.Let ,Sn denote the sum ofthe first n tems ofa geometric paogression.

$-q'= q'-t-, lrn?x.' 3r. 3{ - JLtr?{.t - F: 3 r-rr - 31r'-r

Given 81 is the fifth tem; Lt5 = q74Ttrts u, = a76 =11361=129

S,= A+er +qrL...4r^-t t ar^-iMultiply ( I ) throughout by /, we have :fsn= q( ^+ A.\ +4.3 . 4ah-l + qrr\(l) - (2): Sn-rS,=a-sv' or. (l-r)S, = a(l - i"n)

/)\(2)- (t): rS,-5,=ar'-a

". (r - 1)S, = a(r' -l)Hence, for a geometric series, sum ofthe fint r? te1ms ofthe series is given by

a(l r') o* r, = '(tiir) t, + D1rc4-8


9/17

ffi: Find Lhe.umof rhe lirsrright term. of Lhe geometnc series f *z*{'! '...Soln: 11 / "' 'G.P. *ith o:3./=?3o!

I

?./ t

l- ( /Sc=l-:

to 15rn inclusive. Prove that

-r- (',).) ., +r?a -zg

f . .,, 2.,lEt\.{,

t, z( e,'-? j,,

L2(2(J.Since (x l) is a factor (why?),:> (x -i)ex2 +2xl\ = o...15 =x=trrejecrrorL{ '/3 tttr"j""rto'. l,,A tt

't,q - Stta

,t ^l.1 -(it- ( _)l. ( - ) f !- .)

)(

c4-9tOGi*;,, i :-' - )E)\. c,

[!]il: A geometric series has first term I and the common ratio r, where r * 1, is positive. Thesum of the first five tems is twice the sum of the terms from the 6r'

"=jrf -rr.

?'t;, '' 2 'r,' [ '5!;-"'; I'L t' ' -i1 -'\)ct' ')a/,? r ll 3( l- , t)' a ?r'Let x=15

=2.:r3-3x+1=0

(proven) r + t){2 t )Y-l -. - 2 i

&14:Sum of frrst 5 tems : 2(Sum of first l5 tems Sum ollirst 5 terms)

2r'5 - grt +l = a>(r')t- 36rt)+l=

ulte:Factlsation by'inspection'Example \r'3 -:'2 I I '+b.

:r- L-r i,. -/r ),,i r:- , *.,--Q:?:4_2r, L _ r

z(.., t;


10/17

2.J lplinite Gcometric Seriestz4\ct q.11I t-consider rhe geomelric series: l-t ; ; r (-tnFd n1.,",,t )

Therefore sum ol the tirstlr tems is S,, a(l r")l-r l- aa , ,,r'rhu\ .. rl --) l1* | ]-'', rs1,,r,s* = 2lI l-'*I

LI

. _.2047 _.'' 2048Note that as we add more and more terms, the sum ofthe above series gets closer and closer to 2.Since we can make S, as close to 2 by choosing n large enough, we say that

When r = 0,1,12 = 0.01rr = 0.001ra = 0.0001" = o.ooool

Asn >a,r" -+0

When r= -0.1:r'? = 0.01rr = 0.001ra = 0.000115 = o-ooool

As n - ca,r' --+ 0

When / = 2,

Asn+a,ro +a

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11/17

lngcncril.wlrrr L Ir.c. l, I ", h'".Iior the general gcornetric series: a + ar'+ ar2 +-.., ,tll t')"'- 1-' ;;],\'hen .1


12/17

EF. D: A geomet ic series has firsl term a and common ratio r. ,S is the sum to infinity of the series, risthe sum to infinity of the even-numbered terms (i.e. u2 + u4 + u6 + ... ) ofthe series. Given that S is fourtimes the value of Z, find the value ofr.

nr.e


13/17

@l: Thc first two tems of a geometric progression are 3 and 2- Find the least value of n for whichthe difference between sum ofthe firs1n tems and sum to rnfinity is within 2% ofthe sum lo infurity.Soln:

o:1, ,: =?3

1",, ,-l; /;o !^a(t r'\t - o l-, - - tt It;, L '-")-'l

r-3!^\, r./

( - r,)

i"(a,( :,)

?./ oDh6i . -v"nlrI cL q,v.

/(:t'l .' f"s,^c (?'1" 't''rri .. \n .a .-2!3 bb( tl'n ''a-"'., ..,, :l 'i',t-,.fl>-n t' '"-.)tt^ - l

I - -rl/ - r ' ' | . ?-,!- tlIt u,"'il ,',-I r '"'r '.1"1 ',.1to , )r 'l;-, I I t uo

'*- |t,! :l

l" )...,,.;o 1 --1,,.

l o.4

r 2 LGalln ( -r, - 3.5

Smallest distance the first blow must drive the nail in = 1.2cm

. r- J 6/ar F.&,/ , '{,

chaptefl: ar;thmetic and geometric series

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