Form 3 Mathematics – SEQUENCE AND SERIES msomimaktaba, November 13, 2018August 17, 2024 SEQUENCE AND SERIES SEQUENCEIs a set of numbers written in a definite order such that there is a rule by which the terms are obtained.OrIs a set of number with a simple pattern.Example 1. A set of even numbers• 2, 4, 6, 8, 10 ……2. A set of odd numbers• 1, 3, 5, 7, 9, 11….Knowing the pattern the next number from the previous can be obtained.Example 1. Find the next term from the sequence• 2, 7, 12, 17, 22, 27, 32The next term is 37.2. Given the sequence• 2, 4, 6, 8, 10, 12………What is i)The first term =2ii) The 3rd term =6 iii)The 5th term =10iv)The nth term [ the general formula]2=2×14=2×26=2×38=2×410=2×512=2×6Nth =2xnTherefore nth term =2nFind the 100th term, general formula =2n100th term means n=100100th term =2×100100th term =2003. Find the nth term,Solution,4. Given the general term 3[2n]a) Find the first 5termsb) Find the sum of the first 3 termsSolution 3[2n] when;n=1, 3 [21] =6n=2, 3 [22] =12n=3, 3 [23] =24n=4, 3 [24] =48n=5, 3 [25] =96First 5 terms = 6, 12, 24, 45, 96Sum of the three termsSum of the first three = 6+12+24Sum of the first three = 42 Exercise 1. 1. Find the nth term of the sequence 1, 3, 5, 7….. 2. Find the nth term of the sequence 3, 6, 9, 12…… 3. The nth term of a sequence is given by 2n+1 write down the 10th term. 4. The nth term of a certain sequence is 2n-1 find the sum of the first three terms. 5. If the general term of a certain sequence is Find the first four terms increasing or decreasing in magnitudesSolution1. 1, 3, 5, 7…nthFrom the sequence the difference between the consecutive terms is 2 thusnth =2+n 2. 3, 6, 9, 12……..nth The difference between the consecutive terms is 3 thusnth =3n3. 10th = 2[10+1]10th =20+110th =214. Whenn=1, 21-1n=2, 22-1n=3, 23-1Sum of the first three terms =1, 2, 4Sum = 1+2+4Sum = 7 5. When;n = 1, = 1n = 2, =n = 3, = 1n = 4, = The first four terms are SERIESDefination: When the terms of a sequence are considered as the sum, the expression obtained is called a serier or a propagation Example (a). 1+2+3+4+5+………. (b). 2+4+6+8+10+………+100. (c). -3-6-9-…….The above expression represent a series. There are two types of series 1.Finite series Finite series is the series which ends after a finite number of terms e.g. 2+4+6+8+………….+100. -3-6-9-12-………….-27. 2. infinite Series Is a series which does not have an end. e.g. 1+2+3+4+5+6+7+8………….. 1-1+1-1+1-1+1-1+1…………..Exercise 5.2.1 1. Find the series of a certain sequence having 2(-1)n as the general term 2. Find the sum of the first ten terms of the series -4-1+2+……. 3. The first term of a certain series is k. The second term is 2k and the third is 3k. Find (a). The nth term (b). The sum of the first five termsExercise 5.2.1 Solution1. n=1, 2(-1)1 = -2 n=2, 2(-1)2 = 2 n=3, 2(-1)3 = -2 n=4, 2(-1)4 = 2 n=5, 2(-1)5 = -2 The series is -2+2-2+2-2+2-2+2-2+2+………………..+2(-1)n2. The sum of the first n terms of the series -4-1+2+5+8+11+14+17+20+23 = 95 3. (a) k + 2k + 3k + 4k +………. + nk The nth term of the series is nk (b) k + 2k + 3k + 4k + 5k = 15k ... The sum of the first 5 terms = 15kARITHMETIC PROGRESSION [A.P]An arithmetic progression is a series in which each term differ from the preceding by a constant quantity known as the common difference which is denoted by “d“.For instance 3, 6, 9, 12….. is an arithmetic progression with common difference 3.The nth term of an arithmetic progression If n is the number of terms of an arithmetic progression, then the nth term is denoted by An Therefore An+1=An + d e.g. First term = A1 Second term = A1+ d = A2 Third term = A2 + d = A3Consider a series 3+6+9+12+….. A1 = 3, d= 3 A2 = A1 + d A3 = A2 + d =A1+d+d = A1+ 2d A4 = A3 + d = A1+ 2d+ d = A1 + 3d A5 = A4 + d = A1+ 3d+ d = A1 + 4d A6 = A5 + d = A1+ 4d+ d = A1 + 5dAn = A[n-1] + d = A1+ (n-1)d... The general formula for obtaining the nth term of the series is An = A1+ (n-1)dThe general formula for obtaining the nth term in the sequence is also given by An =A1+[n-1]dQuestion1. A7 = A2 and A4 =16. Find A1 and d.SolutionA7= A1+6d =A1+6d= ( A1+d)2A1+12d=5A1+5d3A1=7d A1=dA 4 =16 A4=A1+3d = 16 d +3d= 16d = 16d= 3But A1 +3d= 16A1 +9=16 A1= 7 Exercise 2 1. The pth term of an A.P is x and the qth term of this is y, find the rth term of the same A.P 2. The fifth term of an A.P is 17 and the third term is 11. Find the 13th term of this A.P. 3. The second term of an A.P is 2 and the 16th term is -4 find the first term. 4. The sixth term of an A.P is 14 and the 9th of the same A.P is 20 find 10th term. 5. The second term of an A.P is 3 times the 6th term. If the common difference is -4 find the 1st term and the nth term 6. The third term of an A.P is 0 and the common difference is -2 find; (a) The first term (b) The general term7. Find the 54th term of an A.P 100, 97, 94 8. If 4, x, y and 20 are in A.P find x and y 9. Find the 40th term of an A.P 4, 7, 10…….. 10.What is the nth term of an A.P 4, 9, 1411.The 5th term of an A.P is 40 and the seventh term of the same A.P. is 20 find the a) The common difference b) The nth term 12. The 2nd term of an A.P is 7 and the 7th term is 10 find the first term and the common difference Exercise 2 Solution 1.d = y-x rth = A1+[n-1]d= A1+nd-d = A1+n[y-x]-[y-x] = A1+[ny-nx-[y-x] =A1+[ny-y]-[nx-x] A1+g[n-1]-x[n-1] rth=A1+[y-x][n-1]2.A5=17A3=11 A13=?A5 =A1+4d = 17 A3=A1+2d =17 A1+2d=11Solve the simultaneous equation by using equilibrium method.A1+4d=17 A1+2d=11 = d = 3A1+4[3]=17 A1+12=17-12 A1=5A13=A1+12d A13=5+12[3] A13=5+36 A13 =41 3.A2=2 A16=-4 A1=?A2=A1+d A1+d=2A16=A1+15d A1 +15d= -4Solve the two simultaneously equations by using the elimination method A1 + d=2A1 +15d=-4 d= From the 1st equationA1 += 2A1 = 2+ A1 = 4.A6= 14A9=20A6= A1+5d = 14A9 = A1+8d=20Solve the simultaneous equation by using the elimination methodA1+5d= 14A1+8d=20 ( difference between two equations)Then d = 2From 1st equationA1 +5[2] = 14 A1 +10= 14-10 A1=4A10= A1+9d A10=4+9[2] A 10=4+18 A10=22An =A1+[n-1]d An =4[n-1]2 An =4+2n-2 An= 2n+4-2 An=2n+25. A2= 3 x Ab D= -4 A1=? An=?Az = 3 x A6 A 1 +d = 3 x A1+5d A1+d =3A1+15d 2A1+14d=0 d= -42A1+14[-4]=0 2A1+-56= 0 2A1= 56 A1= 28An = A1+[n-1]d An= 28+[n-1]-4 An=32-4n An =32-4n6.(a) A3= 0 d= -2From the formulaA1+2d= 0 A1+2[-2]= 0 A1-4=0 A1= 4The first term is 4.b) The general termAn =A1+ [n-1] d A n =4+[n-1]-2 An = 4+ -2n+2 An = 6-2nThe general term is 6-2n.7.A54 =? 100, 97, 94 = A1, A2, A3 A54= A1+53d d= A2-A1= A3-A2 d= 97-100= 94-97 d= -3A54= 100+53[-3] A 54=100 + -159 A54= -59 8.4, x, y, 20 A 4= 20 A1 +3d=20 , but A1=44+3d= 20 3d=16 d = 16/3A 1, A2, A3, A4A2 =A1+d4+X = A3 = A1+2d4+ 2x A3 = Hence x= and y= 9.A40 =? A1 =4 A2= 7 A 3= 10 A1+39d=?d= 7-4= 10-7 d= 3A1 +39[3] A1+117 4+117 A40 =12110.A1 =4 A2=9 A3=14 d= 9-4= 14-9 d= 5An =A1+[n-1]d An =4+[n-1]5 An =4+5n-5 An =5n-1The nth term is 5n-1.11.a) the common difference A5= 40 A7= 20A1+ 4d= 40 ………… (1) A1+ 6d= 20 ………….(2)Subtracting equation (2) from equation (1) we obtain-2d=20 d= -10b) the tenth termA10= A1+9d, But A1 +4d=40 A1=80∴A10=A1+9d =80-90 A 10=-10 12.A2= A1+d A7= A1+6d A1 +6d = 10 A1+d=7Solving the simultaneous equations by using the elimination method;-5d= -3 d = 3/5 A1 += 7 A1 =SUM OF THE FIRST n TERMS OF AN ARITHMETIC PROGRESSION Consider a series with first term A1, common difference d. If the sum n terms is denoted by Sn, thenSn = A1 + (A1+ d) + (A1 + 2d)+ …+ (A1 – d) + An + Sn = An + (An- d) + (An – 2d)+ …+ (A1 + d) + A12Sn = (A1+ An) +(A1+ An) + (A1+ An)+ …+ (A1+ An)+ (A1+ An)There are n terms of (A1+ An) then 2Sn = n(A1+ An) Sn = n(A1+ An) 2 ... The sum of the first n terms of an A.P with first termA1 and the last termAn is given by Sn = (A1+ An)But An =A1+ (n-1) d Thus, from Sn = (A1+ An) Sn = [A1+ A1+ (n-1) d] Sn = [2A1+ (n-1) d]... therefore, the sum of the first n term of an A.P with the first A1 and the common difference d in given by Sn = [2A1+ (n-1) d]Wheren =number of terms A1= first term An =last term d= common differenceExamplei) Find sum of the first 5th term where series is 2, 5, 8, 11, 14 first formula SolutionS5 =[2 + 14] S5 = 40(ii)Sn =[2A1 + [n-1] d] S5 =[2 x 2 + [5-1] (3)]S5= 40 Arithmetic MeanIf a, m and b are three consecutive terms of an arithmetic. The common difference d= M – a Therefore M- a = b – M 2M = a+ b M= a + b 2 M is called the arithmetic mean of and b E.g. Find the arithmetic mean of 3 and 27 M = 3+ 27 2 = 30 = 15 2GEOMETRIC PROGRESSION (G.P) Definition: Geometric progression is a series in which each term after the first is obtained by multiplying the preceding term by the fixed number. The fixed number is called the common ratio denoted by r. E.g. 1+2+4+8+16+32+… 3+6+12+24+48The nth term of Geometric Progression If n is the number of terms of G.P, the nth term is denoted by Gn and common ratio by r. Then G n+1 = rGn for all natural numbers. G1 = G1 G2 = G1r G3 =G2r = G1r.r = G1r2 G4 =G3r = G1r2.r = G1r3 G5 =G4r = G1r3.r = G1r4 G6 =G5r = G1r4.r = G1r5 G7 =G6r = G1r5.r = G1r6 G8 =G7r = G1r6.r = G1r7 The nth term is given by Gn = G1rn-1Example1: Write down the eighth term of each of the following. (a). 2+ 4+ 8+… (b). 12+ 6+ 3+ …Solution (a) The first term G1 =2, the common ratio r=2 and n=8, Then from Gn = G1rn-1 G8 =(2)(2)8-1 G8 = (2). 27 G8 = 256 (b) The first term G1 =12, the common ratio r=1/2 and n=8, Then from Gn = G1rn-1 G8 =(12)(1/2)8-1 G8 = (12). (1/2)7 G8 = 12/128 = 3/32Example2: Find the numbers of terms in the following 1+2+4+8+16+…+512.Solution The first term G1 =1, the common ratio r=2 and Gn= 512, Then from Gn = G1rn-1 512 =(1)(2)n-1 512 = 2n-1 512 = 2n.2-1 512 x 2= 2n 1024= 2n 210 = 2n n = 10The sum of the first n terms of a geometrical progression. Let the sum of first n terms of a G.P be denote by Sn Sn = G1+ G2+ G3+ G4+…+ Gn Since the common ratio is r From G2 = G1r G3 = G1r2 G4 = G1r3 ...Gn = G1rn-1 Sn = G1+ G1r+ G1r2+ G1r3+…+ Gnrn-1 Multiplying by common ratio r both sides we haverSn = rG1+ + G1r3+ G1r4+…+ GnrnSubstract Sn from rSnrSn – Sn = – G1 Sn(r- 1) = (rn– 1) Sn = where r ≥ 1 for r≠1G1= first term of G.P r= common ratio Sn = sum of the first nth term n= number of termsfor r< 1 the formular is given bysn – rsn = G1 – G1rnSn(1-r) = G1(1- rn)Sn= for r<1When r= 1, the sum is simply given by Sn= G1+G1+G1+G1+G1+G1+…+G1sn=nG1 for r=1Example1. Sum of the first 5th term of G.P where series is 2+4+8+16+32S5 = S5 =S5 = 62Sn = 2. The sum of the first n terms of a certain series is given by Sn= 3n-1 show that this series is a G.P Solution Whenn= 1; S1 = 31-1 = 3 – 1 = 2n= 2; S2 = 32-1 = 9 – 1 = 8n= 3; S3 = 33-1 = 27 – 1 = 26n= 4; S4 = 34-1 = 81 – 1 = 80 2 + 6 + 18 + 54… r= r= 3Exercise1. An arithmetic progression has 41 terms. The sum of the first five terms of this A.P is 35 and the sum of the last five terms of the same A.P is 395 find the common difference and the first term. SolutionS5 = 35 A5 =395 d= ? A1 =?S5 =[2A1 + [5-1] d] S5 = [2A1+4d]S5= 5A1+10d35= 5A1+10d … (1) 395=5 A1+190d… (2)Solve the simultaneous equationsThen the value of d = 2From the 1st equation5A1+10d=35 5A1+10[2]=35 5A1+ 20 =35 A1 = 3Therefore the first term is = 32. An arithmetic progression has the first term of 4 and n th term of 256 given that the sum of the nth term is 1280. Find the value of the nth term and common differenceSolution A1 =4 An =256 Sn =1820 n=? d=? Sn =[A1 + An]1820 = [4 + 256]1820 = n [130]n= n= 14Therefore the n term = 14An = A1+ [n-1]d An = 4+[14-1]d 256= 4+13d 256-4=13d 252 = 13d d= ∴ Common difference = 3. The 4th, 5th and6th terms of an A.P are (2x +10), (40x-4) and (8x+40) respectively. Find the first term and the sum of the first 10A4= 2x+10 A5= 40x-4 A6=8x+40SolutionA4= A1+3d = (2x + 10)…i A5 = A1+4d = (40x – 4)…iiSolve the equations by using elimination methodA1+3d= 2x+10 A1+4d= 40x-4– d = 2x + 10 – 40x + 4 d= 38x – 14From 1st equationA1+3[38x-14] =2x+10 A1+114x – 42=2x +10 A1 = 2x+10 – 114x + 42 A1= -112x +52Therefore the first terms = -112x+52S10= [ 2(-112x+52) +[10-1] 38x-14]S10=5[-224x+104] +9 [38x-14]S10=5[-224x+104+342x-126]S10=5[118x-22]S10=590x-110 4. The sum of the first n terms of an A.P is given by sn=n[n+3] for all integral values of n. write the first four terms of the seriesSolutionWhen; n = 1 then sn=n[n+3] =1[1+3]= 4n = 2 then =2[2+3]= 10n = 3 then =3[3+3]= 18n = 4 then =4[4+3]= 28The first four terms of the series is 4+6+8+105. The sum of the first and fourth terms of an A.P is 18 and the fifth terms is 3 more than the third term. Find the sum of the first 10 terms of this A.PSolution A1+A4=18 A5 = ( A1 + 4d )=3 +A3= 3 + A1 + 2d S10 =? From,A1+ A4= 18 A1+ A1+ 3d =182A1+ 3d = 18………..(i) and A1 + 4d =3 + A1 + 2d……..(ii) Take equation (ii)A1 + 4d = 3 + A1 + 2d2d= 3 d = 3/2Substitute the value of d into equation …….. (i)2A1+ 3d = 182A1+ 3(3/2) = 182A1 + 9/2 = 182A1= 18 – 9/2 = 27/22A1= (27/2) Divide by 2 both sidesA1= 27/4 So, The sum of the first ten terms Sn = n/2 (A1 + An) Sn = n/2 (A1+ A1 + (n – 1)d) Sn = n/2 (2A1+ (n – 1)d) S10 = 10/2 ((27/4)2+ (10 – 1) x 3/2) S10 = 135Therefore the value of ten terms will be,S10 = 135 6. How many terms of the G.P 2+4+8+16…… must be taken to give the sum greater than 10,430?SolutionG1+G2+G3+G4…………. 2+4+8+16Sn =Sn =10430 =5215 =2n-15216 =2n thenn= Then more than term should be taken to provide the sum greater than 10430 7. In a certain geometrical progression, the third term is 18 and the six term is 486, find the first term and the sum of the first 10 terms of this G.PSolutionG3= G1 r2 = 18 ………… (1) G6= G1 r5 = 486……….. (2)Take equation (2) divide by equation (1) = = 27r= 3But = 18G1 = 2Sn = thenS10 == 2046The first term is 2The sum of the first 10 terms is 2046 8. Given that p-2, p-1 and 3p-5 are three consecutive terms of geometric progression find the possible value of pSolution R== r=[p-1][p-1]=[p-2][3p-5]r=p2-2p+1=3p2-5p-6p+10p2-2p+1=3p2-11p+10p2-2p+1=3p2-11p+100=3p 2-p2-11p+2p+10-12p2-9p+9=0From the general quadratic formulap= p=P=3 or p= Geometric mean If a, m and b are consecutive terms of a geometric progression then the common ratio r= M/a = b/M M2 = ab M = Example: Find the geometric mean of 4 and 16. from G. M =G.M= G.M = G.M = 12. 5.7. APPLICATION OF A.P AND G.P.Simple interest is an application of arithmetic progression which is given by;I = Where I=simple interestP=principalR= rate of interestT= period of interestCompound interest is an application of geometric progression it is given by;An=p+1 or An =p)n Where An =an amount at the end of the New Year R= rate of interest n= number of years T=period of interest p= principalExample1. Find the simple interest on Tshs 10,000/= deposited in a bank at the rate of 10% annually for 4 yearsSolutionI = I = I=1000×4I=4000The interest for 4 years will be 4000/= ALL NOTES FOR ALL SUBJECTS QUICK LINKS:AGRICULTURE O LEVEL PURE MATHEMATICS A LEVELBAM NOTES A LEVELBASIC MATH O LEVELBIOLOGY O/A LEVELBOOK KEEPING O LEVELCHEMISTRY O/A LEVELCIVICS O LEVELCOMPUTER(ICT) O/A LEVELECONOMICS A LEVELENGLISH O/A LEVELCOMMERCE O/A LEVELACCOUNTING A LEVELGENERAL STUDIES NOTESGEOGRAPGY O/A LEVELHISTORY O/A LEVELKISWAHILI O/A LEVELPHYSICS O/A LEVELMOCK EXAMINATION PAPERSNECTA PAST PAPERS Basic Mathematics Study Notes Form 3 Basic Mathematics Study Notes FORM 3MATHEMATICSPost navigationPrevious postNext postRelated Posts Basic Mathematics Study Notes Form 3 Mathematics – FUNCTION November 13, 2018August 17, 2024FUNCTION A function is a set of ordered pairs which relates two sets such that to each element of one set there is only one element of the second set Example Represent the following set of ordered pairs in a pictorial diagram. (1,4), (2,3), (-1,2), (-2,-3). Solution A function whose… Read More Basic Mathematics Study Notes MATHEMATICS FORM 1 – NUMBERS November 11, 2018August 17, 2024NUMBERS What is numbers? Numbers are symbols or words which represent quantity of something for example in form 1B there are forty four students. i.e. 44 students The numbers are represented by symbols called numerals. Each symbol in a numeral is called a digit. E.g. In 256 there are three… Read More Basic Mathematics Study Notes MATHEMATICS FORM 1 – UNITS November 11, 2018August 17, 2024UNITS UNITS OF LENGTH Basic unit of length is meter. Other unit are such as :- Unit Symbol Millimeter mm Centimeter cm Decimeter dm Meter m Decameter dam Hectametre hm Kilometer km Km hm dam m dm cm mm 1 0 0 0 0 0 0 1… Read More Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment *Name * Email * Website Save my name, email, and website in this browser for the next time I comment. Δ
Basic Mathematics Study Notes Form 3 Mathematics – FUNCTION November 13, 2018August 17, 2024FUNCTION A function is a set of ordered pairs which relates two sets such that to each element of one set there is only one element of the second set Example Represent the following set of ordered pairs in a pictorial diagram. (1,4), (2,3), (-1,2), (-2,-3). Solution A function whose… Read More
Basic Mathematics Study Notes MATHEMATICS FORM 1 – NUMBERS November 11, 2018August 17, 2024NUMBERS What is numbers? Numbers are symbols or words which represent quantity of something for example in form 1B there are forty four students. i.e. 44 students The numbers are represented by symbols called numerals. Each symbol in a numeral is called a digit. E.g. In 256 there are three… Read More
Basic Mathematics Study Notes MATHEMATICS FORM 1 – UNITS November 11, 2018August 17, 2024UNITS UNITS OF LENGTH Basic unit of length is meter. Other unit are such as :- Unit Symbol Millimeter mm Centimeter cm Decimeter dm Meter m Decameter dam Hectametre hm Kilometer km Km hm dam m dm cm mm 1 0 0 0 0 0 0 1… Read More