Part  I. Mathematics


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1 Part  I Mathematics
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3 CHAPTER Set Theory. Objectives. Itroductio. Set Cocept.. Sets ad Elemets. Subset.. Proper ad Improper Subsets.. Equality of Sets.. Trasitivity of Set Iclusio.4 Uiversal Set.5 Complemet of a Set.6 Uio of Sets.6. Properties of Uio Operatio.7 Itersectio of Sets.7. Disjoit Sets.7. Properties of Itersectio Operatio.7. Relative Complemet of a Set.7.4 Illustrative Example.8 De Morga's Laws.9 Distributive Laws of Uio ad Itersectio. Illustrative Examples. Summary. Check your Progress  Aswers. Questios for Self  Study.4 Suggested Readigs. OBJECTIVES After studyig the cocept of a set ad its fudametal operatios you ca explai the followig : Studets ca verify : Cocept of a set Complemet of a set Uio ad itersectio of sets. De Morga's Laws. Distributive laws of uio ad itersectio. Studets ca solve problems ivolvig umerical data.. INTRODUCTION The theory of sets forms the basis of the moder mathematics. The begiig of the theory ca be traced from the Germa mathematicia Cator at the ed of the 8th cetury. At preset it covers a very extesive part of mathematics. However, we restrict ourselves to a very elemetary part of it.. SET  CONCEPT I our everyday life we ofte make use of collective phrases. Like, (i) (ii) Studets wearig spectacles, Wagos attached to a trai, Set Theory /
4 (iii) (iv) (v) Flowerig trees i a garde, Studets i a class, A pack of cards. Such collectios are called 'sets' i mathematics. To describe a set we eed the followig cosideratios. (a) (b) set or collectio elemets i a set (c) rule of property which eables us to say whether a give object is i a set or ot. e.g. I a set "studets wearig spectacles" we ote that a elemet is "every studet who wears spectacles" ad the rule is "oly those studets who wear spectacles.".. Sets ad Elemets Cosider a set cosistig of umbers,,, 4, 5. Each of these umbers is a elemet of the set. We deote this set by a letter A. A = {,,, 4, 5} The fact that a umber is i a set A is writte as " belogs to the set A". This cocept of belogig to the set is symbolically deoted as A ad read as " belogs to A". While the fact that the elemet 6 is ot i the set A; i.e. 6 does ot belog to the set A is symbolically writte as 6 A ad read as "6 does ot belog to A". There are various ways of represetig a set. Oe such method is idicated above to describe a set. Aother method is to state all the elemets of a set. There is oe more method, which we shall study ow. The set is described by usig the property of the elemets. e.g. The fact that the elemet x of the set A havig a property P is described as P(x) ad the set is writte as A = {x P(x)} The above set A = {,,, 4, 5} ca ow be writte as i this otatio as A = {x x is a positive iteger betwee ad 6} A elemet of the set is couted oce oly, i.e. {,,, } is the same as {,, }. Also set is regarded as the same eve if its elemets are writte i differet order. e.g. {p, q, r, s} = {r, p, s, q} = {s, r, p, q}. Defiitio : Two sets are said to be equal if they cotai the same elemets. e.g. A = {,4}, B = { x x 6x + 8 = } The elemets of the set B are the roots of the equatio x 6x + 8 =. We kow from Elemetary Algebra that the roots of this equatio are ad 4. Thus by the defiitio two sets A ad B are equal; ad we write A = B. Defiitio : A set havig oly oe elemet is called a sigleto set. e.g. {prime miister of Idia}, {a}, {}, {b} are. all sigleto sets of : prime miister of Idia, a,, b. Next, cosider a set { x x =, x = }. Mathematics & Statistics /
5 We kow that there is o umber which is equal to ad at the same time. Thus this set has o elemet. Such a set havig o elemet is called a empty set. Defiitio : A set havig o elemet is called a empty (or ull or void) set ad is deoted by. Thus = {x x x} All empty sets are equal.. Check your progress.. Use appropriate symbols i the blak. (a).. {,,,4} (b) 5.. {,6,7,8}. Are the followig sets equal? A = {x x is a positive iteger, 5} B = {,,, 4, 5} C = {x x is a root of the equatio x x + = }. State whether true or false. A = {x x is a positive iteger, 4}, B = ad A = B. C = (x x is a root of the equatio 4x + = }, ad C is a sigleto set. Cosider two sets.. SUBSET A = {x x is a studet i st year class} B = {y y is a studet wearig spectacles i st year class} All studets who wear spectacles are i a st year class. I other words every y is x. Obviously every studet i the class may ot wear spectacles i.e. every x is ot y. i.e. Every elemet of the set B is a elemet of A, but every elemet of A is ot a elemet of B. I this case, we say that B is a subset of A We itroduce the followig otatio. If P ad Q are two statemets such that, if P is true the Q must be true; we say that P implies Q. We express this i a symbolic form as P Q. (read as P implies Q) e.g. (x = ) ( x 6x + 9 = ) If i additio to P Q, we have Q P, we have both sided implicatios, ad we write it as P Q. read as "P implies ad is implied by Q." e.g. x = x =8, whe x is real. For the two sets C, D defied as C = {x x is a triagle} D = {x x is a equilateral triagle} We see that x D x C. Defiitio : B is called a subset of A if x B x A. The statemet that B is a subset of A is symbolically writte as BA. If x A x B, the A is a subset of B. i.e. A B Set Theory /
6 ad is read as "A is cotaied i B" or "B cotais A" or "A is a subset of B". To illustrate the relatioship betwee gets, we use differet diagrams called Ve diagrams... Proper ad Improper Subsets We shall ow cosider two extreme cases of subsets. I the illustratio of sets A = {x x is a studet i st year class} B = {y y is a studet wearig spectacle i st year class} (i) It may happe that o studet i st year class wears spectacles. i.e. B = Hece ca be regarded as a subset of A. I fact A for every set A. (ii) Aother possibility is, every studet i st year class may wear spectacles. i.e. B = A. I this case B A A A. I fact every set A ca be regarded as a subset of itself. Both ad A are called improper subsets of A. Except ad A all other subsets of A are called proper subsets of A. e.g. Let A = {a, b, c} The possible subsets of A are, A = {a}, A = {b}, A = {c}, A 4 = {a, b}, A 5 = {b. c}, A 6 = {c, a), A = a, b, c} Thus A, A, A, A 4, A 5, A 6 are proper subsets of A. The set of all subsets of A is called a power set of A ad is deoted by P(A). The total umber of elemets of P(A) is give by a formula : (P(A)) = (A) where (A) deote the umber of elemets i the set A. Here (A) = ad ( P (A) ) = = 8. We easily cout the umber subsets of A as 8. Next cosider, S = {,,,5,7}, T = {,5,7} Here T is a proper subset of S. T S. Sometimes S is called superset of T... Equality of Sets The equality of two sets ca be defied as : Mathematics & Statistics / 4
7 Defiitio : Two sets A ad B are said to be equal if each of A ad B is a subset of other. A = B A B ad B A. This ca be proved as follows : Let A B ad B A. x A x B ad x B x A. As all these relatios hold at the same time, it is obvious that A ad B have the same elemets. i.e. A = B. Coversely, let us suppose that A = B. Hece x A x B i.e. A B ad x B x A i.e. B A. This provides us with a very importat tool to prove the equality of two sets (especially whe they cotai ifiite umber of elemets).. Trasitivity of set iclusio Now cosider, A = {p, q, r, s} B = {o, p, q, r, s, t} C = {o, p, q, r, s, t, u, v} We clearly see the followig relatios. i.e. A B, B C. Also A C. (AB, BC) AC. This property kow as set iclusio is said to be trasitive. The geeral proof of the above property is give below. A B B C meas x A x B meas x B x C Hece A B ad B C meas i.e. x A x B x C x A x C i.e. A C. We have the followig ve diagram to illustrate the above two properties.. Check your progress.. State whether true or false. (a) {,,} (b) {, 4, 6} Set Theory / 5
8 (c) {, } {x x is a positive iteger, x } (d) ad A are improper subsets of A. (e) The proper subsets of A = {, 5, 6} are A = {}, A = {5}, A = {6}, A 4 = {, 5}, A 5 = {, 6}, A 6 = {5, 6}.. Write dow all possible subsets of the set A = {,, }.. (a) Is A = {a, b, c} a super set of B = {a, b, c, d}? (b) Write dow all possible subsets of B = {p, q, r, s} each cotaiig two  elemets oly. Cosider the followig sets..4 UNIVERSAL SET A = {x x is a studet i st year class} B = {x x is a studet i a college i Idia} C = {x x is a studet i a college i the world} U = {x x is a studet} We see that A B,B C, C U. I fact A, B, C are all subsets of a fixed set U. We may fid umber of sets which are subsets of a fixed set like U. This fixed set is called the uiversal set. Let P, Q, R, S, T,... be the set of books writte i Marathi, Gujarathi, Begali, Hidi Frech, Germa, laguages. All these are subsets of "a set of books' say X. I this case X is the uiversal set. May other illustratios of uiversal sets ca be costructed..5 COMPLEMENT OF A SET Cosider a uiversal set X = {x x is a tree} ad a set A = {x x is a tree havig height more tha meters}. We ca form aother set B = {x x is a tree havig height less tha meters}. We observe that sets A ad B together form a uiversal set X. I this case, B is called complemet of the set A. The elemets of B are the elemets of X but are ot elemets of A. The complemet of a set A is deoted by A. Defiitio: The complemet of a set A has elemets i X which are ot i A. i.e. A = {x x X, x A} e.g. Let X = {,,,, 4, 5} ad A = {,, 4} The A = {,, 5} The complemet of A is {,, 4}, which is A agai. (A) = A This ca be proved as follows : x (A) xa xa i.e. (A')' = A We shall represet X, A, A' by the followig ve diagram. Mathematics & Statistics / 6
9 We ote the followig poits regardig complemet of a set. (i) ' = X (ii) X' = (iii) If A B the B' A' We shall prove it as: Let x B' x B, But A B x A x A' Thus x B' x A' B' A'. The represetatio by a ve diagram is as :.4 &.5 Check your progress: Solve the Followig Give a uiversal set X = {,,, 4, 5, 6, 7, 8, 9} Obtai the complemet of the followig sets usig uiversal X (above) (a) A = {,, 5, 7, 9} (b) B = {, 4, 6, 8} (c) C = {,, 9} (d) D = {,,,4, 5,6,7,8,9} (e).6 UNION OF SETS We shall ow begi with operatios o sets or what is usually called as Algebra of sets. Let A = {a, b, c, d, e}, B = {a, c, f, g, h} ad C = {a, b, c, d, e, f, g, h} We fid here that every elemet of A is a elemet of C ad every elemet of B is a elemet of C. I other words, if a elemet belogs to A, it belogs to C ad similarly if a elemet belogs to B, it belogs to C. i.e. A elemet of C is either a elemet of A or a elemet of B or it is a commo elemet of both A ad B. Such a set Set Theory / 7
10 C is called uio of two sets A ad B. This set is deoted by A B, ad is read as "A uio B". Defiitio: The uio of two sets A ad B is A B = {x xa or xb or x both A ad B } or A B = {x x A ad or x B} Thus A B is the set of all elemets of A ad B, the commo elemets, if ay, beig take oce oly. The followig Vediagram will illustrate the statemet..6. Properties of Uio Operatio We shall state some properties of the uio operatio (i) AB = BA (commutativity) Proof.: xab x A ad or xb x B ad or x A x BA i.e. we get the same set, i whatever order the sets are take for uio. I this case we say that uio is commutative. (ii) (iii) (iv) (v) (vi) A = A A X = X A A' = X Proof.: A A = A Proof.: AAB ad BAB x A A' xa ad or x A' x A ad x A' x X x A A xa ad or x A xa Both these follow from the Ve diagram of A B. (Fig..8) (vii) If A B, the A B = B Proof.: Sice A B, x A x B X A B x A ad or x B x B (i either case x B) Mathematics & Statistics / 8
11 Similarly, we ca show that BAB A B B A B = B. (viii) A (BC) = (AB) C (associativity) Proof. : x A ( B C ) x A ad or x (BC) x A ad or x B ad or x C x (A B) ad or x C x (A B) C I this case we say that uio operatio is associative. i.e. if we remove the brackets o either side we get the same result, (ix) viz. A B C. A B C = {x x A ad or x B ad or x C} = {x x at least oe of A, B, C} If AC ad B C the (AB) C C Proof.: A C meas x A x C B C meas x B x C x (AB) xa ad or x B x C (A B) C. Here we say that uio operatio is trasitive. (x) Priciple of iclusio ad exclusio. Let A ad B be fiite subsets of a uiversal set. If A ad B are disjoit sets ( i.e. A B = ) Set Theory / 9
12 ( A B) = ( A ) + ( B ) I particular (A) + ( A') = (X).6 Check your progress: Give a uiversal set X = {5,, 5,, 5,, 5} A = {,, }, B = {5,, 5, 5}, C = {}. Write dow the followig sets (a) A B (b) B C (c) A C (d) A' B' (e) A' C'. State whether true or false. (a) C B (b) A C = A (c) B C = B (d) A B A C. Verify the associative law A (B C) = (A B) C.7 INTERSECTION OF SETS This is aother operatio with sets. Let A = {,,,4,6} B = {4, 6, 8, } C = {4, 6} We fid here that every elemet of C is a elemet of A as well as a elemet of B. I other words, if a elemet belogs to C; it belogs to both A ad B. i.e. a elemet of C is commo to both A ad B. Such a set C is called itersectio of two sets A ad B. This set is deoted by A B; read as "A itersectio B". Defiitio: The itersectio of two sets A ad B is A B = {x x A ad x B} Mathematics & Statistics /
13 Thus A B is a set of all commo elemets of A ad B. This has bee show above i Fig Disjoit Sets Now cosider the two sets, A = {x x is a eve iteger} B = {x x is a odd iteger) We kow that there is o iteger which is both eve ad odd at the same time. Thus these two sets A ad B have o commo elemet. Their itersectio is a empty set. A B = Two such sets are called disjoit sets. Defiitio : Two sets are disjoit, if they have o elemet i commo. e.g. A ad A' are always disjoit. A A' = for ay set A..7. Properties of Itersectio Operatio We shall list some properties of itersectio. (i) A B = B A (commutativity) Proof.: xab xa ad x B x B ad x A x ( B A ) i.e. A B = B A We say that itersectio operatio is commutative. i.e. We get the same set i whatever order the sets are take for itersectio. (ii) A = (iii) (iv) (v) A X = A A A' = A A = A Proof.: x A A x A ad x A x A A A = A Set Theory /
14 (vi) We ca easily see from Ve diagram of Ve diagram represetatio is as follows: A B that (a) (A B) A (b) (A B) B (vii) If A B the A B = A Proof.: Sice A B, x A x B x A B x A ad x B x A i.e. A B = A. (i either case) (viii) A ( B C ) = ( A B ) C (associativity) Proof.: x A (B C) x A ad x ( B C) x A ad x B ad x C (x A ad x B ) ad x C x (AB) ad x C x (AB)C We say that itersectio operatio is associative. i.e. if we remove the brackets o either side we get the same result. viz. A B C = {x x A ad x B ad x C} = {x x all the sets A, B, C}. Mathematics & Statistics /
15 (ix) If C A ad C B the C ( A B ) Proof.: C A meas x C x A C B meas x C x B x C x A ad x B x ( A B). C ( A B )..7. Relative Complemet of a Set We have see the cocept of a complemet of a set i relatio to a uiversal set. Now we shall cosider complemet of a set A relative to aother set B. Defiitio: Relative complemet of A with respect to B is defied as A B' ad is usually deoted by A B. A B = {x x A ad x B} i.e. A B is a set of all elemets of A except the commo elemets of A ad B. A B is also called the differece of two sets A ad B. Similarly, relative complemet of B with respect to A is B A' = B A. B A = {x x B ad x A} i.e. B A is a set of all elemets of B except the commo elemets of A ad B. The Vediagram represetatio i as : (ix) Icidetally we observe the followig from the above Ve diagram Fig..7. (a) sets A B, B A ad A B are pair wise disjoits sets. i.e. every pair of sets A B, B A ad AB have o elemet is commo. (b) i.e. A B is uio of disjoit sets viz A B, B A, AB. A B = (A B) (B A) (AB) e.g. Let A = {, 4, 6, 8,, } B = {4, 8,, 6} We have A B = {, 6, } B A = {6} A B = {8,4, } Set Theory /
16 (x) AB = {,4,6,8,,, 6} = {,6, } {6} {4, 8, }. If X is the uiversal set ad A is ay of its subset the A' = X A. Geeralized Priciple of Iclusio ad Exclusio Se shall state various results without proof. The utility of them will be see i illustrative examples. (i) (ii) Let A, B, C be ay fiite subsets of a uiversal set. We have the followig rules. (AB) = (A) + (B) (AB) (ABC) = (A) + (B) + (C) (AB) (BC) (AC) + (ABC) (iii) From the Ve diagram of AB, we have the followig: (A B) = (A) (AB) = (AB') (iv) (B A ) = (B ) (A B ) = (A' B ) (v) Similarly from the Ve diagram of ABC. it follows that Number of elemets of A oly = (A) (AB) (AC) + (ABC)..7.4 Illustrative Example We shall illustrate the various operatios that we have studied so far by meas of a example. Let X= {,,,, 4, 5, 6, 7, 8, 9} A= {,, 4, 6, 8} B= {,,7,8,9} C= {,5,7,9} We have the followig : (i) AB= {,,,, 4, 6, 7, 8, 9} =X {5} (ii) B C= {,,, 5, 7, 8, 9} (iii) A C= {,,4, 5, 6,7,8,9} (iv) A B= {8} (v) B C= {7,9} (vi) A C= {} (vii) A'= X A = {,, 5, 7, 9} (viii) B'= X B = {,, 4, 5, 6} (ix) C'= X C = {,,, 4, 6, 8} Mathematics & Statistics / 4
17 (x) (xi) (xii) A B= {,, 4, 6} = A (AB) B C= {,, 8} = B (BC) C A= {5, 7, 9} =C (CA) (xiii) A' B = {5} (xiv) (xv) A B C = {,,,, 4, 5, 6, 7, 8, 9} = X A B C = (xvi) B A = {,,7, 9} (xvii) Symmetric differece A B = (A B ) (B A) = {,,,, 4, 5, 6, 7, 9} The Vediagram represetatio of some of these operatios is give below..7 Check your progress: Solve the followig Give X = {a, b, c, d, e, f, g, h, i} A = {a, c, d, f, g}, B = {c, d, e, f, h}, C = {a, i}. Write dow the followig sets. (i) A B (ii) B C (iii) A C (v) A' B (iv) A' B' (vi) B' A (vii) Symmetric differece A B. Draw Vediagram to represet the followig sets. (i) A B (ii) A B (iii) A C (v) A B (iv) A'C Set Theory / 5
18 .8 DE MORGAN'S LAWS We shall study two results kow as De Morga's laws. We shall offer theoretical proofs ad their verificatios for certai sets. (I) (A B)' = A'B' Proof.: x(ab)' x(ab) x A ad/or x B xa' ad x B' x (A'B') (AB)' = A'B' (II) (A B)' = A'B' Proof. : x (A'B') x A' ad/ o r x B' x A ad xb x (AB) x(ab)' (AB)' =A'B' We shall verify these laws for the followig sets. X = {x x is a positive iteger ad x } A = {,, 7}, B = {, 4, 6, 8, } We have AB = {} AB = {,, 4, 6, 7, 8, } A' = {, 4, 5, 6, 8, 9, } B' = {,,5,7,9} (AB)' = {,,4,5,6,7,8,9,} (AB)' = {,5,9} A'B' = {,5,9} A'B' = {,, 4, 5, 6, 7, 8, 9, } (AB)' = {, 5, 9} = A'B' Ad (AB)' = {,, 4, 5, 6, 7, 8, 9, } = A'B' This completes the verificatio. Check your progress .8 X = {a, b, c, d, p, q, r, s} A = {a, c, d, p, s} B = {b, c, d, p, q} verify: (i) (AB )' = AB' (ii) ( A B )' = A'B'.9 DISTRIBUTIVE LAWS OF UNION AND INTERSECTION We are familiar with the followig simple operatios. (4 + 6) = (4) + ( 6) The two operatios used are multiplicatio ad additio. The above relatio is expressed as "multiplicatio is distributive over additio." I the above illustratio, we ca ot iterchage multiplicatio ad additio. + (46) ( + 4) ( + 6) The laws for uio ad itersectio are as: (I) A (B C) = (A B) (A C) By iterchagig ad i (I), we get aother distributive law: Mathematics & Statistics / 6
19 (II) A(BC) = (AB)(AC) We shall verify these laws by the followig Ve diagram: Let Hece Ad A = {I, II, III, IV} B = {II, III, VI, VII} C = {III, IV, V, VI} AB = {I, II, III, IV, VI, VII} AC = {I, II, III, IV, V, VI} BC= {II, III, IV, V, VI, VII} AB = {II, III} A C = {III, IV} B C = {III, VI} A(BC) = {I, II, III, IV, VI} = (AB) (AC) A(BC) = {II, III, IV} = (AB)(AC). Check your progress.9 For the sets A = {a, b, c, d}, B = {c, d, e, f}, C = {a, d, f, g} Verify: (i) A(BC) = (AB)(AC) (ii) A(BC) = (AB)(AC). ILLUSTRATIVE EXAMPLES () Show that each of the followig coditio is equivalet to A B. (i)ab = A (ii)ab=b (iii)b'a' (iv)ba' = X (v)ab' = Solutio: (i) A B = A meas all the elemets that are commo to A ad B are i A. i.e. all the elemets i A are elemets i B while all elemets i B are ot i A. i.e. AB. (ii) We kow that A (A B) for ay two sets A ad B. (iii) B A' AB. i.e. x B' x A' i.e. x B x A i.e. x A x B Set Theory / 7
20 i.e. AB (iv) We kow that for ay set B, B B'= X We are give that B A' = X B'A' ad hece from (iii), it follows that A B. (v) We kow that for ay set A, AA'= We are give that A B' = B'A' ad the result follows from (iii). () Prove that for ay two set A ad B (AB)A(AB). Solutio : x(ab)xa ad xb I particular let x A x (A B ) x A. i.e. (AB) A...(i) Further, if x A, the x A ad/or x B. i.e. x(ab)...(ii) From (i) ad (ii), it follows that (AB)A(AB). () Give A = {a, b, c, d, e, f}, B = {d, e, f, g, h, i} C = {b, e, h, i}, D = {d, e}, E = {b, d}, F = {b}. Let x be a ukow set. Determie which sets A, B, C, D, E or F ca be equal to X, if we are give the followig iformatio. (i) X A ad X B (ii) X B ad X C (iii) X A ad X C (iv) X B ad X C. Solutio : (i) The set which is subset of A ad B both, is A B. Here this set is D. (ii) (iii) X = D. The set is ot a subset of B, but it is a subset of C. The oly elemet of C which is ot cotaied i B is b. Hece X = F. The set is ot a subset of A ad C. The oly set which satisfies the coditio is B. X= B. (iv) The set is a subset of B, but it is ot a subset of C. Hece the set may be B or D. X = B, D. (4) Prove the followig : (i) If A B = X, the A' B (ii) A (A'B) = AB. Solutio : Mathematics & Statistics / 8
21 (i) We kow that X A' = A' ad we are give that A B = X. (ii) (A B ) A' = A' By distributive law, we get (AA')(BA') = A' i.e. (BA'). = A' i.e. B A' = A' i.e. A' B By distributive law, we have A (A'B) = (A A') (AB) = (AB) = AB A(A'B) = AB. (5) The studets i a hostel were asked whether they had a TV set or a computer i their rooms. The result showed that 65 studets had a TV set, 5 did ot have a TV set, 75 had a computer ad 5 had either a TV set or a computer. Fid the umber of studets who, (a) live i the hostel. (b) have both a TV set ad a computer. (c) have oly a computer. Solutio : We shall draw a Vediagram,. Let C = set of studets havig a computer, T = set of studets havig a TV set, X = set of studets who live i the hostel. Let x be the umber of studets havig TV ad computer both. By data (CT) = x (T) = 65 (C) = 75 (CT)' = 5 Also there are 5 studets who do ot have a TV set. 75 x + 5 = 5 i.e. 5 5 = x i.e. x = 75. i.e. umber of studets havig both TV & computer is 75. (b) Thus the umber of studets havig both TV set ad a computer is 75. (a) Total umber of studets stayig i the hostel = (X) = (CT) + (CT)' = (C) + (T) (CT) + 5 Set Theory / 9
22 (c) = = 8. Number of studets havig oly a computer = (C) (CT) = =. (6) Amog studets, study Mathematics, study Physics, 45 study Biology. 5 study Mathematics ad Biology. 7 study Mathematics ad Physics. study Physics ad Biology. do ot study ay of the three subjects. The fid, (a) umber of studets who study all the three subjects. (b) umber of studets who study Mathematics oly. Solutio : Here also we ca draw a Vediagram to represet the data. Let X = set of studets P = set of studets who study Physics M = set of studets who study Mathematics B = set of studets who study Biology x = umber of studets who study all the three subjects. By data (P) = (M) = (B) = 45 (PM) = 7 (MB) = 5 (PB) = (PMB)'= (X) = ad (P M B ) = x. Number of studets who are studyig at least oe of the subjects = (PMB). (PMB) = (P) + (M) + (B) (PM) (M B ) (P B ) + (P M B) i.e. (X) (PMB)'= x i.e. = 97 + x i.e. x = 5. (a) Number of studets who study all the three subjects is 5. (b) Number of studets who study Mathematics oly = (M) (MP) (MB)+(PMB) = = 5. Mathematics & Statistics /
23 a) AB b) AB c) A ', B' d) (AB)' = B' A' e) A exits f) (AB)' = A' B' g) (AB)' = A' B'.SUMMARY A set is a well defied collectio of objects. There are types of sets. For ay two sets A ad B we ca write h) (AB) = ( A) + (B) i) (AB)= (A) + (B) (AB) j) (AB) = (A)+(B) Whe AB =. CHECK YOUR PROGRESS ANSWERS.. (a) {,,, 4} (b) 5 {, 6,,7 8}. A = B, A C, B C. (a) False (b) True.. (a) False (b) False (c) True (d) True (e) True.4 ad.5. Φ{}, {}, {}, {, }, {, }, {, } A.. (a) o, A B (b) {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s} (a) A' ={, 4, 6, 8} (b) B' = {,, 5, 7, 9} (c) C = {, 4, 5, 6, 7, 8} (d) (f) Φ X.6. (a) A B = {5,,, 5,, 5} (b) B C = {5,, 5,, 5} (c) A C = {,, } (d) A'B'={5, 5,, 5,, 5} (e) A'C'= {5, 5,,, 5, 5}. (a) False (b) True (c) False (d) True. Commo set {5,, 5,, 5, 5}.7. (i) A B = {c, d, e, f} (ii) B B =Φ (iii) A C = {a} (iv) A' B' = {a, b, g, h, i} (v) A' B = {b, i} (vi) B' A = {b, i} (vii) A B = {a, e, g, h} Set Theory /
24 . QUESTIONS FOR SELF STUDY Problems For Practice () 9 studets appeared for two papers i Mathematics at the first year examiatio. Exactly 74 ad 66 studets passed i papers I ad II respectively. 64 studets passed i both the papers. Draw a Vediagram to idicate these results ad hece or otherwise fid the umber of studets who have failed i both papers. () Amog 6 families, families have o childre, 4 families have oly boys ad have oly girls. How may families have both boys ad girls? () I a group of studets, 8 are takig a Mathematics class, 6 are takig a Chemistry class ad are takig both classes. (i) How may studets are takig either a Mathematics class or a Chemistry class? (ii) How may studets are takig either class? (4) Suppose that studets at a college take at least oe of the laguages Frech, Germa ad Russia. 65 studets study Frech, 45 study Germa ad 4 study Russio. Also studets study Frech ad Germa, 5 studets study Frech ad Russia, 5 studets study Germa ad Russia. Fid the umber of studets who study (i) all the three laguages, (ii) exactly oe laguage. (5) It is kow that at the uiversity 6% of professors play Teis, 5% play Cricket, 7% play Hockey, % play Teis ad Cricket. % play Teis ad Hockey, 4% play Cricket ad Hockey. Assumig that each professor play at least oe of the games, determier % of professors playig all the three games. (6) A computer compay must hire 5 programmers to hadle systems programmig tasks ad 4 programmers for applicatios programmig. Of those hired 5 will be expected to perform tasks of each type. How may programmers must be hired?.4 SUGGESTED READINGS. Mathematics ad Statistics by M. L. Vaidya, M. K. Kelkar. Pre degree Mathematics by Vaze, Gosavi Mathematics & Statistics /
25 Set Theory / NOTES
26 NOTES Mathematics & Statistics / 4
27 CHAPTER Fuctios. Objectives. Itroductio. Number System.. Basic Operatios i Mathematics.. Divisibility Test. Prelimiary Cocepts.4 Correspodece.5 Fuctios.6 Types of Fuctios.7 Graph of Fuctio.8 Summary.9 Check your Progress Aswers. Questios for Self  Study. Suggested Readigs. OBJECTIVES After studyig the umber system ad certai prelimiary cocepts, studets ca explai the followig * Fuctios i various otatios * Types of fuctios * Fuctios of fuctios * Graph of fuctios * Formula of a fuctio * Fuctio as a correspodece * Studets ca solve problems ivolvig the all above cocepts.. INTRODUCTION There is o permaet place i the world for ugly mathematics. It may be very hard to defie mathematical beauty but that is just as true of beauty of ay kid  G.H. Hardy The cocept of term relatio i mathematics has bee draw from the meaig o relatio i Eglish laguage. Accordigly two objects or quatities are related if there is a coectio or lik betwee the two objects or quatities.. NUMBER SYSTEM N = Set of all atural Numbers. = {,,,..} W = Set of whole umbers. = {,,,,..} I = Set of all Itegers. = {. , , ,,,,,..} Fuctios / 5
28 Q = Set of ratioal Numbers p = p, q,, I, q q.. Basic operatios i Mathematics ) Additio i) a, b are two umbers the a + b is called additio of two umbers. a c ii) ad two fractios the b d a c ad bc b d bd e.g ) Subtractio a c ad bc b d bd e.g ) Multiplicatio a c a c b d b d 4 4 e.g ) Divisio a c a d b d b c a d e.g. b c where c d is called Multiplicative Iverse e.g. 4 5 (Now use rule of Multiplicatio) ) Order a ad b are two give umbers the possible relatio betwee these two are, i) a < b a is less tha b ii) a > b a is greater tha b iii) a = b a is equal to b Mathematics & Statistics / 6
29 e.g. a ad b are two studets i TMV. the a ad b must be admitted to ay of the 5 coerces ru by TMU. If a take admissio to B.C.A. b take admissio to M.C.A the by age a smaller tha b a ot bigger tha b ad a is ot same age as b. 6) Number lie All umbers we ca preset o a lie is called umber lie.  ve Numbers O + ve Numbers origi e.g. Represet the umbers o lie ,,, 5, Meas All positive umbers are o right had side of Number lie whereas all ve umber o Left had side... Divisibility Test Number a is called divisible by b whe we divide o. a by o. b ad remaider equal to zero If remai is ot equal to zero, we say that a is ot divisible by b Test of Divisibility ) Test of : A umber is said to be divisible by whe its uit place digit is oe of,, 4, 6, ad 8. T U e.g. ) 5 6 uit place digit = give umber is divisible by U ) uit place digit = 9 give umber is ot divisible by ) Test of : A umber is said to be divisible by, whe additio of digits of give umber is divisible by. e.g. ) = 5 5 = 5 Remaider = give umber is divisible by ) = = is ot Iteger =.  give o is ot divisible by ) Test of 4 : A umber is said to be divisible by 4, whe last two digits T ad U Fuctios / 7
30 is divisible by 4. e.g. ) 4 5 last two digits ot divisible by 4 give o. is ot divisible by 4 ) last two digits 4 4 is divisible by 4 give o. is divisible by 4 4) Test of 5 : A umber is said to be divisible by 5, whe the last digit is or 5. e.g. ) last digit = 5 give o. is divisible by 5 ) last digit = give o. is ot divisible by 5 5) Test of : A umber is said to be divisible by, whe last digit is. e.g. ) 5 last digit = give o. is ot divisible by ) 4 4 last digit = give o. is ot divisible by. Check your Progress State True/False ) is divisible by 5 ) is divisible by ) is divisible by 5 4) is ot divisible by 5) is divisible by, 5,, ad 4.,,, Natural Numbers.. PRELIMINARY CONCEPTS takes always value oe i ay case, cocept, example. so the umbers are called as costats. But whe value is ot costat e.g. Age of studet i F.Y. Probable year is after th std. meas above 7 years. But it may be 8, 9 also. So the age of studet is a variable ad deoted by small alphabets as, , y, z, p, q, That meas the age of studet lies betwee 7 ad 9 ad above. These values are called itervals of variables. There are four types of itervals. ) Ope Ope iterval e.g. (,5) meas our variable takes values betwee ad 5. But ot ad 5. Mathematics & Statistics / 8
31 < x < ) Ope Closed < x 5 e.g. ( 5) meas variable takes values betwee ad 5 ad 5 also but ot. ) Closed Closed e.g. (, 4) meas Variable takes values ad 4 betwee all values ad 4. 4) Closed Ope (5, 7) meas Variable takes values betwee 5 ad 7 ad 5 but ot 7... Absolute Value x is a Variable. The x = x if x > o = x if x < o x ca be read as modx Imp. Results ) x y x y ) x y x y ) xy x y 4) x x if y y y.4 CORRESPONDENCE Defiitio If A ad B are two sets such that by some rule to a elemet a A takes oe or more elemets i B, the the rule is called a correspodece There are Four types of correspodece ) Oe Oe correspodece e.g. Marks i examiatio of studets. A = Set of Number of studets. B = Set of marks obtaied by studets. A = { a, b, c, d, e } B = { 9, 8, 6, 68, 7} 5 x< 7 Fuctios / 9
32 Ve diagram a b c d e A B ) May Oe Correspodece A = Number of subjects for F. Y. B = Number of subjects i F. Y. i TMU A = { Computer Network, VB Net, Math & stats office Automatio, Operatig system, C++. Accoutig} B = {a, b, c, d, e, f} C++ CN VB M&S OA OS ACC A Subjects of studet A take ) M & S ) OA ) C++ 4) OS 5) VB 5 papers (subject) for each studet. a b c d e f B ) May May A = {x, y, z} B = {a, b, c, d} x y z A B C d A B Mathematics & Statistics /
33 4) Oe May ) A = {,, } B = {a, b, c, d} A a b c d B ) A = Set of states B = Set of cities R : A B R : called as Rule of correspodece from oe set to aother set. Check your Progress.4. State the correspodece. Draw ve diagrams. i) A = {,, } B = {a, b, c, d} R : A B R = { (,a) (,b), (, c)} ii) A = {, 4, 6, 8} B = {p, q, r, s} R : A B R = {(,p), (,q), (4,r) (6,s), (8,r)} iii) A = {,, 5, 7} B = {p, q} R : A B R = {(,p), (,p), (5,q), (7,q)} iv) A = {, 4} B = {r, s} R = A B R = {(, r), (,s), (4,r), (4,s)}.5 FUNCTIONS Defiitio Let A ad B are two o empty sets A correspodece from the set A to set B is called fuctio if it is either oetooe or maytooe. Every fuctio is correspodece but every correspodece is ot to a fuctio. Fuctio is a relatio R : A B (Relatio from A to B) for every x A there is uique y B such that x R y (x related to y). A fuctio is deoted by letters f, g, h. We write f : A  B g : C D h : C B if f : A B the xry x is related to y y = f(x) The set A is called as the domai of the fuctio The set B is called as the codomai of the fuctio y = f(x) is called as Rule of correspodece y B is called as image of x uder. The set of images i B is called as the Rage of the fuctio. Rage of r = {y f(x) = y, where x A, y B} The Rage of a fuctio is subset of its co domai Fuctios /
34 Note If A B, the, ) Every elemet of set A is related to oe ad oly oe elemet of set B. ) More tha oe elemet ca be related to set A to oe elemet of set B ) Set B may cotai elemets which are ot related to ay elemet of set A. e.g. Let A = {,,,} B = {4, 5, 6, 7,} f : AB f = { (, 5), (,6), (,7)} the ve diagram A Set A is domai Set B is Codomai Rage = {5, 6, 7} Images f() = 5 f() = 6 f() = 7 Rule f(x) = x + 4 = y y B Set of order pair = {(,5), (,6), (,7)} B Tabular form X y = f(x) Check your Progress. Fid the image of followig fuctios a) f(x) = x²  x + 4 fid f(), f(), f(), f(), f() b) f(x) = x²  5 Fid f (), f(), f(), f(), f() ) Oto fuctio ) Ito fuctio ) OeOe fuctio 4) May Oe fuctio 5) Eve fuctio 6) Odd fuctio 7) Composite fuctio.6 TYPES OF FUNCTIONS ) Oto fuctio (subjective fuctio) Defiitio A fuctio F : A B is said to be a Oto fuctio if every elemet of set B is the image of some elemet of set A. R = B (Rage = codomai) symbolically, we write, oto Mathematics & Statistics /
35 f : A B e.g. A = {, , ,, } B = {, 4, 9} f : A B f(x) = x² A B ) Ito fuctio (subjective fuctio) Defiitio A fuctio f : A B is said to bea ito fuctio if there exits at least oe elemet i B, which is ot the image of ay elemet of A. RCB (Rage is proper subset of B) e.g A = {, ,,, } B = {,,,, 4} f : A B f(x) = x² = y A B Rage = R = {,, 4} B = {,,,, 4} RCB Note : The fuctio is Oto or Ito which deped o Rage of that fuctio. ) Oe Oe fuctio (Ijective fuctio) Defiitio A fuctio f : A B is said to be Oetooe fuctio if distict elemets of A have differet images i B uder f. e.g. A = set of studets. B = set of Roll Numbers R : A B Roll Number is fixed with studet. Amit s Roll No. is 8. meas No. 8 is Asie to Amit oly ad ot other, ay studet. oetooe correspodece. oeto oe fuctio. 4) Mayoe fuctio Defiitio A fuctio f : A B is said to be maytooe fuctio if two or more elemets of A have the same image B i.e. there is at least elemet i B, which has more tha oe oe preimage i A. 5) Eve fuctio Defiitio A fuctio f : A B is said to that meas f(x) does ot chage with x ad x replacemet. e.g. ) f : A B Fuctios /
36 f (x) = y = x² + Now put x = x f(x) = y = (x)² + = x² + x² = (x)² f(x) is eve fuctio. ) f(x) = x + 5 put x = x f(x) = (x) + 5 = x + 5 f(x) f(x) Give fuctio is ot eve fuctio. 6) Odd fuctio Defiitio A fuctio f : A B is said to be odd fuctio if f(x) = f(x) e.g. ) A fuctio f(x) = x³ + x f(x) = (x)³ + (x) = (x)³ x = [x³ + x] =  f(x) Give fuctio is A Odd fuctio. ) f(x) = x² + 4x f(x) = (x)² + 4(x) = x²  4x =  f(x) Give fuctio is ot Odd fuctio. 7) Composite fuctio Defiitio A fuctio f : A B ad g : B C be two fuctios. x is ay elemet of A. The y = f(x) B. Sice B is the domai of fuctio g ad C is its codomai, g(y) C so z = g(y) C This fuctio is called as composite fuctio of f ad g. Let it deoted by h. Thus gof = h : A C such that h(x) = g [f(x)] A B C f X Xy = f ( x ) Z = g ( y ) g h e.g. ) Let f : {,, 4, 5} {, 4, 5, 9} ad g : {4,, 5, 9} {7,, 5} fuctios defied as f() =, f() = 4, f (4) = = f(5) = 5 ad g () = g(4) =7, g(5) = g(9) = fid g of. Solutio ) Mathematics & Statistics / 4
37 give ) f(x) = g of () gof () gof (4) gof (5) = g[f()] = g() = 7 = g[f()] = g(4) = 7 = g[f(4)] = 9(5) = = g[f(5)] = 9(5) = A B C g f A h C h g of (x) = g[f(x)] g of (x) = g of () = g[f(x)] = 9[f()] = g() = 7 g of () = g[f()] = 9(4) = 8 gof() = g[f()] = g(5) = 9.6 Check your progress Fill i the blaks i) The fuctio is either oto or Ito which depeds o the give ii) a) Rage b) domai c) codoma d) value A oeoe fuctio is also called a) Ijective b) Bijective c) Oto d) Ito Fuctios / 5
38 iii) Oeoe ad Oto fuctio is called a) Bijective b) Ijective c) Ito d) eve. iv) Mayoe correspodece is called fuctio a) eve b) add c) Bijective d) Mayoe. v) A mayoe fuctio ca be either a) ito or oto b) eve or odd c) eve or bijective d) ito or odd. vi) Nature of fuctio whether it is oetooe or mayoe deped, upo of fuctio. a) Domai b) Codoma c) Rage d) eve..7 GRAPH OF FUNCTION Defiitio A fuctio f : A B, x A ad y B the (x,y) be a elemet of f. We ca plot the poit (x,y) i a plae by choosig a suitable coordiate system. O plottig all such order pair, we get geometrical represetatio (curve) of fuctio f this is called graph of fuctio f. e.g. f(x) = x + X F(x) = y Y 5 4 Y=x X .8 SUMMARY Fuctios meas a actio, the velatio betwee variable ad umber. Correspodeces are of four types. There are also types of fuctios. We ca draw proper ad et Ve diagram for each correspodeces. The et graph is there for each fuctio..9 CHECK YOUR PROGRESS  ANSWERS. ) False ) True ) True 4) True 5) True Mathematics & Statistics / 6
39 .5 ) a ) b c d p q r s A B A B oeoe oemay ) 4) 5 7 P q 4 A r s B A MayOe B May May.6 ) c) ) a) ) a) 4) d) 5) a) 6) a). QUESTIONS FOR SELF STUDY Problems For Practice. If f(x) = x² < x < fid f (), f ( ), f(). Fid g of ad fog if f(x) = x + ad g(x) = x²  x + 4. If f(x) = x x = 6 < x < = x 6 x 4 Stat (a) Domai ad Rage (b) fid f(), f(4), f() 4. f : R R, g : R R are defied as, f(x) = x² x = x x > g(x) = x + x = x² x > fid g of (x), x R 5. Test whether followig fuctios are eve or odd ) f(x) = g  x² ) f(x) = x ) f(x) = x + 4) f(x) = x 5) f(x) = x x ± Fuctios / 7
40 6. Draw the graph of give fuctios ) f(x) = ) f(x) = x x  R ) f(x) = x 4) f(x) =  if  x <  =  if  x < = if x < = if x fid images of x =  = = = 7. Fuctios f ad g are give by the followig i) f = { (, ), (, ), (, 4), (4, 5), (5, 6), (6, 7)} g = { (, 4), (, 6), (4, 8), (5, ), (6, ), (7, 4) } ii) If f = { (, ), (4, 5), (6, 7), (8, 9)} g = { (, ), (, 4), (, 6), (4, 8)} iii) If f = { (, ), (, ), (, 5), (4, 7) ), (5, 9)} g = { (, 5), (, 4), (, ), (4, ), (5, )} a) Express f ad g by formula b) Show that f ad g are oeoe fuctios. c) fid fog ad gof. Problems for practice ) f() = , f( ) = 4, f() does ot exits ) gof (x) = 6 x²  x + 9 fog (x) = x² + x + 8 ) Rage = [,4] f() = 6, f(4) does ot exist, f() = 4) gof (x) = 9 [f(x)] x g[f()] = x² = g[f() = x² = gof (x) = g[f(x)] = g [f()] = g(x²) = (x + )² = 4 gof (x) = 9[f()] = g() = x² = 5) ) eve ) eve ) eve 4) odd 5) odd. Mathematics & Statistics / 8
41 6) ) Y = f(x) = ) (, ). SUGGESTED READINGS. Predegree Mathematics by Vaze, Gosavi. Discrete Mathematical Structures for Computer Sciece by Berard Kolma ad Robert C Busby. Statistical Aalysis: A Computer  Orieted Approach Itroductio to Mathematical Statistics by S. P. Aze & A. A.Afifi Fuctios / 9
42 NOTES Mathematics & Statistics / 4
43 CHAPTER Sequeces, Progressios ad Series. Objectives. Itroductio. Sequece.. Summatio of terms of a sequece. Arithmetic Progressio.. The th term of A.P. (T ).. Sum of first terms of A.P. (S ).4 Geometric Progressio..4. The th term of G.P. (T ).4. Sum of first terms of G.P. (S ).5 Harmoic progressio (H.P.).6 The three Meas.6. Properties of meas.7 Series.7. Stadard series.7. Ifiite Geometric series.8 Summary.9 Check your Progress  Aswers. Questios for Self Study. Suggested Readigs. OBJECTIVES After studyig this chapter, you ca explai ad use various types of sequeces, series give below : Sum of terms of a sequece. Three types of sequeces. Arithmetic progressio. Geometric progressio. Harmoic progressio. The three meas : A.M., G.M. ad H.M. Summatio of series of terms. Summatio of certai ifiite series i G.P.. INTRODUCTION I computer applicatios may cocepts ca be expressed as a ordered umbers usig,. We have already see oe such case of expressig a set or a fuctio formed by various combiatios of,. Briefly the ordered set of umbers forms a sequece. We shall the cosider three types of sequeces ad summatio of terms of sequeces formig a series.. SEQUENCE We kow that the system of atural umbers is,,,..., +,... This is a collectio of umbers satisfyig the followig properties. (i) It is ordered i.e. each umber of the collectio has a defiite positio. (ii) There exists defiite law, accordig to which every umber ca be writte dow. (iii) Every umber is followed by ext oe. Ay other collectio of umbers which satisfies the first two of the above Sequeces, Progressios ad Series / 4
44 properties is called a sequece. The sequece is further called ifiite sequece, if the third property is also satisfied, otherwise it is called a fiite sequece. Defiitio : A ordered set of umbers formed accordig to a well defied law is called a sequece. I terms fuctio, it is a fuctio f : N R If N, the f() is called th term of the sequece. By givig differet values to, we get the correspodig term of a sequece. e.g. (i),,,...,,... Here f() =, i.e. square of a atural umber. Whe =, f() = is the rd term of a sequece. (ii),,,..., 4.. Here f() = (iii), 5, 8,, 4,... Here f() = or f() = + ( ) or it is a sequece i which each term is obtaied by addig to the previous term. (vi) (iv), 4, 8, 6,,... Here each term is a power of. f() = is the th term. (v).,.,.4... Here f() = ( + ) is the th term. All these are illustratios of ifiite sequeces. Now we shall cosider certai fiite sequeces. The sequece,,,,,,,,,, is a fiite sequece with repeated terms. The digit, for, example occurs as the d, rd, 5 th, 6 th ad 8 th elemets of the sequece. The correspodig set is simply {, }, i which the order of 's ad 's is ot specified. (vii) A ordiary word i Eglish, such as 'Physics' ca be viewed as a fiite sequece. p, h, y, s, i, c, s composed of letters from the ordiary alphabets. If we omit commas, we get the word physics. Such represetatio is referred to as strig. I computer sciece a sequece is sometimes called a liear array or list. A array may be viewed as a "sequece of positios" which we represet below as boxes. The positio form a fiite or ifiite list, depedig o the desired size of array. Elemets from some set may be assiged to the positios of the array. The elemet assiged to positio will be deoted by S() [correspodig to f() i the defiitio] ad the sequece S(l), S(), S(), S(4), will be called the sequece of values of the array S... Summatio of terms of a sequece: Aother problem related with sequeces is to fid out the sum of first terms. It is tedious to add terms oe after aother i case of large umber of terms. It is therefore ecessary to fid a law givig sum of the first terms. Mathematics & Statistics / 4
45 Let T ad S deote the th term ad sum of first terms of a sequece respectively. Thus S = T +T T = T read as "sigma T r, r varyig from to " With this otatio, the sum of first ( ) terms is S = T r = T + T + + T r Hece S S, = T, true for. We shall illustrate it by simple examples. r Example () : If T = , fid T 4, T 9 ad T Solutio : We have T = Example (): If S = Solutio: puttig = 4, 9 ad respectively, T 4 = (4 ) + 4 (4 ) + 7 = 7 T 9 = ( 9 ) + 4 ( 9 ) + 7 = 86 T = ( ) + 4 () + 7 = 566. Fid T 5 ad T 7. We have S S, = T puttig = 5, we get S 5 S 4 = T 5.T 5 = = 5 = 5. Similarly T 7 = S 7 S 6 7(7 ) 6(6 ) = = 8 = 7. Example (): Fid T 5 give that S = ( ). Solutio : We have S S, = T puttig = 5, we get, T 5 = S 5 S 4 = = = 4 = ( 4 ) = 5 Sequeces, Progressios ad Series / 4
46 Check your progress. () The array represets a ifiite sequece Fill i the blak. () Choose correct figure from the bracket to fill i the blak. (i) T = +,T =.. (4, 5, 6, ) (ii) S = 6, T =...(4, 6, 8, ) (iii) S =, T =... (,,, + ). ARITHMETIC PROGRESSION Amogst the three types of sequeces which are very commo, the arithmetic progressio is oe of them. Cosider a sequece, 5, 8,, 4,... Here we see that 5 = 8 5 = 8 =... = i.e. the differece betwee a term ad its precedig term is costat. Such a sequece is called arithmetic progressio; abbreviated as A.P. Defiitio : A arithmetic progressio (A.P.) is a sequece i which the differece betwee ay term ad the immediately precedig term is costat. This costat differece is called the commo differece of the arithmetic progressio ad is usually deoted by d. I the above example the commo differece is ad first term is. The geeral form of a arithmetic progressio is where a, a + d, a + d, a + d,... a = first term, ad d = commo differece... The th term of A.P. (T ) Cosider a geeral A.P. a, a + d, a + d, a + d,... We observe that T = a = a + ()d =a + (l l)d T = a + d = a + ( l)d T = a + d = a + ( ) d etc. We ote that every term is obtaied by addig to 'a' certai multiple of d; ad this multiple is exactly oe less tha that of the suffix of T. T = a + ( l)d. Example () : Fid T for the followig A.P., 5, 9,, 7 Solutio : Here T = a = ad commo differece d = 4 T = a + ( l)d = l+( l)4 = 4. Mathematics & Statistics / 44
47 Example (): Fid the umber of terms i the A.P., 4, 7,..., 8. Solutio: Here a =, d = 7 4 = Let 8 be the th term of A.P. 8 = + ( ) i.e. 8 = ( l) = 7 ad = 8. Thus there 8 terms i the give A.P. Example () : If the th term of a sequece is +, show that it is a A.P., what is its first term ad the commo differece? Solutio : Here T = + Replacig by ( ) we get, T _ = ( l) + = l Cosider, T T l = + ( ) = This is costat (idepedet of ) Hece it is th term of A.P. with d =. st term = T = () + = 5 ad the commo differece is ( = T T _ )... Sum of the first terms of A.P. (S ) Let S deote the sum of st terms of the geeral A.P. with T = a ad T = a + ( l)d Let T = a + ( )d= l The we have T = l d, T,= l d Now S = T +T + T T + T +T Reversig the order of terms i summatio, Addig vertically, we get = a + (a+d) + (a+d) (l d) + (l d) l S = l + (l d) + (l d) (a+d) + (a+d)+a S = (a+l) + (a+l) (a+l) + (a+l) to terms = ( a + l) S = (a + l) Substitutig the value of, we get, S = {a + ( l)d} We shall illustrate the use of these formula i certai simple examples. Example (4) : Fid T ad S for the followig A.P. 9, 5,, 7,... Solutio : Here a = 9, d = 5 ( 9 ) = 9 5 = 4 T = a + ( ) d = 9 + ( ) (4 ) = 4. Sequeces, Progressios ad Series / 45
48 ad S = { a + ( ) d } = { 58 + ( l)4 = ( l). Example (5): Fid the sum of the first odd atural umber Solutio : The first odd atural umbers are,, 5,... T = th odd atural umber = l+( l) (a =, d = ) = = l say S = (a + l) = =. ( + l) Example (6) : If the sum of the first terms of a sequece is + 4, show that it is a A.P. Fid the first term ad the commo differece. Solutio : Here s = + 4 Replace by ( ) to obtai S. S _ = ( ) + 4 ( ) = = T = S S = + 4 ( l) = 6 + l. Replace by ( ) to obtai T _ T = 6( ) + = 6 5. Cosider, T T _ = 6+ (6 5 ) = 6, which is costat. Hece S = +4 is sum of first terms of a A.P. with d = 6. The first term = T = 6() + =7 or S = + 4 = 7 Check your progress. () State whether true or false. If false write correct aswer. (i) Numbers a d, a, a + d are i A.P. (ii) For a A.P., T T _ is ot costat. (iii) For a A.P., T 8 = 6 the S 5 = 54. (iv) I a A.P., s = the a = T = 4. (v) I a A.P., S = the d = 6. (vi) The sum of first eve atural umbers is ( + ). (vii) For a A.P. a =, T 7 =, the commo differece is 7. (viii) For a A.P. S = 86, the T 6 = 6. (ix) For a sequece havig terms, S = Mathematics & Statistics / 46 the the sequece is a A.P.
49 .4 GEOMETRIC PROGRESSION Cosider a sequece, 9, 7, 8,... We observe that the ratio of ay term to its precedig term is = ad is costat. Such a sequece i which ratio of 9 7 a term to its precedig term is costat is called geometric progressio, abbreviated as G.P. Defiitio: A geometric progressio (G.P.) is a sequece i which every term bears a costat ratio to the oe immediately precedig it. This costat ratio is called the commo ratio of the G.P. ad is usually deoted by r. I the above example the commo ratio is. The first term is also here. The geeral form of a geometric progressio is a, ar, ar,... where a is the first term, ad r is the commo ratio..4. The th term of G.P. (T ) Cosider a G.P. a, ar, ar... With the usual otatio, T = a = ar = ar T = a r = a r T = ar = ar We ote that each term is a product of two factors. The first factor is a, ad is commo to all the terms. The secod factor is a certai power of r. This is exactly oe less tha the correspodig suffix of T. T = ar Example () : Fid T for the followig G.P.,, Solutio : Here a =, r = / = / = / = T = a r = =,,... 4 Example (): Give T =, T 7 = of a certai G.P. Fid T 8. Solutio : We have T = a r = ad T 7 = a r 6 = where a is the first term ad r is the commo ratio of G.P. O divisio, we get, a r a r The a r = gives a = 6 i.e. r 4 = 6 r = r 4 = 5.4. Sum of first terms of a G.P. (S ) T 8 = a r 7 = 5 ( 7 ) = 64. Let a be the first term ad r be the commo ratio of a geeral G.P. The T = ar ad Sequeces, Progressios ad Series / 47
50 S = a + ar + ar ar + ar Multiply by r. rs = ar + ar + + ar + ar + ar. (II) Subtractig (II) from (I), we get S rs = a ar Divide by ( r). i.e. S ( r) = a ( r ) a S r = r ar =, if r <, if r > r However if r =, G.P. becomes a, a,... Here S = a + a +... to terms = a We shall illustrate the use of these formula by meas of some simple examples. Example () : Fid T ad S for the followig G.P., 9 7,,, Solutio: Here a =, r = 9 4 T = a r = a ad S r = r 5 Example (4): For a G.P. a = 5, r =, S = 65, fid. a Solutio : Here a = 5, r =, S r = r 5 65 = 7 = 8 = i.e. 7 = = 7 Check you progress.4 () Fill i the blaks by choosig appropriate umber give i the bracket (i) I a G.P. with a =, r =, T =... (8,,,9) b (ii) a, b,... are i G.P. the the commo ratio is... a a b,b,a, b a (iii) If for a sequece, S = (4 ), the the sequece is... (A.P., G.P., oe of these) (iv) I a G.P. a = ad T6 = 6 the the commo ratio r =... (,, 4, 6) Mathematics & Statistics / (I)
51 (v), 9 7,, are i (A.P., G.P., H.P., oe of them) () I a G.P. T = T 7 = fid a ad r. () I a G.P., a = 5, r = 5 fid T 5. (4) State whether true or false ad if false write the correct statemet, (i) I a G.P. S =, the commo ratio is. (ii) I a sequece, 6, 8,... the 4 th term is 6. (iii) I a G.P.,,, the th term is. a a a a (iv) I a G.P. T = ad T5 = the T4 = ± HARMONIC PROGRESSION (H.P.) Cosider a sequece,,,,...(a) 4 The reciprocal of the terms of the sequece, form aother sequece viz.,,, 4,..(B) The terms of the sequece (B) are i A.P. Such a sequece (A), i which the reciprocals of the terms form A.P. is kow as Harmoic progressio, ad abbreviated as H.P. Defiitio : The terms a, a,a,..., a,... are said to be i harmoic progressio (H.P.) if.. a, a, a,... a be i arithmetic progressio. To fid th term of H.P., we have to fid th term of the correspodig A.P. Example: Fid T ad hece T 5 ad T 7 i the followig harmoic progressio.,,,... 7 Solutio: The correspodig A.P. obtaied by takig reciprocal of the terms i the give sequece is 7,,,... Here a = 7, d= Hece T of A.P. is a + ( ) d The th term of give H.P. is, = 7+( ) T = Hece puttig = 5 ad 7, we get T 5 = T 7 = There is o coveiet formula that ca be developed to obtai sum of first terms of H.P. ad Sequeces, Progressios ad Series / 49
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