Chapter 15: Progressions PDF
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This document explains progressions, focusing on arithmetic progressions. It covers definitions, formulas, and examples. The document provides clear explanations making it useful for high school math students learning about sequences and progressions.
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60 100 Progressions 3.1 Introduction. (1) Sequence : A sequence is a function whose domain is the set of natural numbers, N. E3 If f : N C is a sequence, we usually denote it by f (n) f (1), f (2), f (3),.... It is not necessary that the terms of a sequence always follow a certain pattern...
60 100 Progressions 3.1 Introduction. (1) Sequence : A sequence is a function whose domain is the set of natural numbers, N. E3 If f : N C is a sequence, we usually denote it by f (n) f (1), f (2), f (3),.... It is not necessary that the terms of a sequence always follow a certain pattern or they are described by some explicit formula for the nth term. Terms of a sequence are connected by commas. Example : 1, 1, 2, 3, 5, 8, …………. is a sequence. ID (2) Series : By adding or subtracting the terms of a sequence, we get a series. If t1 , t 2 , t 3 ,..... t n ,..... is a sequence, then the expression t1 t 2 t 3 ..... t n.... is a series. Example : 1 U A series is finite or infinite as the number of terms in the corresponding sequence is finite or infinite. 1 1 1 1 .... is a series. 2 3 4 5 D YG (3) Progression : A progression is a sequence whose terms follow a certain pattern i.e. the terms are arranged under a definite rule. Example : 1, 3, 5, 7, 9, …….. is a progression whose terms are obtained by the rule : Tn 2n 1 , where Tn denotes the nth term of the progression. Progression is mainly of three types : Arithmetic progression, Geometric progression and Harmonic progression. Arithmetic progression Geometric progression ST U However, here we have classified the study of progression into five parts as : Arithmetico-geometric progression Harmonic progression Miscellaneous progressions Arithmetic progression(A.P) 3.2 Definition. A sequence of numbers t n is said to be in arithmetic progression (A.P.) when the difference t n t n 1 is a constant for all n N. This constant is called the common difference of the A.P., and is usually denoted by the letter d. Progressions 101 If ‘a’ is the first term and ‘d’ the common difference, then an A.P. can be represented as a, a d , a 2d , a 3 d ,........ Example : 2, 7, 12, 17, 22, …… is an A.P. whose first term is 2 and common difference 5. Algorithm to determine whether a sequence is an A.P. or not. 60 Step I: Obtain a n (the nth term of the sequence). Step II: Replace n by n – 1 in a n to get a n 1. Step III: Calculate a n a n 1. If a n a n 1 is independent of n, the given sequence is an A.P. otherwise it is not an A.P. An E3 arithmetic progression is a linear function with domain as the set of natural numbers N. t n An B represents the nth term of an A.P. with common difference A. 3.3 General Term of an A.P.. ID (1) Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Then its nth term is a (n 1)d. Tn a (n 1)d U (2) pth term of an A.P. from the end : Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. having n terms. Then pth term from the end is (n p 1)th term from the beginning. D YG p th term from the end T(n p 1) a (n p)d Important Tips General term (Tn) is also denoted by l (last term). Common difference can be zero, +ve or –ve. n (number of terms) always belongs to set of natural numbers. If Tk and Tp of any A.P. are given, then formula for obtaining Tn is If pTp = qTq of an A.P., then Tp + q = 0. If pth term of an A.P. is q and the qth term is p, then Tp + q = 0 and Tn = p + q – n. If the pth term of an A.P. is If Tn =pn + q, then it will form an A.P. of common difference p and first term p + q. U T p Tk Tn Tk . nk p k ST 1 1 and the qth term is , then its pqth term is 1. q p Example: 1 Let Tr be rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, m n, Tm Solution: (d) (a) 1 1 m n Tm 1 1 a (m 1) d n n and Tn 1 1 and Tn , then a – d equals n m (b) 1 1 1 a (n 1) d m m Subtract (ii) from (i), we get (m n) d (c) 1 mn [AIEEE 2004] (d) 0 …..(i) …..(ii) 1 1 (m n) 1 (m n) d d , as m – n 0 n m mn mn 102 Progressions 1 1 n 1 1 (n 1) d d. Therefore a – d = 0 m m mn mn a The 19th term from the end of the series 2 + 6 + 10 + …. + 86 is (a) 6 (b) 18 (c) 14 86 2 (n 1) 4 n 22 Example: 2 Solution: (c) (d) 10 19th term from end tn 19 1 t22 19 1 t4 2 (4 1) 4 14 (a) 0 (b) – 1 (c) – 12 We have 5 T5 8 T8 Solution: (a) Let a and d be the first term and common difference respectively (d) – 13 E3 5{a (5 1)d} 8{a (8 1) d} [AMU 1991] 60 In a certain A.P., 5 times the 5th term is equal to 8 times the 8th term, then its 13th term is Example: 3 3a 36 d 0 a 12d 0 , i.e. a (13 1) d 0. Hence 13th term = 0 If 7th and 13th term of an A.P. be 34 and 64 respectively, then its 18 th term is Example: 4 (a) 87 Solution: (c) (b) 88 (c) 89 (d) 90 Let a be the first term and d be the common difference of the given A.P., then T7 34 a 6d 34 ID …..(i) T13 64 a 12d 64 …..(ii) From (i) and (ii), d = 5, a = 4 Trick: U T18 a 17 d 4 17 5 89 Tn Tk Tp Tk T 34 64 34 T T7 T13 T7 18 18 T18 89 18 7 11 13 7 6 n k p k am If an is an arithmetic sequence, then m 1 an n 1 (a) 1 (c) 0 D YG Example: 5 (b) –1 ap p 1 equals (d) None of these Let a be the first term and d the common difference. Then ar a (r 1) d Solution: (c) a (m 1) d m 1 a (n 1)d n 1 a ( p 1) d a a a m 1 n 1 p 1 p m n p d m n p 1 1 1 1 1 1 1 U 1 1 1 m n p a m n p d m n p a.0 d.0 0 1 1 1 1 1 1 ST The nth term of the series 3 + 10 + 17 + ….. and 63 + 65 + 67 + …… are equal, then the value of n is Example: 6 (a) 11 [Kerala (Engg.) 2002] (b) 12 (c) 13 (d) 15 nth term of 1st series 3 (n 1)7 7 n 4 Solution: (c) nth term of 2nd series 63 (n 1) 2n 61 we have, 7n 4 2n 61 n = 13 3.4 Selection of Terms in an A.P.. When the sum is given, the following way is adopted in selecting certain number of terms : Number of terms 3 Terms to be taken a – d, a, a + d Progressions 103 4 a – 3d, a – d, a + d, a + 3d 5 a – 2d, a – d, a, a + d, a + 2d In general, we take a – rd, a – (r – 1)d, ……., a – d, a, a + d, ……, a + (r – 1)d, a + rd, in case we have to take (2r + 1) terms (i.e. odd number of terms) in an A.P. And, a (2r 1)d , a (2r 3)d ,......., a d , a d ,......., a (2r 1)d , in case we have to take 2r terms in 60 an A.P. When the sum is not given, then the following way is adopted in selection of terms. Number of terms Terms to be taken a, a d , a 2 d 4 a, a d , a 2d , a 3 d 5 a, a d , a 2d , a 3 d , a 4 d Sn sum of n terms of the series n [2a (n 1) d ] 2 n (a l) , where l = last term = a (n 1) d 2 U Also, S n The ID Sum of n terms of an A.P. : a (a d ) (a 2d ) ....... {a (n 1) d } is given by E3 3 D YG Important Tips The common difference of an A.P is given by d S 2 2S1 where S 2 is the sum of first two terms and S 1 is the sum of first term or the first term. , when d 0 The sum of infinite terms . , when d 0 If sum of n terms S n is given then general term Tn S n S n 1 , where S n 1 is sum of (n – 1) terms of A.P. Sum of n terms of an A.P. is of the form An 2 Bn i.e. a quadratic expression in n, in such case, common difference is n 1 A B T An B S 2 If for two A.P.’s n then n Tn Cn D S n n 1 C D 2 ST U twice the coefficient of n 2 i.e. 2A. T f (2n 1) S f If for the different A.P’s n n , then n Tn (2n 1) S n n Some standard results Sum of first n natural numbers 1 2 3 ........ n n r r 1 n (n 1) 2 n Sum of first n odd natural numbers 1 3 5 ..... (2n 1) (2r 1) n 2 r 1 n Sum of first n even natural numbers 2 4 6 ...... 2n 2r n (n 1) r 1 If for an A.P. sum of p terms is q and sum of q terms is p, then sum of (p + q) terms is {–(p + q)}. If for an A.P., sum of p terms is equal to sum of q terms, then sum of (p + q) terms is zero. 104 Progressions If the pth term of an A.P. is Example: 7 1 1 1 and qth term is , then sum of pq terms is given by S pq ( pq 1) 2 q p 7th term of an A.P. is 40, then the sum of first 13 terms is (a) 53 (b) 520 [Karnataka CET 2003] (c) 1040 (d) 2080 13 {2a 12 d } 13 {a 6 d } 13 T7 13 40 520 2 S 13 Example: 8 The first term of an A.P. is 2 and common difference is 4. The sum of its 40 terms will be [MNR 1978; MP PET (a) 3200 (b) 1600 60 Solution: (b) (c) 200 (d) 2800 n 40 [2a (n 1) d ] [2 2 (40 1)4 ] 3200 2 2 S Example: 9 The sum of the first and third term of an A.P. is 12 and the product of first and second term is 24, the first term is (a) 1 (b) 8 (c) 4 Let a d, a, a d,........ be an A.P. (a d) (a d) 12 a 6. Also, (a d) a 24 6 d First term a d 6 2 4 (b) 2r + 1 D YG U If the sum of the first 2n terms of 2, 5, 8…. is equal to the sum of the first n terms of 57, 59, 61…., then n is equal to We have, (d) 13 2n n {2 2 (2n 1)3} {2 57 (n 1)2} 6n 1 n 56 n 11 2 2 (b) 2 : 1 (c) 2 : 3 (d) 3 : 2 a 1 10 5 {{a (10 1)d } 4 {2a (5 1)d } 2a 9d 4 a 8d d 2a ,a:d=1:2 d 2 2 2 150 workers were engaged to finish a piece of work in a certain number of days. 4 workers dropped the second day, 4 more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is [Kurukshetra CEE 1996] (a) 15 Solution: (c) [IIT Screening 2001] (c) 11 Let a be the first term and d the common difference Then, Example: 13 (b) 12 If the sum of the 10 terms of an A.P. is 4 times to the sum of its 5 terms, then the ratio of first term and common difference is [Rajasthan PET 1986] (a) 1 : 2 Solution: (a) (d) 2r + 3 d {8 r 2 2} (2r 1)a d (4 r 2 1) 2 2r 1 a (2r 1)d a (2r 1)d ST Example: 12 (c) 4r + 1 (2r 1)a (a) 10 Solution: (c) S 3 r S r 1 is equal to S 2 r S 2 r 1 3r (r 1) d {2a (3r 1)d } {2a (r 1 1) d } (2r 1)a {3r(3r 1) (r 1)(r 2)} S 3 r S r 1 2 2 2 S 2 r S 2 r 1 T2r a (2r 1) d Example: 11 24 4 d2 6 If S r denotes the sum of the first r terms of an A.P., then (a) 2r – 1 Solution: (b) (d) 6 U Example: 10 [MP PET 2003] ID Solution: (c) E3 Solution: (a) (b) 20 (c) 25 (d) 30 Let the work was to be finished in x days. Work of 1 worker in a day 1 150 x Progressions 105 Now the work will be finished in (x + 8) days. Work done = Sum of the fraction of work done 1 1 1 1 150 (150 4 ) (150 8 ) ....... to (x + 8) terms 150 x 150 x 150 x 1 150 4 (x 8 1) 150 x (x 8 ){150 2(x 7)} (x 8 )(x 7) 600 0 2 150 x 150 x x 8 2 60 (x 8 )(x 7) 25 24 , x 8 25 Hence work completed in 25 days. If the sum of first p terms, first q terms and first r terms of an A.P. be x, y and z respectively, then x y z (q r) (r p) (p q) is p q r (a) 0 Solution: (a) (b) 2 (c) pqr (d) We have a, the first term and d, the common difference, x {2a ( p 1) d } z d y d a (q 1) and a (r 1) r 2 q 2 p x d a ( p 1) 2 p 2 ID Similarly, 8 xyz pqr E3 Example: 14 x y z d d d (q r) (r p) ( p q) a ( p 1) (q r) a (q 1) (r p) a (r 1) ( p q) p q r 2 2 2 d {( p 1)(q r) (q 1)(r p) (r 1)( p q)} 2 U a{(q r) (r p) ( p q)} d d [{ pq pr rq pq pr qr {(q r) (r p) ( p q)} 0 {0 0} 0 2 2 The sum of all odd numbers of two digits is [Roorkee 1993] (a) 2475 (b) 2530 (c) 4905 (d) 5049 Required sum, S 11 13 15 ....... 99 Let the number of odd terms be n, then 99 11 (n 1)2 n 45 Example: 15 Solution: (a) D YG a.. 0 S Example: 16 If sum of n terms of an A.P. is 3 n 2 5 n and Tm 164 , then m = (a) 26 (b) 27 Example: 17 The sum of n terms of the series 2n 1 ST (a) Solution: (d) 1 Sn 1 3 Example: 18 (c) 28 [Rajasthan PET 1991, 95; DCE 1999] (d) None of these Tm S m S m 1 164 (3m 2 5m) {3(m 1)2 5(m 1)} 164 3(2m 1) 5 m 27 U Solution: (b) n S 2 (a l) 45 (11 99 ) 45 55 2475 2 (b) 1 3 5 3 1 ( 3 1)( 3 1) 1 2 1 5 7 1 3 1 3 5 2n 1 ...... 1 5 7 (c) ....... is 2n 1 [UPSEAT 2002] (d) 1 ( 2n 1 1) 2 1 2n 1 2n 1 5 3 7 5 ..... 2 2 2n 1 2 n 1 2 1 1 [ 3 1 5 3 7 5 ..... ( 2n 1 2n 1 )] [ 2n 1 1] 2 2 If a1 , a 2 ,......, an1 are in A.P., then (a) 1 n 1 a1 a n 1 (b) 1 a1 a n 1 1 1 1 ..... is a1 a 2 a 2 a 3 a n a n 1 (c) n 1 a1 a n 1 [AMU 2002] (d) n a1 a n 1 106 Progressions Solution: (d) 1 1 1 1 1 1 a a a a a a 1 1 1 1 2 2 3 n n 1 S .... ...... (a2 a1 ) (a3 a2 ) (an 1 an ) a1a2 a2 a3 an an 1 As a1, a2 , a3 ,......., an , an 1 are in A.P., i.e. a2 a1 a3 a2 ......... an 1 an d (say) S 1 1 1 1 1 1 1 1 1 [a (n 1 1) d ] a1 an 1 a1 1 ...... .. a a a a a a d a a d a a d. a1. an 1 1 2 2 3 1 n 1 n 1 n 1 1 n 60 1 d S nd n d a1 an 1 a1 an 1 3.5 Arithmetic Mean. E3 (1) Definitions (i) If three quantities are in A.P. then the middle quantity is called Arithmetic mean (A.M.) between the other two. If a, A, b are in A.P., then A is called A.M. between a and b. ID (ii) If a, A1 , A 2 , A 3 ,....., A n , b are in A.P., then A1 , A 2 , A 3 ,......, A n are called n A.M.’s between a and b. (2) Insertion of arithmetic means D YG U (i) Single A.M. between a and b : If a and b are two real numbers then single A.M. between a ab and b 2 (ii) n A.M.’s between a and b : If A1 , A 2 , A 3 ,......., A n are n A.M.’s between a and b, then b a , n 1 b a An a nd a n n 1 A1 a d a A 2 a 2d a 2 b a , n 1 A3 a 3d a 3 b a , n 1 ……., Important Tips Sum of n A.M.’s between a and b is equal to n times the single A.M. between a and b. If A1 and A2 are two A.M.’s between two numbers a and b, then A1 ST U ab i.e. A1 A2 A3 .......... An n 2 1 1 (2a b), A2 (a 2b ). 3 3 Sum of m A.M.'s m . Sum of n A.M.'s n Between two numbers, n 1 If number of terms in any series is odd, then only one middle term exists which is 2 n If number of terms in any series is even then there are two middle terms, which are given by 2 th term. th th n and 1 2 term. Example: 19 After inserting n A.M.’s between 2 and 38, the sum of the resulting progression is 200. The value of n is [MP PET 2001] (a) 10 Solution: (b) (b) 8 (c) 9 There will be (n + 2) terms in the resulting A.P. 2, A1, A2 ,......, An , 38 (d) None of these Progressions 107 Sum of the progression 3 A.M.’s between 3 and 19 are (a) 7, 11, 15 (b) 4, 6, 10 common difference d Example: 21 (d) None of these 19 3 4.Therefore A1 3 d 7 , A2 3 2d 11 , A3 3 3d 15 3 1 If a, b, c, d, e, f are A.M.’s between 2 and 12, then a b c d e f is equal to (a) 14 Solution: (b) (c) 6, 10, 14 Let A1 , A2 , A3 be three A.M.’s. Then 3, A1 , A2 , A3 , 19 are in A.P. 60 Solution: (a) (b) 42 (c) 84 Since, a, b, c, d, e, f are six A.M.’s between 2 and 12 Therefore, a b c d e f 6 6 (a f ) (2 12 ) 42 2 2 3.6 Properties of A.P.. (d) None of these E3 Example: 20 n2 (2 38 ) 200 (n 2) 20 n 8 2 (1) If a1 , a 2 , a 3..... are in A.P. whose common difference is d, then for fixed non-zero number ID K R. (i) a1 K, a 2 K, a 3 K,..... will be in A.P., whose common difference will be d. (ii) Ka 1 , Ka 2 , Ka 3........ will be in A.P. with common difference = Kd. a1 a 2 a 3 , ,...... will be in A.P. with common difference = d/K. K K K U (iii) D YG (2) The sum of terms of an A.P. equidistant from the beginning and the end is constant and is equal to sum of first and last term. i.e. a1 a n a 2 a n 1 a 3 a n 2 .... (3) Any term (except the first term) of an A.P. is equal to half of the sum of terms 1 equidistant from the term i.e. an (an k an k ) , k < n. 2 (4) If number of terms of any A.P. is odd, then sum of the terms is equal to product of middle term and number of terms. U (5) If number of terms of any A.P. is even then A.M. of middle two terms is A.M. of first and last term. (6) If the number of terms of an A.P. is odd then its middle term is A.M. of first and last ST term. (7) If a1 , a 2 ,......a n and b1 , b 2 ,......b n are the two A.P.’s. Then a1 b1 , a 2 b 2 ,...... a n b n are also A.P.’s with common difference d 1 d 2 , where d 1 and d 2 are the common difference of the given A.P.’s. (8) Three numbers a, b, c are in A.P. iff 2b a c. (9) If Tn , Tn 1 and Tn 2 are three consecutive terms of an A.P., then 2Tn 1 Tn Tn 2. (10) If the terms of an A.P. are chosen at regular intervals, then they form an A.P. Example: 22 If a1, a2 , a3 ,....., a24 are in arithmetic progression and a1 a5 a10 a15 a20 a24 225 , then a1 a2 a3 ..... a23 a24 (a) 909 [MP PET 1999; AMU 1997] (b) 75 (c) 750 (d) 900 108 Progressions Solution: (d) a1 a5 a10 a15 a20 a24 225 (a1 a24 ) (a5 a20 ) (a10 a15 ) 225 3(a1 a24 ) 225 a1 a24 75 (∵ In an A.P. the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and last term) Example: 23 24 (a1 a24 ) 12 75 900 2 If a, b, c are in A.P., then (a) A.P. 1 1 1 will be in , , bc ca ab (b) G.P. [DCE 2002; MP PET 1985; Roorkee 1975] (c) H.P. 1 1 1 , , will be in A.P. bc ca ab Solution: (a) a, b, c are in A.P., Example: 24 If log 2, log( 2n 1) and log( 2n 3) are in A.P., then n = Solution: (b) (d) None of these [Dividing each term by abc] [MP PET 1998; Karnataka CET 2000] E3 (a) 5/2 60 a1 a2 ..... a24 (b) log 2 5 (c) log 3 5 (d) 3 2 As, log 2, log( 2n 1) and log( 2n 3) are in A.P. Therefore ID 2 log(2 n 1) log 2 log(2 n 3) (2 n 5)(2 n 1) 0 As 2 n cannot be negative, hence 2 n 5 0 2 n 5 or n log 2 5 Geometric progression(G.P.) U 3.7 Definition. D YG A progression is called a G.P. if the ratio of its each term to its previous term is always constant. This constant ratio is called its common ratio and it is generally denoted by r. 12 36 108 Example: The sequence 4, 12, 36, 108, ….. is a G.P., because ..... 3 , which is 4 12 36 constant. Clearly, this sequence is a G.P. with first term 4 and common ratio 3. The sequence 3 1 1 3 9 1 1 1 and common ratio , , , ,.... is a G.P. with first term 2 3 2 3 2 4 8 3 3.8 General Term of a G.P.. U (1) We know that, a, ar, ar 2 , ar 3 ,.....arn 1 is a sequence of G.P. Here, the first term is ‘a’ and the common ratio is ‘r’. ST The general term or nth term of a G.P. is Tn ar n 1 It should be noted that, r T2 T3 ...... T1 T2 (2) pth term from the end of a finite G.P. : If G.P. consists of ‘n’ terms, pth term from the end (n p 1)th term from the beginning ar n p. 1 Also, the p term from the end of a G.P. with last term l and common ratio r is l r th Important Tips If a, b, c are in G.P. b c or a b b 2 ac n 1 Progressions 109 If Tk and Tp of any G.P. are given, then formula for obtaining Tn is Tn T k 1 1 n k Tp p k Tk If a, b, c are in G.P. then b c ab b c ab a ab a or or a b ab b c b c b bc b Let the first term of a G.P be positive, then if r > 1, then it is an increasing G.P., but if r is positive and less than 1, i.e. 0< r < 1, then it is a decreasing G.P. Let the first term of a G.P. be negative, then if r > 1, then it is a decreasing G.P., but if 0< r < 1, then it is an increasing G.P. a b c 1 .... If a, b, c, d,… are in G.P., then they are also in continued proportion i.e. b c d r Example: 25 The numbers ( 2 1), 1, ( 2 1) will be in (a) A.P. Solution: (b) (b) G.P. (c) H.P. Clearly (1) ( 2 1).( 2 1) 2 Example: 26 E3 2 1, 1, 2 1 are in G.P. [AMU 1983] (d) None of these ID 60 If the pth, qth and rth term of a G.P. are a, b, c respectively, then aq r b r p c p q is equal to [Roorkee 1955, 63, 73; Pb. CET 1991, 95] Solution: (b) (b) 1 2 (c) abc U (a) 0 3 Let x , xy , xy , xy ,.... be a G.P. a xy p 1 , b xy q 1 , c xy r 1 (d) pqr D YG Now, aq r. b r p. c p q (xy p 1 )q r (xy q 1 )r p (xy r 1 )p q x (q r)(r p )( p q ). y ( p 1)(q r)(q 1)(r p )(r 1)(p q ) x 0. y p (q r) q (r p ) r( p q )(q r r p p q ) x 0. y 0 0 (xy )0 1 Example: 27 If the third term of a G.P. is 4 then the product of its first 5 terms is (a) 4 Solution: (c) 3 (b) 4 4 (c) 4 5 [IIT 1982; Rajasthan PET 1991] (d) None of these Given that ar 4 2 Then product of first 5 terms a(ar)(ar 2 )(ar 3 )(ar 4 ) a5 r10 [ar 2 ]5 4 5 If x, 2 x 2, 3 x 3 are in G.P., then the fourth term is (a) 27 (b) – 27 Given that x, 2 x 2, 3 x 3 are in G.P. U Example: 28 Solution: (d) (c) 13.5 [MNR 1980, 81] (d) – 13.5 Therefore, (2 x 2)2 x (3 x 3) x 2 5 x 4 0 (x 4 )(x 1) 0 x 1, 4 ST Now first term a = x, second term ar 2(x 1) r 3 8 2( x 1) 2(x 1) 2 (x 1)3 , then 4th term ar 3 x x x x Putting x 4 , we get 8 27 T4 (3)3 13.5 16 2 3.9 Sum of First ‘n’ Terms of a G.P.. If a be the first term, r the common ratio, then sum S n of first n terms of a G.P. is given by Sn a(1 r n ) , 1r |r|< 1 110 Progressions a(r n 1) , r 1 S n na , Sn |r|> 1 r=1 3.10 Selection of Terms in a G.P.. Number of terms 60 (1) When the product is given, the following way is adopted in selecting certain number of terms : Terms to be taken a , a, ar r 4 a a , , ar, ar 3 3 r r 5 a a , , a, ar, ar 2 2 r r ID E3 3 U (2) When the product is not given, then the following way is adopted in selection of terms Number of terms 3 a, ar, ar 2 D YG Example: 29 Terms to be taken 4 a, ar, ar 2 , ar 3 5 a, ar, ar 2 , ar 3 , ar 4 100 Let a n be the nth term of the G.P. of positive numbers. Let 100 a 2 n and a then the common ratio is (b) U (a) Solution: (a) , such that , [IIT 1992] (c) (d) Let x be the first term and y, the common ratio of the G.P. 100 a ST Then, 2n 100 a2 a4 a6 .... a200 and n 1 a 2 n 1 a1 a3 a5 ...... a199 n 1 xy xy 3 xy 5 ..... xy 199 xy x xy 2 xy 4 ..... xy 198 x Example: 30 2 n 1 n 1 n 1 1 y 200 1 (y 2 )100 xy 2 2 1y 1y 1 y 200 1 (y 2 )100 x 1 y2 1 y2 y. Thus, common ratio The sum of first two terms of a G.P. is 1 and every term of this series is twice of its previous term, then the first term will be [Rajasthan PET 1988] Progressions 111 1 4 (a) Solution: (b) (b) 1 3 (c) 2 3 (d) an 2 a n 1 We have, common ratio r = 2; 1 1 1 1r 12 3 60 Let a be the first term, then a ar 1 a(1 r) 1 a 3.11 Sum of Infinite Terms of a G.P.. (or 1 r 1) (1) When |r|< 1, a 1r E3 S (2) If r 1, then S doesn’t exist The first term of an infinite geometric progression is x and its sum is 5. Then (a) 0 x 10 Solution: (b) According to the given conditions, 5 Now, |r|< 1 i.e. 1 r 1 n lim r 1 n e is n D YG Example: 32 n r 1 (a) e + 1 n Solution: (b) x x , r being the common ratio r 1 1r 5 1 1 x 2 , 0 x 10 5 lim n n e 1 (b) e – 1 r/n r 1 lim n 1 n n e r/n r 1 lim n [IIT Screening 2004] (d) x 10 x 1 5 2 x 0 5 U 0 (c) 10 x 0 (b) 0 x 10 ID Example: 31 3 4 (c) 1 – e 2 x 0 5 i.e. [AIEEE 2004] (d) e 1 1/n 1 (e e 2 / n e 3 / n ..... e n / n ) lim [e 1 / n (e 1 / n )2 (e 1 / n )3 ..... (e 1 / n )n ] n n n 1 (e 1) 1 1 / n 1 (e 1 / n )n 1 1/n 1 e (1 e )(e 1 / n 1 1) (e 1) lim e lim e lim lim lim 1 / n n n n n n n n e n n (1 e 1 / n ) 1 e1 / n 1 e 1 / n n 1 1 h, we get h 0 n U Put ST 0 (e 1) lim Example: 33 h 0 (e 1) lim h 0 .2 3 4 .234343434.... 1 17 1 5 500 1 1 100 Example: 34 0 0 form 1 (e 1).1 e 1. eh The value of.2 3 4.234 is 232 232 (a) (b) 9990 990 Solution: (a) h eh 1 [MNR 1986; UPSEAT 2000] 0. 232 (c) 990 2 34 34 34 2 34 ...... 10 1000 100000 10 7 10 1000 1 17 100 1 232 17 116 1 5 500 99 5 99 495 990 If a, b, c are in A.P. and |a|, |b|, |c| < 1, and 232 (d) 9909 1 1 1 ....... 2 100 (100 ) 112 Progressions x 1 a a 2 .... y 1 b b 2 .... z 1 c c 2 ..... Then x, y, z shall be in (a) A.P. (b) G.P. x 1 a a 2 .... 1 1a y 1 b b 2 .... 1 1b z 1 c c 2 ..... 1 1c (d) None of these 60 Solution: (c) [Karnataka CET 1995] (c) H.P. 1 – a, 1 – b, 1 – c are in A.P. E3 Now, a, b, c are in A.P. 1 1 1 , , are in H.P. Therefore x, y, z are in H.P. 1a 1b 1c 3.12 Geometric Mean. ID (1) Definition : (i) If three quantities are in G.P., then the middle quantity is called geometric mean (G.M.) between the other two. If a, G, b are in G.P., then G is called G.M. between a and b. (ii) If a, G1 , G 2 , G 3 ,.... Gn , b are in G.P. then G1 , G 2 , G 3 ,.... G n are called n G.M.’s between a and U b. (2) Insertion of geometric means : (i) Single G.M. between a and b : If a and b are two real D YG numbers then single G.M. between a and b ab (ii) n G.M.’s between a and b : If G1 , G 2 , G 3 ,......, G n are n G.M.’s between a and b, then 1 2 3 n b n 1 b n 1 b n 1 b n 1 G1 ar a , G 2 ar 2 a , G 3 ar 3 a , ……………….., Gn ar n a a a a a Important Tips Product of n G.M.’s between a and b is equal to nth power of single geometric mean between a and b. i.e. G1 G2 G3...... Gn ( ab )n G.M. of a1 a2 a3...... an is (a1 a2 a3..... an )1 / n If G1 and G2 are two G.M.’s between two numbers a and b is G1 (a 2b)1 / 3 , G2 (ab 2 )1 / 3. The product of n geometric means between a and b n 1 If n G.M.’s inserted between a and b then r a ST U 1 is 1. a 1 3.13 Properties of G.P.. (1) If all the terms of a G.P. be multiplied or divided by the same non-zero constant, then it remains a G.P., with the same common ratio. (2) The reciprocal of the terms of a given G.P. form a G.P. with common ratio as reciprocal of the common ratio of the original G.P. (3) If each term of a G.P. with common ratio r be raised to the same power k, the resulting sequence also forms a G.P. with common ratio r k. Progressions 113 (4) In a finite G.P., the product of terms equidistant from the beginning and the end is always the same and is equal to the product of the first and last term. i.e., if a1 , a 2 , a 3 ,...... an be in G.P. Then a1 an a 2 an1 a3 an2 an an3 .......... ar. anr1 (5) If the terms of a given G.P. are chosen at regular intervals, then the new sequence so formed also forms a G.P. If a1 , a 2 , a 3 ,....., a n...... is a G.P. of non-zero, non-negative terms, then 60 (6) log a1 , log a 2 , log a 3 ,..... log an ,...... is an A.P. and vice-versa. (7) Three non-zero numbers a, b, c are in G.P. iff b 2 ac. E3 (8) Every term (except first term) of a G.P. is the square root of terms equidistant from it. i.e. Tr Tr p Tr p ; [r > p] (9) If first term of a G.P. of n terms is a and last term is l, then the product of all terms of ID the G.P. is (al)n / 2. (10) If there be n quantities in G.P. whose common ratio is r and S m denotes the sum of the Example: 35 The two geometric mean between the number 1 and 64 are (a) 1 and 64 (b) 4 and 16 [Kerala (Engg.) 2002] (c) 2 and 16 (d) 8 and 16 Let G1 and G2 are two G.M.’s between the number a 1 and b 64 D YG Solution: (b) r S n S n 1. r 1 U first m terms, then the sum of their product taken two by two is 1 1 1 1 G1 (a 2b) 3 (1.64 ) 3 4 , G2 (ab 2 ) 3 (1.64 2 ) 3 16 Example: 36 The G.M. of the numbers 3, 3 2 , 3 3...... 3 n is 2 (a) 3 n (b) 3 n 1 2 [DCE 2002] n (c) 3 2 1 2 3 .... n n n(n 1) 2n (d) 3 n 1 2 Solution: (b) G.M. of (3. 3 2. 3 3...... 3 n ) (3. 3 2. 3 3...... 3 n )1 / n (3) Example: 37 If a, b, c are in A.P. b – a, c – b and a are in G.P., then a : b : c is U (a) 1 : 2 : 3 Solution: (a) (b) 1 : 3 : 5 3 3 n 1 2 (c) 2 : 3 : 4 (d) 1 : 2 : 4 Given, a, b, c are in A.P. 2b = a + c ST b – a, c – b, a are in G.P. So (c b)2 a (b a) (b a)2 (b a) a 2b a c b b a c b a c b b 2a [∵ b a] Put in 2b = a + c, we get c = 3a. Therefore a : b : c = 1 : 2 : 3 Harmonic progression(H.P.) 3.14 Definition. A progression is called a harmonic progression (H.P.) if the reciprocals of its terms are in A.P. 114 Progressions 1 1 1 .... a a d a 2d 1 1 1 1 Example: The sequence 1, , , , ,... is a H.P., because the sequence 1, 3, 5, 7, 9, ….. is an 3 5 7 9 Standard form : A.P. 60 3.15 General Term of an H.P.. 1 1 1 , , ,.... then corresponding A.P. is a, a d , a 2d ,..... a a d a 2d Tn of A.P. is a (n 1) d If the H.P. be as E3 1 a (n 1) d Tn of H.P. is In order to solve the question on H.P., we should form the corresponding A.P. 1 3 and 8th term is then its 6th term is 5 3 (a) 1 6 Let 1 1 1 , , ,....... be an H.P. a a d a 2d (b) 4th term 3 7 (c) 1 7 U Solution: (b) The 4th term of a H.P. is 5 a 3d 3 3 5 …..(i) From (i) and (ii), d …..(ii) 2 1 , a 3 3 1 1 3 2 5 7 a 5d 3 3 U 6th term ST If the roots of a(b c) x 2 b(c a) x c (a b) 0 be equal, then a, b, c are in (a) A.P. Solution: (c) (d) 3 1 1 5 a 3d a 3d Similarly, 3 a 7 d Example: 39 [MP PET 2003] D YG Example: 38 1 1 or Tn of H.P. a (n 1) d Tn of A.P. ID Thus, General term : Tn (b) G.P. (c) H.P. [Rajasthan PET 1997] (d) None of these As the roots are equal, discriminate = 0 {b(c a)}2 4 a(b c) c(a b) 0 b 2 c 2 a 2b 2 2ab 2 c 4 a 2bc 4 a 2 c 2 4 ab 2 c 4 abc 2 0 (b 2 c 2 2ab 2 c a 2 b 2 ) 4 ac{ab bc ac} (ab bc )2 4 ac (ab bc ac) {b(a c)}2 4 abc(a c) 4 a2c 2 b 2 (a c)2 2b(a c) 2ac (2ac)2 0 [b(a c) 2ac]2 0 b 2 ac ac Thus, a, b, c are in H.P. Example: 40 If the first two terms of an H.P. be (a) 6th term 2 12 and then the largest positive term of the progression is the 5 23 (b) 7th term (c) 5th term (d) 8th term Progressions 115 For the corresponding A.P., the first two terms are Common difference The A.P. will be 30 23 23 5 and i.e. and 12 12 12 2 7 12 30 23 16 9 2 5 , , , , , ,...... 12 12 12 12 12 12 The smallest positive term is 2 , which is the 5th term. The largest positive term of the H.P. will 12 60 Solution: (c) be the 5th term. 3.16 Harmonic Mean. D YG U ID E3 (1) Definition : If three or more numbers are in H.P., then the numbers lying between the first and last are called harmonic means (H.M.’s) between them. For example 1, 1/3, 1/5, 1/7, 1/9 are in H.P. So 1/3, 1/5 and 1/7 are three H.M.’s between 1 and 1/9. Also, if a, H, b are in H.P., then H is called harmonic mean between a and b. (2) Insertion of harmonic means : 2ab (i) Single H.M. between a and b ab 1 1 1 ..... a a2 an 1 1 (ii) H, H.M. of n non-zero numbers a1 , a 2 , a 3 ,...., an is given by. H n (iii) Let a, b be two given numbers. If n numbers H 1 , H 2 ,...... H n are inserted between a and b such that the sequence a, H 1 , H 2 , H 3...... H n , b is an H.P., then H 1 , H 2 ,...... H n are called n harmonic means between a and b. Now, a, H 1 , H 2 ,...... H n , b are in H.P. 1 1 1 1 1 , , ,...... , are in A.P. a H1 H 2 Hn b U Let D be the common difference of this A.P. Then, 1 (n 2)th term Tn 2 b ab 1 1 (n 1) D D (n 1) ab b a ST Thus, if n harmonic means are inserted between two given numbers a and b, then the ab common difference of the corresponding A.P. is given by D (n 1) ab Also, 1 1 1 1 1 ab 1 nD where D 2 D ,……., D, H2 a Hn a H1 a (n 1) ab Important Tips 2 ac. ac If a, b, c are in H.P. then b If H 1 and H 2 are two H.M.’s between a and b, then H1 3 ab 3 ab and H 2 a 2b 2a b 116 Progressions 3.17 Properties of H.P.. (1) No term of H.P. can be zero. (2) If a, b, c are in H.P., then ab a . b c c 1 1 1 1 H a H b a b Example: 41 (ii) (H 2a)(H 2b) H 2 H a H b 2 H a H b The harmonic mean of the roots of the equation (5 2 )x 2 (4 3 ) x 8 2 3 0 is (a) 2 Solution: (b) (iii) (b) 4 (c) 6 Let and be the roots of the given equation a 4 3 5 2 , 82 3 5 2 (d) 8 2(8 2 3 ) 4 (4 3 ) 4 4 3 4 3 ID 82 3 2 5 2 2 Hence, required harmonic mean 4 3 [IIT 1999] E3 (i) 60 (3) If H is the H.M. between a and b, then 1 1 1 1 1 1 If a, b, c are in H.P., then the value of is b c a c a b (a) Solution: (c) 2 1 bc b 2 (b) a, b, c are in H.P. 3 2 c 2 ca D YG Example: 42 U 5 2 (c) 3 2 b 2 ab [MP PET 1998; Pb. CET 2000] (d) None of these 1 1 1 , , are in A.P. a b c 1 1 2 a c b U 2 1 1 1 1 1 1 1 1 1 2 2 1 1 2 2 1 1 3 2 3 Now, 2 ab b c a c a b b a c a b b b b a b b b a b Example: 43 If a, b, c are in H.P., then which one of the following is true 1 1 1 b a b c b ST (a) Solution: (d) (b) a, b, c are in H.P. b b a ac b ac (c) ba bc 1 b a b c [MNR 1985] (d) None of these 2 ac , option (b) is false ac 2ac c (a c) a (c a) a b c ac ac ca 1 1 a c 1 1 a c a c a c a c 2 2 , option (a) is false b a b c a c a c a c ac ac 2ac b b a b c (c a)(b a) (b c)(a c) a c b a b c a c b b a c (a c) b b a b c a (c a) c (a c) ac a c ac c a ac ac ac ac 1 b 2b 2b 2 ac 2ac b option (c) is false. Progressions 117 Arithmetico-geometric progression(A.G.P.) 3.18 nth Term of A.G.P.. a1 , a2 , a3 ,......, an ,...... is an A.P. and b1 , b2 ,......, bn ,...... is a G.P., then the sequence 60 If a1 b1 , a 2 b 2 , a 3 b 3 ,......, an b n ,..... is said to be an arithmetico-geometric sequence. Thus, the general form of an arithmetico a, (a d ) r, (a 2d ) r , (a 3d ) r ,..... 3 geometric sequence is E3 2 From the symmetry we obtain that the nth term of this sequence is [a (n 1) d ] r n 1 Also, let a, (a d ) r, (a 2d ) r 2 , (a 3d ) r 3 ,..... be an arithmetico-geometric sequence. Then, 3.19 Sum of A.G.P.. ID a (a d ) r (a 2d ) r 2 (a 3d ) r 3 ... is an arithmetico-geometric series. (1) Sum of n terms : The sum of n terms of an arithmetico-geometric sequence U a, (a d ) r, (a 2d ) r 2 , (a 3 d ) r 3 ,..... is given by D YG a (1 r n 1 ) {a (n 1) d }r n dr , when r 1 1 r 1r (1 r) 2 Sn n [2a (n 1) d ], when r 1 2 (2) Sum of infinite sequence : Let |r|< 1. Then r n , r n 1 0 as n and it can also be shown that n. r n 0 as n . So, we obtain that S n a dr , as n . 1 r (1 r)2 In other words, when |r|< 1 the sum to infinity of an arithmetico-geometric series is a dr 1 r (1 r)2 U S ST 3.20 Method for Finding Sum. This method is applicable for both sum of n terms and sum of infinite number of terms. First suppose that sum of the series is S, then multiply it by common ratio of the G.P. and subtract. In this way, we shall get a G.P., whose sum can be easily obtained. 3.21 Method of Difference. If the differences of the successive terms of a series are in A.P. or G.P., we can find nth term of the series by the following steps : Step I: Denote the nth term by T n and the sum of the series upto n terms by S n. Step II: Rewrite the given series with each term shifted by one place to the right. Step III: By subtracting the later series from the former, find T n. 118 Progressions Step IV: From T n , S n can be found by appropriate summation. Example: 44 1 3 5 7 ...... is equal to 2 22 23 (a) 3 [DCE 1999] (b) 6 (c) 9 (d) 12 3 5 7 ....... 2 22 23 1 1 3 5 S ....... 2 2 22 23 1 2 2 2 S 1 2 3 ........ (on subtractin g) 2 2 2 2 Example: 45 S S 1 1 1 1/2 1 2 2 3 .... 1 2 3. Hence S = 6 2 2 2 2 2 1 1 / 2 Sum of the series 1 2.2 3.2 2 4.2 3 ..... 100.2 99 is (a) 100. 2 Solution: (b) 100 1 (b) 99. 2 100 1 1.2 2.2 3.2 ..... 99.2 2S = 3 100.2 100 U S 1 2100 2 100.2100 1 99. 2100 D YG 1 (10 n 1 9 n 28 ) 27 1 (10 n 1 9 n 10 ) 27 (b) Tn 3(1 10 100 ....... to n terms) 3 1 n n 1 (c) [Rajasthan PET 2000] 1 (10 n 1 10 n 9 ) (d) None of these 27 10 n n 1 1 3 n 10 n 1 1 (10 n 1) 10 1 3 1 10 n 1 1 3 10 10 1 3 n 1 n 1 1 (10 n 1 9 n 10 ) 27 ST S n 1 1 (10 n 1) 3 3 U Sn The sum of n terms of the following series 1 (1 x ) (1 x x 2 ) .... will be (a) Solution: (c) 2(2 99 1) 100.2100 2 1 S 3 33 333 ...... to n terms 3 33 ....... S 0 3 30 300 ......... to n terms Tn (on subtractin g) Example: 47 (d) 100. 2100 1 The sum of the series 3 + 33 + 333 + ….. + n terms is (a) Solution: (b) 1 …..(ii) S 1 (1.2 1.2 2 1.2 3 ..... upto 99 terms) 100.2100 1 Example: 46 100 …..(i) 99 Equation (i) – Equation (ii) gives, [IIIT (Hydrabad) 2000; Kerala (Engg.) 2001] (c) 99. 2 Let S 1 2.2 3.2 2 4.2 3 .... 100.2 99 2 E3 ID Solution: (b) 60 S 1 1 xn 1 x (b) x (1 x n ) 1 x (c) n (1 x ) x (1 x n ) (d) None of these (1 x )2 S 1 (1 x ) (1 x x 2 ) ...... S 1 (1 x ) ....... 0 (1 x x 2 ..... to n terms) Tn Tn 1 xn 1 x n Sn (on subtractin g) n 1 n Tn n 1 1 xn 1 1 x 1 x n n 1 1 1 1 x n x n 1 n 1 xn 1 1 n x 1 x 1 x 1 x [IIT 1962] Progressions 119 The sum to n terms of the series 1 3 7 15 31 ....... is (b) 2 n 1 n 2 Solution: (b) (c) 2 n n 2 S 1 3 7 15 31 ...... 1 3 7 15 ....... S 0 (1 2 4 8 16 ..... to n terms) Tn (on subtractin g) Tn 1 2 4 8 ........... to n terms 1 n Sn n Tn n 1 n (2n 1) n 1 2n 1 2n 1 2 1 2n 1 n 2n n 1 (d) None of these 60 (a) 2 n 1 n [IIT 1963] 1 2 2 1 n 2 n 1 n 2 E3 Example: 48 n x (1 x n ) n (1 x ) x (1 x n ) 1 x (1 x )2 (1 x )2 n 1 Miscellaneous series 3.22 Special Series. ID There are some series in which nth term can be predicted easily just by looking at the series. Then S n n n Tn n 1 n n U If Tn n 3 n 2 n ( n 3 n 2 n ) n 1 n3 n 1 2 n2 n 1 n n n 1 n 1 n 1 Note D YG n (n 1) n (n 1)(2n 1) n (n 1) n 6 2 2 : 1 2 2 2 3 2 ....... n 2 Sum n r 2 r 1 Sum U 1 3 2 3 3 3 4 3 ....... n 3 squares of first n natural cubes n r 3 n (n 1) 2 of first n natural 2 ST 3.23 Vn Method. (1) To find the sum of the series 1 1 1 ..... a1 a 2 a 3.....ar a 2 a 3.....ar 1 an an 1.....an r 1 Let d be the common difference of A.P. Then an a1 (n 1) d. Let S n and Tn denote the sum to n terms of the series and nth term respectively. Sn 1 1 1 ..... a1 a 2.....ar a 2 a 3.....ar 1 an an 1.....an r 1 Tn 1 an an 1.....an r 1 numbers n (n 1)(2n 1) 6 of r 1 of numbers 120 Progressions 1 ; an 1 an 2.....an r 1 Vn Vn 1 Tn 1 an an 1.....an r 2 an an r 1 1 1 an 1 an 2.....an r 1 an an 1.....an r 2 an an 1.....an r 1 [a1 (n 1) d ] [a1 {(n r 1) 1}d ] d (1 r) Tn an an 1.....an r 1 1 {Vn 1 Vn } , S n d (r 1) n T n n 1 1 (V0 Vn ) d (r 1) 1 1 1 (r 1)(a 2 a1 ) a1 a 2....ar 1 an 1 an 2......an r 1 E3 Sn Vn 1 60 Let Vn Example: If a1 , a 2 ,.....a n are in A.P., then ID 1 1 1 1 1 1 ... a1 a 2 a 3 a 2 a 3 a 4 an an 1 an 2 2(a 2 a1 ) a1 a 2 an 1 an 2 (2) If S n a1 a 2.....a r a 2 a 3.....a r 1.... a n a n 1...a n r 1 Tn a n a n 1.....a n r 1 U Let Vn a n a n 1....a n r 1 a n r , Vn 1 a n 1 a n 1......a n r 1 Vn Vn 1 an an 1 an 2.....an r 1 (an r an 1 ) Tn {[a1 (n r 1) d ] [a1 (n 2) d ]} Tn (r 1) d Sn n n 1 Vn Vn 1 (r 1) d D YG Tn Tn 1 (r 1)d n (V n Vn 1 ) n 1 1 1 (Vn V0 ) {(a n a n 1....a n r ) (a 0 a1....a r )} (r 1) d (r 1) d 1 {an an 1....an r a0 a1.....ar } (r 1)(a 2 a1 ) 1 {n (n 1)(n 2)(n 3) 0.1.2.3} 5.1 1 {n (n 1)(n 2)(n 3)} 5 U Example: 1.2.3.4 2.3.4.5 ...... n (n 1)(n 2)(n 3) The sum of 13 2 3 3 3 4 3 .... 15 3 is (a) 22000 (b) 10000 [MP PET 2003] ST Example: 49 (c) 14400 2 (d) 15000 2 Solution: (c) n (n 1) 15 16 S 13 2 3 3 3 ...... 15 3 ; For n 15 , the value of 14400 2 2 Example: 50 n A series whose nth term is y , the sum of r terms will be x r (r 1) (a) ry 2x r Solution: (a) Sr n 1 Example: 51 r tn n 1 r (r 1) (b) 2x n 1 y x x r n 1 r ny r (r 1) (c) ry 2x 1 r (r 1) r (r 1) yr ry 2 2x 1 x n 1 1 If (1 2 t1 ) (2 2 t2 ) .... (n 2 tn ) n (n 2 1) , then tn is 3 [UPSEAT 1999] r (r 1) (d) rx 2x Progressions 121 (a) n 2 (c) n 1 (b) n 1 (d) n 1 n (n 2 1) (1 2 2 2 .... n 2 ) (t1 t2 ..... tn ) 3 t1 t 2 ..... tn 12 2 2 3 2 ....... n 2 tn S n S n 1 Example: 52 n (n 1) n (n 1) Sn 2 2 n (n 1) (n 1) n n 2 2 The sum of the series (a) 60 t1 t2 t3 ..... tn 1 n (n 1)(2n 1) 1 n (n 1) n (n 2 1) n (n 2 1) [2n 1 (2n 2)] 3 6 3 6 1 1 1 .... is 3 7 7 11 11 15 1 3 (b) 1 6 [MNR 1984; UPSEAT 2000] E3 Solution: (d) (c) 1 9 1 12 (d) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 S .... ....... 0 3 7 7 11 11 15 4 3 7 7 11 11 15 4 3 4 3 12 Example: 53 The sum of the series 1.2.3 + 2.3.4 + 3.4.5 + ….. to n terms is ID Solution: (d) (a) n (n 1)(n 2) (b) (n 1)(n 2)(n 3) 1 n (n 1)(n 2)(n 3) 4 (d) Tn n (n 1)(n 2) n 3 3n 2 2n Solution: (c) U (c) 1 (n 1)(n 2)(n 3) 4 n S 1.2.3 2.3.4 3.4.5 ...... to n terms (n 3 3n 2 2n) D YG [Kurukshetra CEE 1998] n 1 n n 1 n n3 3 n n2 2 n 1 n n 1 2 n (n 1)(2n 1) n (n 1) 1 n (n 1) S 2 n (n 1)[n (n 1) 2 (2n 1) 4 ] 3 6 2 4 2 1 1 n (n 1)[n 2 5 n 6] n (n 1)(n 2)(n 3) 4 4 3.24 Properties of Arithmetic, Geometric and Harmonic means between Two given Numbers. ST U Let A, G and H be arithmetic, geometric and harmonic means of two numbers a and b. ab 2ab Then, A , G ab and H 2 ab These three means possess the following properties : (1) A G H A ab 2ab , G ab and H 2 ab AG ab ( a b )2 ab 0 2 2 AG G H ab …..(i) a b 2 ab 2ab ab ( a b ) 2 0 ab ab ab ab 122 Progressions GH …..(ii) From (i) and (ii), we get A G H Note that the equality holds only when a = b AH a b 2ab ab ( ab )2 G 2 2 ab Hence, G 2 AH 60 (2) A, G, H from a G.P., i.e. G 2 AH (3) The equation having a and b as its roots is x 2 2 Ax G 2 0 ab A 2 and G ab The roots a, b are given by A A 2 G 2 ID x 2 2 Ax G 2 0 E3 The equation having a and b its roots is x 2 (a b)x ab 0 (4) If A, G, H are arithmetic, geometric and harmonic means between three given numbers 3G 3 x G3 0 H U a, b and c, then the equation having a, b, c as its roots is x 3 3 Ax 2 D YG 1 1 1 1 ab c 1/3 a b c and A , G (abc) H 3 3 a b c 3 A, abc G 3 and 3G 3 ab bc ca H The equation having a, b, c as its roots is x 3 (a b c)x 2 (ab bc ca)x abc 0 x 3 3 Ax 2 3G 3 x G3 0 H 3.25 Relation between A.P., G.P. and H.P.. ST U A when n 0 a n 1 b n 1 G when n 1 / 2 (1) If A, G, H be A.M., G.M., H.M. between a and b, then an b n H when n 1 (2) If A1 , A 2 be two A.M.’s; G1 , G 2 be two G.M.’s and H 1 , H 2 be two H.M.’s between two numbers a and b then G1 G 2 A A2 1 H1 H 2 H1 H 2 (3) Recognization of A.P., G.P., H.P. : If a, b, c are three successive terms of a sequence. Then if, If, ab a , then a, b, c are in A.P. b c a ab a , then a, b, c are in G.P. b c b Progressions 123 If, ab a , then a, b, c are in H.P. b c c (4) If number of terms of any A.P./G.P./H.P. is odd, then A.M./G.M./H.M. of first and last terms is middle term of series. 60 (5) If number of terms of any A.P./G.P./H.P. is even, then A.M./G.M./H.M. of middle two terms is A.M./G.M./H.M. of first and last terms respectively. (6) If pth, qth and rth terms of a G.P. are in G.P. Then p, q, r are in A.P. (7) If a, b, c are in A.P. as well as in G.P. then a b c. If the A.M., G.M. and H.M. between two positive numbers a and b are equal, then [Rajasthan PET 2003] (a) a = b Solution: (a) (b) ab = 1 (c) a > b A.M. = G.M. ( a )2 2 a b ( b )2 ( a b )2 ab 0 ab 0 ab 2 2 2 G.M. = H.M. ab 2ab a b 2 ab 0 ( a b )2 0 ab a b ab U (d) a < b ID Example: 54 E3 (8) If a, b, c are in A.P., then x a , x b , x c will be in G.P. (x 1) Thus A.M. =(G.M.) (H.M.) So a b Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation [AIEEE 2004] D YG Example: 55 (a) x 2 18 x 16 0 Solution: (b) (b) x 2 18 x 16 0 (c) x 2 18 x 16 0 (d) x 2 18 x 16 0 A = 9, G = 4 are respectively the A.M. and G.M. between two numbers, then the quadratic equation having its roots as the two numbers, is given by x 2 2 Ax G 2 0 i.e. x 2 18 x 16 0 If a b c , , are in H.P., then b c a (b) a 2 b, b 2 c, c 2 a are in H.P. (c) a 2 b, b 2 c, c 2 a are in G.P. (d) None of these a b c , , are in H.P. b c a ST Solution: (a) [UPSEAT 2002] (a) a 2 b, c 2 a, b 2 c are in A.P. U Example: 56 b c a b c a , , are in A.P. abc , abc , abc are in A.P. b 2 c, ac 2 , a 2 b are in A.P. a b c a b c a 2 b, c 2 a, b 2 c are in A.P. Example: 57 If a, b, c are in G.P., then log a x, log b x, log c x are in (a) A.P. Solution: (c) (b) G.P. (c) H.P. a, b, c are in G.P. log x a, log x b, log x c are in A.P. log a x, log b x, log c x are in H.P. 1 1 1 , , are in A.P. log a x log b x log c x [Rajasthan PET 2002] (d) None of these 124 Progressions Example: 58 If A1 , A2 ; G1, G2 and H1 , H 2 be two A.M.’s, G.M.’s and H.M.’s between two quantities, then the value of G1G2 is H1 H 2 [Roorkee 1983; AMU 2000] A1 A2 (a) H1 H 2 Let a and b be the two numbers b a a 2b b a 2a b A1 a , A2 a 2 3 3 3 3 H1 b 1 / 3 , G2 a a1 / 3 b 2 / 3 a 1 3 3 ab 3 ab , H2 2 1 1 1 1 1 2a b a 2b a b a b a 3 G1G2 (a 2 / 3b1 / 3 )(a1 / 3b 2 / 3 ) (a 2b)(2a b) 3 ab 3 ab H1 H 2 9 ab a 2b 2 a b H1 H 2 2a b a 2b 3ab 3ab 9 ab (a b) 3ab a 2b 2a b (a 2b)(2a b) (a 2b)(2a b) U 2 a b a 2b ab 3 3 A1 A2 (a 2b)(2a b) G1G2 H1 H 2 9 ab H1 H 2 D YG Solution: (d) b 1/3 A1 A2 Example: 59 a 2/3 ID 2 1/3 E3 b G1 a a A1 A2 (d) H1 H 2 60 Solution: (a) A1 A2 (c) H1 H 2 A1 A2 (b) H1 H 2 If the ratio of H.M. and G.M. of two quantities is 12 : 13, then the ratio of the numbers is[Rajasthan PET 1990] (a) 1 : 2 (b) 2 : 3 (c) 3 : 4 (d) None of these Let x and y be the numbers 2 xy H.M. , G.M. xy x y 2 x /y 12 2r H.M. 2 xy , ( r x 13 r 2 1 G.M. x y 1 y x ) 12 r 2 26 r 12 0 6 r 2 13 r 6 0 y U 13 13 2 4.6.6 13 5 18 8 3 2 , , 26 12 12 12 2 3 ST r Ratio of numbers Example: 60 4 x 9 : 1 9 : 4 or 4 : 9 r 2 : 1 : 1 or 9 y 4 If the A.M. of two numbers is greater than G.M. of the numbers by 2 and the ratio of the numbers is 4 : 1, then the numbers are [Rajasthan PET 1988] (a) 4, 1 Solution: (c) (b) 12, 3 Let x and y be the numbers Also, (c) 16, 4 A.M. = G.M. + 2 (d) None of these x y 2 xy 2 x 4 :1 x 4y y 4y y 5y 4 y.y 2 2 y 2 y 4 x 4 4 16 2 2 The numbers are 16, 4. Example: 61 If the ratio of A.M. between two positive real numbers a and b to their H.M. is m : n, then a : b is Progressions 125 n m n (b) m n n n m n a 1 m (a b ) m (a b) / 2 b We have, a n 4 ab n 2ab /(a b) 4 b a 2 m r2 , r (1 r 2 ) 2 m r n n r 2 b n r 2 m 4m 4n n m m n ( m m n )( m m n ) n( m m n) n n a b. Hence, m m n m m n m m n m (m n) n( m m n) n m m n. ID 3.26 Applications of Progressions. a a 1 b b n r2 2 m r n 0 n Considering +ve sign, r m m n 2 m a a 1 2 n b b m m n 2 n r2 m n 4 E3 Let (d) None of these m m n 2 2 Solution: (c) m m n (c) 60 m n n (a) Example: 62 If x log 5 3 log7 5 log 9 7 then 3 2 (b) x 1 3 D YG (a) x Solution: (c) U There are many applications of progressions is applied in science and engineering. Properties of progressions are applied to solve problems of inequality and maximum or minimum values of some expression can be found by the relation among A.M., G.M. and H.M. (c) x 2 (d) None of these 2 x log 5 3 log7 5 log 9 7 log 5 3 log 7 5 log 9 7 (log 5 3. log 7 5. log 9 7)1 / 3 3 Example: 63 3 3 x 1 (log 9 3)1 / 3 x 3(log 9 91 / 2 )1 / 3 x 3 3 2 [A.M. G.M.] 1/3. Hence x 3 3 2 If a, b, c, d are four positive numbers then U a a b c d (a) 4 e b c d e a b c d e 5 b c d e a ST (c) Solution: (a,b,c) (d) a b 1/2 b c a b ; We have 2 b c a b a 2 b c c Similarly, a a c b d (b) 4 e b d c e b c d e a 1 a b c d e 5 ( A.M. G.M.) …..(i) c d c 2 d e e …..(ii) Multiplying (i) by (ii), a a b c d 4 c b c d e c a a b c d 4 , (a) is true e e b c d e 1/2 a c b d a c Next, 2 b d c e b d 1/2 b d 2 c e a a c b d 4 , (b) is true e b d c e 126 Progressions a b c d e 1/5 a b c d e b c d e a a b c d e 5 , (c) is true b c d e a 5 b c d e a 1/5 Now, b c d e a 5 , (d) is false a b c d e Let an product of first n natural numbers. Then for all n N n (a) n n an n 1 (b) n! 2 (c) n n an 1 Solution: (a,b) We have an 1.2.3.............n n ! , nn n.n.n........ to n times 60 Example: 64 b c d e a b c d e a 5 a b c d e a b c d e (d) None of these n n n.(n 1)(n 2)....{ n (n 1)} nn n.(n 1)(n 2)......... 2.1 n n n ! n n an. So (a) is true E3 nn (n 1)! n n an 1. So (c) is false n 1 2 3 ..... n n(n 1) n 1 n 1 (n ! )1 / n. (n ! )1 / n (1.2.3...... n)1 / n n !. So (b) is true. 2 2n n 2 Example: 65 In the given square, a diagonal is drawn and parallel line segments joining points on the adjacent ID sides are drawn on both sides of the diagonal. The length of the diagonal is n 2 cm. If the distance 1 cm then the sum of the lengths of all possible line segments between consecutive line segments be 2 D YG U and the diagonal is (a) n (n 1) 2 cm Solution: (d) (b) n 2 cm (c) n (n 2) cm Let us consider the diagonal and an adjacent parallel line Length of the line PQ = RS = AC – (AR + SC) = AC – 2AR = AC – 2.PR ( AR = PR) 1 n 2 2 (n 1) 2 cm =n 2 2 2 (d) n 2 2 cm ( AR = SC) C P 45 U Length of line adjacent to PQ, other than AC, will be ((n 1) 1) 2 (n 2) 2o cm Sum of the lengths of all possible line segments and the diagonal ST 2 [n 2 (n 1) 2 (n 2) 2 .....] n 2 , 2 2 [n (n 1) (n 2) ...... 1] n 2 2 2 Example: 66 Let f (x ) (a) A 1/2 cm. S R nN n (n 1) n 2 n 2 {n 1 1} n 2 2 cm 2 1 x n 1 2 3 n 1 and g(x ) 1 2 .... (1)n n. Then the constant term in f (x ) g(x ) is equal to x x 1 x x n (n 2 1) when n is even (b) 6 n (n 1) (d) when n is odd 2 Solution: (b,c) f (x ) Q n (n 1) n when n is odd (c) (n 1) when n is even 2 2 1 x n 1 (1 x )(1 x x 2 ..... x n ) 1 x x 2 ....... x n ; f (x ) 1 2 x 3 x 2 ...... n x n 1 1 x (1 x ) n 1 2 3 f (x ).g(x ) (1 2 x 3 x 2 .... n x n 1 ) 1 2 ...... (1)n x x xn Progressions 127 constant term in f (x ) g(x ) is c 1 2 2 2 3 2 4 2 ....... n 2 (1)n 1 [1 2 2 2 3 2 4 2 ...... n 2 ] 2 [2 2 4 2 6 2 ....] when n is 2 n(n 1)(2n 1) n 1 c [1 2 2 2 ..... n 2 ] 2 [2 2 4 2 6 2 ...... (n 1)2 ] 2.2 2 [1 2 2 2 3 2 ...... 6 2 2 n (n 1)(2n 1) n 2. 2 2 1 2 2 2 .... 6 2 E3 when n is even, c [1 2 2 2 ..... n 2 ] 2 [2 2 4 2 ..... n 2 ] 60 n 1 n 1 n 1 1 2 1 n (n 1)(2n 1) 2 2 2 n (n 1)(2n 1) n (n 1)(n 1) 8 6 6 6 3 n (n 1) n (n 1) n (n 1) (2n 1 2(n 1)) 3 6 6 2 ID n n n 1 2 1 n (n 1)(2n 1) 2 2 2 n (n 1)(2n 1) 1 n (n 1)(n 2) 8 6 6 6 3 1 1 n (n 1)(2n 1 2(n 2)) n (n 1) 6 2 ST U D YG U *** odd,
