Chapter-3-PROGRESSION-(E)-01-Theory

3.1 Introduction. (1) Sequence : A sequence is a function whose domain is the set of natural numbers, N. If f : N  C is a sequence, we usually denote it by 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. (2) Series : By adding or subtracting the terms of a sequence, we get a series. If is a sequence, then the expression is a series. A series is finite or infinite as the number of terms in the corresponding sequence is finite or infinite. Example : is a series. (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 : , where denotes the nth term of the progression. Progression is mainly of three types : Arithmetic progression, Geometric progression and Harmonic progression. However, here we have classified the study of progression into five parts as : • Arithmetic progression • Geometric progression • Arithmetico-geometric progression • Harmonic progression • Miscellaneous progressions Arithmetic progression(A.P) 3.2 Definition. A sequence of numbers is said to be in arithmetic progression (A.P.) when the difference 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. If ‘a’ is the first term and ‘d’ the common difference, then an A.P. can be represented as 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. Step I: Obtain (the nth term of the sequence). Step II: Replace n by n – 1 in to get . Step III: Calculate . If is independent of n, the given sequence is an A.P. otherwise it is not an A.P. An arithmetic progression is a linear function with domain as the set of natural numbers N.  represents the nth term of an A.P. with common difference A. 3.3 General Term of an A.P.. (1) Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Then its nth term is . (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 term from the beginning. 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 and the qth term is , then its pqth term is 1.  If Tn =pn + q, then it will form an A.P. of common difference p and first term p + q. Example: 1 Let 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, and , then a – d equals (a) (b) 1 (c) (d) 0 Solution: (d)  …..(i) and  …..(ii) Subtract (ii) from (i), we get   , as m – n  0 . Therefore a – d = 0 Example: 2 The 19th term from the end of the series 2 + 6 + 10 + …. + 86 is (a) 6 (b) 18 (c) 14 (d) 10 Solution: (c)  19th term from end Example: 3 In a certain A.P., 5 times the 5th term is equal to 8 times the 8th term, then its 13th term is (a) 0 (b) – 1 (c) – 12 (d) – 13 Solution: (a) We have Let a and d be the first term and common difference respectively    , i.e. . Hence 13th term = 0 Example: 4 If 7th and 13th term of an A.P. be 34 and 64 respectively, then its 18th term is (a) 87 (b) 88 (c) 89 (d) 90 Solution: (c) Let a be the first term and d be the common difference of the given A.P., then  …..(i)  …..(ii) From (i) and (ii), d = 5, a = 4  Trick:    Example: 5 If is an arithmetic sequence, then equals (a) 1 (b) –1 (c) 0 (d) None of these Solution: (c) Let a be the first term and d the common difference. Then Example: 6 The nth term of the series 3 + 10 + 17 + ….. and 63 + 65 + 67 + …… are equal, then the value of n is (a) 11 (b) 12 (c) 13 (d) 15 Solution: (c) nth term of 1st series nth term of 2nd series  we have,  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 Terms to be taken 3 a – d, a, a + d 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, , in case we have to take 2r terms in 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 3 4 5 Sum of n terms of an A.P. : The sum of n terms of the series is given by Also, , where l = last term = Important Tips  The common difference of an A.P is given by where is the sum of first two terms and is the sum of first term or the first term.  The sum of infinite terms .  If sum of n terms is given then general term , where is sum of (n – 1) terms of A.P.  Sum of n terms of an A.P. is of the form i.e. a quadratic expression in n, in such case, common difference is twice the coefficient of i.e. 2A.  • If for the different A.P’s , then • If for two A.P.’s then  Some standard results • Sum of first n natural numbers • Sum of first n odd natural numbers • Sum of first n even natural numbers  • 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. • If the pth term of an A.P. is and qth term is , then sum of pq terms is given by Example: 7 7th term of an A.P. is 40, then the sum of first 13 terms is (a) 53 (b) 520 (c) 1040 (d) 2080 Solution: (b) Example: 8 The first term of an A.P. is 2 and common difference is 4. The sum of its 40 terms will be (a) 3200 (b) 1600 (c) 200 (d) 2800 Solution: (a) 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 (d) 6 Solution: (c) Let be an A.P.  . Also,    First term Example: 10 If denotes the sum of the first r terms of an A.P., then is equal to (a) 2r – 1 (b) 2r + 1 (c) 4r + 1 (d) 2r + 3 Solution: (b) Example: 11 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 (a) 10 (b) 12 (c) 11 (d) 13 Solution: (c) We have,   Example: 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 (a) 1 : 2 (b) 2 : 1 (c) 2 : 3 (d) 3 : 2 Solution: (a) Let a be the first term and d the common difference Then,    ,  a : d = 1 : 2 Example: 13 150 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 (a) 15 (b) 20 (c) 25 (d) 30 Solution: (c) Let the work was to be finished in x days.  Work of 1 worker in a day Now the work will be finished in (x + 8) days.  Work done = Sum of the fraction of work done to (x + 8) terms     ,  Hence work completed in 25 days. Example: 14 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 is (a) 0 (b) 2 (c) pqr (d) Solution: (a) We have a, the first term and d, the common difference,  Similarly, and  Example: 15 The sum of all odd numbers of two digits is (a) 2475 (b) 2530 (c) 4905 (d) 5049 Solution: (a) Required sum, Let the number of odd terms be n, then   Example: 16 If sum of n terms of an A.P. is and , then m = (a) 26 (b) 27 (c) 28 (d) None of these Solution: (b)    Example: 17 The sum of n terms of the series is (a) (b) (c) (d) Solution: (d) Example: 18 If are in A.P., then is (a) (b) (c) (d) Solution: (d) As are in A.P., i.e. (say)  3.5 Arithmetic Mean. (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. (ii) If are in A.P., then are called n A.M.’s between a and b. (2) Insertion of arithmetic means (i) Single A.M. between a and b : If a and b are two real numbers then single A.M. between a and b (ii) n A.M.’s between a and b : If are n A.M.’s between a and b, then , , , ……., 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. i.e.  If and are two A.M.’s between two numbers a and b, then .  Between two numbers, .  If number of terms in any series is odd, then only one middle term exists which is term.  If number of terms in any series is even then there are two middle terms, which are given by and 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 (a) 10 (b) 8 (c) 9 (d) None of these Solution: (b) There will be (n + 2) terms in the resulting A.P. Sum of the progression   Example: 20 3 A.M.’s between 3 and 19 are (a) 7, 11, 15 (b) 4, 6, 10 (c) 6, 10, 14 (d) None of these Solution: (a) Let be three A.M.’s. Then are in A.P.  common difference .Therefore , , Example: 21 If a, b, c, d, e, f are A.M.’s between 2 and 12, then is equal to (a) 14 (b) 42 (c) 84 (d) None of these Solution: (b) Since, a, b, c, d, e, f are six A.M.’s between 2 and 12 Therefore, 3.6 Properties of A.P.. (1) If are in A.P. whose common difference is d, then for fixed non-zero number K  R. (i) will be in A.P., whose common difference will be d. (ii) will be in A.P. with common difference = Kd. (iii) will be in A.P. with common difference = d/K. (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. (3) Any term (except the first term) of an A.P. is equal to half of the sum of terms equidistant from the term i.e. , k < n. (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. (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 term. (7) If and are the two A.P.’s. Then are also A.P.’s with common difference , where and are the common difference of the given A.P.’s. (8) Three numbers a, b, c are in A.P. iff . (9) If and are three consecutive terms of an A.P., then . (10) If the terms of an A.P. are chosen at regular intervals, then they form an A.P. Example: 22 If are in arithmetic progression and , then (a) 909 (b) 75 (c) 750 (d) 900 Solution: (d)    (∵ 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 If a, b, c are in A.P., then will be in (a) A.P. (b) G.P. (c) H.P. (d) None of these Solution: (a) a, b, c are in A.P.,  will be in A.P. [Dividing each term by abc] Example: 24 If log 2, and are in A.P., then n = (a) 5/2 (b) (c) (d) Solution: (b) As, log 2, and are in A.P. Therefore  As cannot be negative, hence  or Geometric progression(G.P.) 3.7 Definition. 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. Example: The sequence 4, 12, 36, 108, ….. is a G.P., because , which is constant. Clearly, this sequence is a G.P. with first term 4 and common ratio 3. The sequence is a G.P. with first term and common ratio 3.8 General Term of a G.P.. (1) We know that, is a sequence of G.P. Here, the first term is ‘a’ and the common ratio is ‘r’. The general term or nth term of a G.P. is It should be noted that, (2) pth term from the end of a finite G.P. : If G.P. consists of ‘n’ terms, pth term from the end term from the beginning . Also, the pth term from the end of a G.P. with last term l and common ratio r is Important Tips  If a, b, c are in G.P.  or  If Tk and Tp of any G.P. are given, then formula for obtaining Tn is  If a, b, c are in G.P. then   or or  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.  If a, b, c, d,… are in G.P., then they are also in continued proportion i.e. Example: 25 The numbers will be in (a) A.P. (b) G.P. (c) H.P. (d) None of these Solution: (b) Clearly  are in G.P. Example: 26 If the pth, qth and rth term of a G.P. are a, b, c respectively, then is equal to (a) 0 (b) 1 (c) abc (d) pqr Solution: (b) Let be a G.P.  Now, Example: 27 If the third term of a G.P. is 4 then the product of its first 5 terms is (a) (b) (c) (d) None of these Solution: (c) Given that Then product of first 5 terms Example: 28 If are in G.P., then the fourth term is (a) 27 (b) – 27 (c) 13.5 (d) – 13.5 Solution: (d) Given that are in G.P. Therefore,    Now first term a = x, second term  , then 4th term Putting , we get 3.9 Sum of First ‘n’ Terms of a G.P.. If a be the first term, r the common ratio, then sum of first n terms of a G.P. is given by , |r|< 1 , |r|> 1 , r = 1 3.10 Selection of Terms in a G.P.. (1) When the product is given, the following way is adopted in selecting certain number of terms : Number of terms Terms to be taken 3 4 5 (2) When the product is not given, then the following way is adopted in selection of terms Number of terms Terms to be taken 3 4 5 Example: 29 Let be the nth term of the G.P. of positive numbers. Let and , such that   , then the common ratio is (a) (b) (c) (d) Solution: (a) Let x be the first term and y, the common ratio of the G.P. Then, and   . Thus, common ratio Example: 30 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 (a) (b) (c) (d) Solution: (b) We have, common ratio r = 2; Let a be the first term, then   3.11 Sum of Infinite Terms of a G.P.. (1) When |r|< 1, (or (2) If r  1, then doesn’t exist Example: 31 The first term of an infinite geometric progression is x and its sum is 5. Then (a) (b) (c) (d) Solution: (b) According to the given conditions, , r being the common ratio  Now, |r|< 1 i.e.    i.e. ,  Example: 32 is (a) e + 1 (b) e – 1 (c) 1 – e (d) e Solution: (b) Put we get . Example: 33 The value of .234 is (a) (b) (c) (d) Solution: (a) Example: 34 If a, b, c are in A.P. and |a|, |b|, |c| < 1, and Then x, y, z shall be in (a) A.P. (b) G.P. (c) H.P. (d) None of these Solution: (c) Now, a, b, c are in A.P.  1 – a, 1 – b, 1 – c are in A.P.  are in H.P. Therefore x, y, z are in H.P. 3.12 Geometric Mean. (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 are in G.P. then are called n G.M.’s between a and b. (2) Insertion of geometric means : (i) Single G.M. between a and b : If a and b are two real numbers then single G.M. between a and b (ii) n G.M.’s between a and b : If are n G.M.’s between a and b, then , , , ……………….., 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.  G.M. of is  If and are two G.M.’s between two numbers a and b is .  The product of n geometric means between a and is 1.  If n G.M.’s inserted between a and b then 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 . (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 be in G.P. Then (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. (6) If is a G.P. of non-zero, non-negative terms, then is an A.P. and vice-versa. (7) Three non-zero numbers a, b, c are in G.P. iff . (8) Every term (except first term) of a G.P. is the square root of terms equidistant from it. i.e. ; [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 the G.P. is . (10) If there be n quantities in G.P. whose common ratio is r and denotes the sum of the first m terms, then the sum of their product taken two by two is . Example: 35 The two geometric mean between the number 1 and 64 are (a) 1 and 64 (b) 4 and 16 (c) 2 and 16 (d) 8 and 16 Solution: (b) Let and are two G.M.’s between the number and , Example: 36 The G.M. of the numbers is (a) (b) (c) (d) Solution: (b) G.M. of 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 (a) 1 : 2 : 3 (b) 1 : 3 : 5 (c) 2 : 3 : 4 (d) 1 : 2 : 4 Solution: (a) Given, a, b, c are in A.P.  2b = a + c b – a, c – b, a are in G.P. So   [∵ 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. Standard form : Example: The sequence is a H.P., because the sequence 1, 3, 5, 7, 9, ….. is an A.P. 3.15 General Term of an H.P.. If the H.P. be as then corresponding A.P. is of A.P. is  of H.P. is In order to solve the question on H.P., we should form the corresponding A.P. Thus, General term : or Example: 38 The 4th term of a H.P. is and 8th term is then its 6th term is (a) (b) (c) (d) Solution: (b) Let be an H.P. 4th term   …..(i) Similarly, …..(ii) From (i) and (ii), ,  6th term Example: 39 If the roots of be equal, then a, b, c are in (a) A.P. (b) G.P. (c) H.P. (d) None of these Solution: (c) As the roots are equal, discriminate = 0         Thus, a, b, c are in H.P. Example: 40 If the first two terms of an H.P. be and then the largest positive term of the progression is the (a) 6th term (b) 7th term (c) 5th term (d) 8th term Solution: (c) For the corresponding A.P., the first two terms are and i.e. and Common difference  The A.P. will be The smallest positive term is , which is the 5th term.  The largest positive term of the H.P. will be the 5th term. 3.16 Harmonic Mean. (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 : (i) Single H.M. between a and b (ii) H, H.M. of n non-zero numbers is given by . (iii) Let a, b be two given numbers. If n numbers are inserted between a and b such that the sequence is an H.P., then are called n harmonic means between a and b. Now, are in H.P.  are in A.P. Let D be the common difference of this A.P. Then,  Thus, if n harmonic means are inserted between two given numbers a and b, then the common difference of the corresponding A.P. is given by Also, , ,……., where Important Tips  If a, b, c are in H.P. then .  If and are two H.M.’s between a and b, then and 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 . (3) If H is the H.M. between a and b, then (i) (ii) (iii) Example: 41 The harmonic mean of the roots of the equation is (a) 2 (b) 4 (c) 6 (d) 8 Solution: (b) Let  and  be the roots of the given equation  , Hence, required harmonic mean Example: 42 If a, b, c are in H.P., then the value of is (a) (b) (c) (d) None of these Solution: (c) a, b, c are in H.P.  are in A.P.  Now, Example: 43 If a, b, c are in H.P., then which one of the following is true (a) (b) (c) (d) None of these Solution: (d) a, b, c are in H.P.  ,  option (b) is false   ,  option (a) is false  option (c) is false. Arithmetico-geometric progression(A.G.P.) 3.18 nth Term of A.G.P.. If is an A.P. and is a G.P., then the sequence is said to be an arithmetico-geometric sequence. Thus, the general form of an arithmetico geometric sequence is From the symmetry we obtain that the nth term of this sequence is Also, let be an arithmetico-geometric sequence. Then, is an arithmetico-geometric series. 3.19 Sum of A.G.P.. (1) Sum of n terms : The sum of n terms of an arithmetico-geometric sequence is given by (2) Sum of infinite sequence : Let |r|< 1. Then as n   and it can also be shown that as n  . So, we obtain that , as n  . In other words, when |r|< 1 the sum to infinity of an arithmetico-geometric series is 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 and the sum of the series upto n terms by . 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 . Step IV: From , can be found by appropriate summation. Example: 44 is equal to (a) 3 (b) 6 (c) 9 (d) 12 Solution: (b)   . Hence S = 6 Example: 45 Sum of the series is (a) (b) (c) (d) Solution: (b) Let …..(i) 2S = …..(ii) Equation (i) – Equation (ii) gives,  Example: 46 The sum of the series 3 + 33 + 333 + ….. + n terms is (a) (b) (c) (d) None of these Solution: (b)  Example: 47 The sum of n terms of the following series will be (a) (b) (c) (d) None of these Solution: (c)  Example: 48 The sum to n terms of the series is (a) (b) (c) (d) None of these Solution: (b)  Miscellaneous series 3.22 Special Series. There are some series in which nth term can be predicted easily just by looking at the series. If Then Note :  Sum of squares of first n natural numbers  Sum of cubes of first n natural numbers 3.23 Vn Method. (1) To find the sum of the series Let d be the common difference of A.P. Then . Let and denote the sum to n terms of the series and nth term respectively.  Let ;   ,  Example: If are in A.P., then (2) If Let ,    Example: Example: 49 The sum of is (a) 22000 (b) 10000 (c) 14400 (d) 15000 Solution: (c) ; For , the value of Example: 50 A series whose nth term is , the sum of r terms will be (a) (b) (c) (d) Solution: (a) Example: 51 If , then is (a) (b) (c) (d) n Solution: (d)    Example: 52 The sum of the series is (a) (b) (c) (d) Solution: (d) Example: 53 The sum of the series 1.2.3 + 2.3.4 + 3.4.5 + ….. to n terms is (a) (b) (c) (d) Solution: (c)  3.24 Properties of Arithmetic, Geometric and Harmonic means between Two given Numbers. Let A, G and H be arithmetic, geometric and harmonic means of two numbers a and b. Then, and These three means possess the following properties : (1) and   …..(i)  …..(ii) From (i) and (ii), we get Note that the equality holds only when a = b (2) A, G, H from a G.P., i.e. Hence, (3) The equation having a and b as its roots is The equation having a and b its roots is  The roots a, b are given by (4) If A, G, H are arithmetic, geometric and harmonic means between three given numbers a, b and c, then the equation having a, b, c as its roots is and  and The equation having a, b, c as its roots is  3.25 Relation between A.P., G.P. and H.P.. (1) If A, G, H be A.M., G.M., H.M. between a and b, then (2) If be two A.M.’s; be two G.M.’s and be two H.M.’s between two numbers a and b then (3) Recognization of A.P., G.P., H.P. : If a, b, c are three successive terms of a sequence. Then if, , then a, b, c are in A.P. If, , then a, b, c are in G.P. If, , then a, b, c are in H.P. (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. (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 . (8) If a, b, c are in A.P., then will be in G.P. Example: 54 If the A.M., G.M. and H.M. between two positive numbers a and b are equal, then (a) a = b (b) ab = 1 (c) a > b (d) a < b Solution: (a) A.M. = G.M.     G.M. = H.M.      Thus A.M. =(G.M.) (H.M.) So Example: 55 Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation (a) (b) (c) (d) Solution: (b) 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 i.e. Example: 56 If are in H.P., then (a) are in A.P. (b) are in H.P. (c) are in G.P. (d) None of these Solution: (a) are in H.P.  are in A.P.  are in A.P.  are in A.P.  are in A.P. Example: 57 If a, b, c are in G.P., then are in (a) A.P. (b) G.P. (c) H.P. (d) None of these Solution: (c) a, b, c are in G.P.  are in A.P.  are in A.P.  are in H.P. Example: 58 If ; and be two A.M.’s, G.M.’s and H.M.’s between two quantities, then the value of is (a) (b) (c) (d) Solution: (a) Let a and b be the two numbers  , , ,   Example: 59 If the ratio of H.M. and G.M. of two quantities is 12 : 13, then the ratio of the numbers is (a) 1 : 2 (b) 2 : 3 (c) 3 : 4 (d) None of these Solution: (d) Let x and y be the numbers  H.M. , G.M.   , )     Ratio of numbers or or 4 : 9 Example: 60 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 (a) 4, 1 (b) 12, 3 (c) 16, 4 (d) None of these Solution: (c) Let x and y be the numbers  A.M. = G.M. + 2  Also,       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 (a) (b) (c) (d) None of these Solution: (c) We have,    Let ,     Considering +ve sign,  . Hence, . 3.26 Applications of Progressions. 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. Example: 62 If then (a) (b) (c) (d) None of these Solution: (c) [A.M.  G.M.]    . Hence Example: 63 If a, b, c, d are four positive numbers then (a) (b) (c) (d) Solution: (a,b,c) We have ; ( A.M.  G.M.)  …..(i) Similarly, …..(ii) Multiplying (i) by (ii),  ,  (a) is true Next,  ,  (b) is true  ,  (c) is true Now,  ,  (d) is false Example: 64 Let product of first n natural numbers. Then for all n  N (a) (b) (c) (d) None of these Solution: (a,b) We have , to n times     . So (a) is true  . So (c) is false   .  . So (b) is true. Example: 65 In the given square, a diagonal is drawn and parallel line segments joining points on the adjacent sides are drawn on both sides of the diagonal. The length of the diagonal is cm. If the distance between consecutive line segments be then the sum of the lengths of all possible line segments and the diagonal is (a) (b) (c) (d) Solution: (d) Let us consider the diagonal and an adjacent parallel line Length of the line PQ = RS = AC – (AR + SC) = AC – 2AR ( AR = SC) = AC ¬¬– 2.PR ( AR = PR) = Length of line adjacent to PQ, other than AC, will be  Sum of the lengths of all possible line segments and the diagonal , n  N Example: 66 Let and . Then the constant term in is equal to (a) when n is even (b) when n is odd (c) when n is even (d) when n is odd Solution: (b,c) ;  constant term in is when n is odd, when n is even,