Sequence & Series

Sequence & Series Sequence A succession of numbers a1, a2 , a3 , ..., an formed Ex.1 Write down the sequence whose nth term is according to some definite rule is called a sequence. 3  ( 1)n 2n A sequence is a function whose domain is the set N (i) (ii) 3n n of natural numbers and range a subset of real numbers or complex numbers. 2n Sol. (i) Let tn = n The different terms of a sequence are usually denoted put n = 1, 2, 3, 4, .............. we get by a1,a2 ,a3 ,.... or by t1,t 2,t 3 ,.... The subscript (always a natural number) denotes the position of 8 t1 = 2, t2 = 2, t3 = 3 , t4 = 4 the term in the sequence. The term at the nth place 8 of a sequence. i.e., tn is called the general term of so the sequence is 2, 2, , 4, ........ 3 the sequence. Real sequence : A sequence whose range is a subset of R is called a real sequence. e.g. (i) 2, 5, 8, 11, ....................... (ii) 4, 1, – 2, – 5, ...................... (iii) 3, –9, 27, – 81, ........................ (ii) Let tn = 3  ( 1)n 3n put n = 1, 2, 3, 4, ...... 2 4 2 4 so the sequence is 3 , 9 , , ,............ 27 81 Series : By adding or subtracting the terms of a sequence, Types of sequence : we get an expression which is called a series. On the basis of the number of terms there are two If a1, a2, a3,........an is a sequence, then the expression types of sequence. a1 + a2 + a3 + ...... + an is a series. (i) Finite sequences : A sequence is said to be finite if it has finite number of terms. e.g. (i) 1 + 2 + 3 + 4 + .................... + n (ii) Infinite sequences : A sequence is said to be infinite (ii) 2 + 4 + 8 + 16 + ................. if it has infinitely many terms. (iii) – 1 + 3 – 9 + 27 – .............. 35 Sequence & Series Note : For any sequence {tn}, whose sum of first r terms Progression is Sr, rth term, tr = Sr – Sr – 1. The sequence which obey the definite rule and its general term is always expressible in terms of n, is called progression. Ex.4 Find the sum of all natural numbers divisible by Arithmetic progression (A.P.) : 5, but less than 100. A.P. is a sequence whose successive terms are obtained by adding a fixed number 'd' to the Sol. All those numbers are 5, 10, 15, 20, ..........., 95. preceding terms. This fixed number 'd' is called the Here a = 5, n = 19 &  = 95 common difference. If a is the first term & d the 19 common difference, then A.P. can be written as so S = (5 + 95) = 950. 2 a, a + d, a + 2 d,....... , a + (n  1) d,........ Ex.5 Find the sum of all the three digit natural numbers e.g. – 4, – 1, 2, 5 ........... which on division by 7 leaves remainder 3. nth term of an A.P. : Sol. All these numbers are 101, 108, 115, ........, 997 Let 'a' be the first term and 'd' be the common 997 = 101 + (n – 1) 7 difference of an A.P., then  n = 129 tn = a + (n – 1) d, where d = tn – tn – 1 so S = 129 [101 + 997] = 70821. 2 Ex.6 The sum of n terms of two A.Ps. are in ratio Ex.2 If t54 of an A.P. is – 61 and t4 = 64, find t10. 7n  1 Sol. Let a be the first term and d be the common difference . Find the ratio of their 11 1th terms. 4n  27 so t54 = a + 53d = – 61 .........(i) Sol. Let a1 and a2 be the first terms and d1 and d2 be the and t4 = a + 3d = 64 .........(ii) common differences of two A.P.s respectively, then equation (i) – (ii) we get n [2a1  (n  1)d1 ]  50d = – 125 7n  1 2 = n 5 4n  27 [2a 2  (n  1)d2 ] d=– 2 2  a= 143 2 so t10 =  5 143 + 9   2  = 49   2 Ex.3 Find the number of terms in the sequence 4, 12, 20, ........,108. Sol. a = 4, d = 8 so 108 = 4 + (n – 1)8  n = 14 , The sum of first n terms of an A.P. : If a is first term and d is common difference, then sum of the first n terms of AP is n Sn = [2a + (n – 1) d] 2 n = [a + ]  nt  n1  , for n is odd. (Where  is the 2    2  last term and t  n 1  is the middle term.)    2   n  1 a1    d1 7n  1  2   =  n  1 4n  27 a2    d2  2  For ratio of 11th terms n 1 = 10 2  n = 21 7(21)  1 so ratio of 11th terms is = 4(21)  27 = 4 148 = 3 111 Ex.7 If sum of n terms of a sequence is given by Sn = 2n2 + 3n, find its 50th term. Sol. Let tn is nth term of the sequence so tn = Sn – Sn – 1. = 2n2 + 3n – 2(n – 1)2 – 3(n – 1) = 4n + 1 so t50 = 201. 36 Sequence & Series Useful Formulae: Sum of natural numbers (  n) 2 Properties of A.P.: (a) If tn = an + b, then the series so formed is an A.P. (b) If Sn = an2 + bn + c, then series so formed is an A.P. where, n  N. (c) If every term of an A.P. is increased or decreased Remarks : by the same quantity, the resulting terms will also be (i) The first term and common difference can be zero, in A.P. positive or negative (or any complex number.) (d) If every term of an A.P. is multiplied or divided by (ii) If a, b, c are in A.P.  2 b = a + c & if a, b, c, d are the same non-zero quantity, the resulting terms will in A.P.  a + d = b + c. also be in A.P. (iii) Three numbers in A.P. can be taken as a  d, a, a +d ; four numbers in A.P. can be taken as a  3d, a  d, (e) If terms a1, a2,....., an, an + 1,....., a2n + 1 are in A.P. Then sum of these terms will be equal to (2n + 1)an+1. a + d, a + 3d; Here total number of terms in the series is (2n + 1) five numbers in A.P. are a  2d, a  d, a, a+d, a +2d ; and middle term is an+1. six terms in A.P. are a  5d, a  3d, a  d, a + d, a + 3d, a + 5d etc. (f) In an A.P. the sum of terms equidistant from the (iv) Middle term: If the number of terms is n, and beginning and end is constant and equal to sum of * * n is odd, then first and last terms. term is the middle term. n is even, then and terms. (g) Sum and difference of corresponding terms of two A.P.’s will form a series in A.P. terms are middle (h) If terms a1, a2, ..., a2n – 1, a2n are in A.P. The sum of these terms will be equal to (2n) Ex.8 The sum of three numbers in A.P. is 27 and the sum of their squares is 293, find them. Sol. Let the numbers be a – d, a, a + d so 3a = 27  a=9 Also (a – d)2 + a2 + (a + d)2 = 293. 3a2 + 2d2 = 293 d2 = 25  d=±5 therefore numbers are 4, 9, 14. Ex.9 If a1, a2, a3, a4, a5 are in A.P. with common 5 difference  0, then find the value of 5 Hence a i = 10. (i) nth term of a series is an = Sn – Sn – 1 (n  2) 1 1 1 Ex.10 If b  c , c  a , a  b are in A.P., prove that a2, b2, c2 are also in A.P. 1 1 1 Sol.  b  c , c  a , a  b are in A.P..  1 c a  bc c a c  aa b = (a  b)(c  a) (c  a)(b  c)  ba bc i when a3 = 2. Sol. As a1, a2, a3, a4, a5 are in A.P., we have a1 + a5 = a2 + a4 = 2a3. = A.M. of middle terms. 2 a , i1 , where 1 1 1 – bc = ab – ca c b  = ab b2 – a2 = c2 – b2  a2, b2, c2 are in A.P. i1 37 bca a Ex.11 If c ab b , 1 1 , abc c are in A.P., ., 1 then prove that a , b , c are also in A.P.. Sol. Given bca a , c ab b , abc c are in A.P.. Add 2 to each term a b c are in A.P.. c 1 1 1 divide each by a + b + c  , , are in A.P.. a b c  bca a , c ab b Sequence & Series Ex.14 If the sum of the first n terms of a sequence is of the form where A, B are constants independent of n, then the sequence is (1) an A.P. (2) a G.P. (3) an H.P. (4) None of these Sol. [1] We have, , Ex.12 If a, b, c in A.P. and    n 0 n0 n0 x   an ,y   bn ,z   c n then x,y,z are in (1) AP (2) GP (3) HP (4) None of these Sol. [3] Here a, b, c in A.P., given Also x  1 1 1 ,y  ,z  1 a 1 b 1 c Now a, b, c in AP  1  a,1  b,1  c in A.P.. 1 1 1  , , 1 a 1 b 1 c 1 1 1 in HP  x,y,z in HP Now = 2A (a constant) Hence the sequence is an A.P. Ex.15 If the sum of the first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k is equal to (1) 1 n (2) (3) n 2n (4) n Sol. [4] Let S1 denotes the sum of the first n even natural are the pth, qth, rth terms respectively numbers then ab p  q  bc q  r  ca r  p Ex.13 If a , b , c of an A.P. equals to (1) 1 (2) 1 (3) 0 (4) None of these Sol. [3] Let x be the first term and y be the c.d. of corresponding A.P., then 1  x  p 1 y a 1  x  q 1 y b 1  x  r 1 y c  Let S 2 denotes the sum of the first n odd natural numbers ..... (1)  ..... (2) ..... (3) Multiplying (1), (2) and (3) respectively by abc q  r ,abc r  p ,abc p  q and then adding we get bc q  r  ca r  p  ab p  q  0 ......... (2) Dividing (1) by (2), S1 Hence 38 Sequence & Series Ex.17 Insert 20 A.M. between 2 and 86. Arithmetic mean (mean or average) (A.M.) : Sol. Here 2 is the first term and 86 is the 22nd term of If three terms are in A.P. then the middle term is A.P. so 86 = 2 + (21)d  d=4 called the A.M. between the other two, so if a, b, c so the series is 2, 6, 10, 14,......., 82, 86 are in A.P., b is A.M. of a & c.  required means are 6, 10, 14,...,82. A.M. for any n numbers a1, a2,..., an is; Ex.18 If m arithmetic means are inserted between 1 and 31 so that the ratio of the 7th and (m–1)th means is A= . 5:9, then the value of m is n - Arithmetic means between two numbers : (1) 9 (2) 11 (3) 13 (4) 14 If a, b are any two given numbers & a, A1, A2,...., An, b are in A.P., then A1, A2,... An are the Sol. [4] Let the means be x1,x 2 ,..., xm so that 1,x1,x 2 ,....,x m ,31 is an A.P. of Now, n A.M.’s between a & b. A1 = a +  , A2 = a +  terms.   Given: ,......, An = a + x7   30 Note : Sum of n A.M.’s inserted between a & b is equal to n times the single A.M. between a & b i.e.  Ar = nA, where A is the single A.M. between  a&b i.e. A= Geometric Progression (G.P.) The sequence {an} in which a1  0 is termed a geometric progression if there is a number r 0 ab 2 such that Ex.16 Between two numbers whose sum is 13 , an even 6 number of A.M.s is inserted, the sum of these means exceeds their number by unity. Find the number of means. Sol. Let a and b be two numbers and 2n A.M.s are inserted between a and b, then 2n (a + b) = 2n + 1. 2  13    6  n = 2n + 1. 13   given a  b  6    1. an = r for all n, then r is called common a ratio. Useful Formulae: If a = first term, r = common ratio and n is the number of term, then (a) nth term denoted by tn is given by tn = arn–1 (b) Sum of first n terms denoted by Sn is given by Sn = or corresponding to r <  1 (or) r > 1, (or) Sn = where  is the last term in the series. (c) Sum of infinite terms S a  n = 6. S  Number of means = 12. Note : If r 1 then S (For |r| <1) . 39 Sequence & Series (iv) If each term of a G.P. be multiplied or divided or raised to power by the same nonzero quantity, the resulting sequence is also a G.P.. Ex19. If the first term of G.P. is 7, its nth term is 448 and sum of first n terms is 889, then find the fifth term of (v) If a1, a2, a3,........ and b1, b2, b3,......... are two G.P’s with common ratio r1 and r2 respectively, then the G.P. sequence a1 b1, a 2b2, a 3b3 , ..... is also a G.P. Sol. Given a = 7 with common ratio r1 r2. tn = arn – 1 = 7(r)n – 1= 448. (vi) If a1, a2, a3,..........are in G.P. where each ai > 0, then log a1, loga2, loga3,..........are in A.P. and its converse  7rn = 448 r is also true. a(r n  1) 7(r n  1) Also Sn = = r 1  889 = r 1 448r  7 r 1  r=2 Hence T5 = ar4 = 7(2)4 = 112. Ex.22 Find three numbers in G.P. having sum 19 and product 216. Ex.20 The first term of an infinite G.P. is 1 and any term Sol. Let the three numbers be is equal to the sum of all the succeeding terms. Find 1  the series. so a  r  1  r  = 19   Sol. Let the G.P. be 1, r, r2, r3, ......... and a3 = 216 2 1 r  r= , 2 1 r 1 1 1 Hence series is 1, , , 8 , .............. 2 4 1 1 1 Ex.21 Let S = 1 + + + + .........., find the sum of 8 2 4 given condition  r = (i) first 20 terms of the series (ii) infinite terms of the series.   1  20  1      2  2 20  1   Sol. (i) S20 = = . 1 219 1 2 (ii) S = 1 1 1 2 = 2. Remarks : (i) If a, b, c are in G.P.  b2 = ac, in general if a1, a2, a3, a4,......... an – 1 , an are in G.P., then a1an = a2an – 1 = a3 an – 2 = .......................... (ii) Any three consecutive terms of a G.P. can be taken as a , a , ar.. r (iii) Any four consecutive terms of a G.P. can be taken a a as 3 , , ar, ar3. r r a , a, ar r .......(i) a=6  2 so from (i) 6r – 13r + 6 = 0.  r= 3 2 , 2 3 Hence the three numbers are 4, 6, 9. Ex.23 Find the product of 11 terms in G.P. whose 6th term is 5. Sol. Using the property a1a11 = a2a10 = a3a9 = .............. = a62 = 25 Hence product of terms = 511 Ex.24 Using G.P. express 0. 3 and 1.2 3 as p form. q Sol. Let x = 0. 3 = 0.3333 ............. = 0.3 + 0.03 + 0.003 + 0.0003 + ............. = 3 10 + 3 100 + 3 1000 + 3 10000 + .............. 3 3 1 10 = 1 = 9 = 3. 1 10 Let y = 1.2 3 = 1.233333 = 1.2 + 0.03 + 0.003 + 0.0003 + ............. = 1.2 + 3 10 2 + 3 10 3 + 3 10 4 + ............ 3 = 1.2 + 10 2 1 1 10 1 37 = 1.2 + 30 = 30 . 40 Ex.25 Evaluate 7 + 77 + 777 + ........... upto n terms. Sol. Let S = 7 + 77 + 777 + ..........upto n terms. = 7 9 [9 + 99 + 999 + .......] = 7 9 [(10 – 1) + (102 – 1) + (103 – 1) = 2 3 n [10 + 10 + 10 + ...........+ 10 – n] 7 81 (3) logb a  logc b (4) None of these Sol. [3] We have, a x  b y  c z   say log  log  log  ,y  ,z  loga logb logc x putting x, y, z in (i), we get [10n + 1 – 9n – 10] 2  log   log  log  .    loga logc  logb  Geometric mean (G) logb (i) G = ab where a, b are two positive numbers. (ii) G = (a1 a2 ...... an)1/n is geometric mean of n positive numbers a1, a2, a3,...... an. * (2) logc b  loga c  xloga  ylogb  zlogc  log  n 7  10(10  1)  n  =  9 9   = (1) logb a  loga c x, y, z are in G.P.  y2  xz + ........ + upto n terms] 7 9 Sequence & Series Ex.27 If x, y, z are in G.P. and a  b y  c z then x n GM’s between two given numbers If in between two numbers ‘a’ and ‘b’, we have to insert n GM G1 , G2...... Gn then a, G1, G2 ..... Gn , b will be in GP. The series consist of (n+2) terms and the last term is b and first term is a. arn+2–1 = b 2 or  loga.logc loga b  logb c  logb a  logc b Ex.28 If a, b, c, d are in G.P., then 1 a3  b3 , b3  c 3 1 , c 3  d3 r= (1) A.P. (2) G.P. (3) H.P. (4) None of these Sol. [2] 1 1 1 1  ,  a3  b3 a3 1  r 3 b3  c 3 a3 r 3 1  r 3 G1 = ar , G2 = ar2 ..... Gn = arn or Gn = b/r Note : Product of n GM’s inserted between ‘a’ and ‘b’ is equal to nth power of the single GM between ‘a’ and ‘b’ and 1 1  c 3  d3 a3r 3 1  r 3 Clearly, a3  b3 1 = (G)n where G = ab Ex.29 If r > 1 1 , b3  c 3 G.P. with common ration and c 3  d3 1 are in 1 r3 and , and Ex.26 Insert 4 G.M.s between 2 and 486. = (243) = 3 xy z ab (1) c Hence four G.M.s are 6, 18, 54, 162. (3) 1  b  n 1 Sol. Common ratio of the series is given by r =   a 1/5 are in Let b  ar,c  ar 2 and d  ar 3 . Then, 1 i.e. 1 , then bc a (2) ac b (4) None of these 41 Sequence & Series (d) If terms are given in H.P. then the terms could be picked up in the following way Sol. [1] Since * 1 For three terms 1 Similarly, y and 1 1 For four terms 1 1 2 1 1 1 ......... (1) z 1 * For five terms 1 2 1 1 1 1 ......... (2) Dividing (2) by (1), we get Note : In general, if we are to take (2r + 1) terms in H.P. we take them as Harmonic Progression (H.P.) 1 1 1 1 1 1 A sequence is said to be a harmonic progression, if and only if the reciprocal of its terms form an (e) Harmonical Mean (H.M.) arithmetic progression. If three or more than three terms are in H.P, then all 1 1 1 For example: , , ........ form a H.P. because 2, the numbers lying between first and last term are 2 4 6 called Harmonical Means between them. i.e. 4, 6,........... are in A.P. The H.M. between two given quantities a and b is H 1 1 1 so that a, H , b are in H.P. If a, b, c are in H.P., then a , b , c forms an A.P.. i.e. 1. Some Useful Formulae & Properties: (a) If a, b are first two terms of an H.P. then 1 tn = * (b) There is no formula for sum of a H.P. (c) Harmonic mean H of any two numbers a and b is given by H= 2 = 2ab where a, b are two non–zero 1 1 1 , , a H b are in A.P.. n H.M’s between two given numbers To find n HM’s between a and b we first find n AM’s between 1/a and 1/b then their reciprocals will be required HM’s Ex.30 If pth, qth and rth terms of H.P. are u, v, w respectively, then the value of the expression q  r vw  r  p wu  p  q uv is numbers. (1) 1 (2) 0 (3) 2 (4) 1 Sol. [2] Also H = the harmonic mean of n non–zero numbers a1, a2, a3,....., an. Let H.P. be 1 1 1    ..... a a  d a  2d u  1 1 1 ,v  ,w  a  p1 d a  q 1 d a  r 1 d 42 1 1 1  a  p 1 d  , a  q 1 d  , a  r 1 d  u v w  q r a  p 1 d  r  p a  q 1 d  .....  1 1 q  r  r  p  ..... u v  q  r vw  .....  0 Sequence & Series Ex.33 If m term of H.P. is n, while nth term is m, find its (m + n)th term. th 1 Sol. Given Tm = n or a  (m  1) d = n; where a is the first term and d is the common difference of the corresponding A.P. 1 n so a + (m – 1)d = Ex.31 If H1,H2 ,H3 ,.....Hn be n harmonic means between and a + (n – 1) d = H1  a Hn  b a and b then H  a  H  b  1 n (1) 0 (2) n or d = (3) 2n (4) 1 1 mn Hence Sol. [3] ab n  1 ab n  1 Here H1  b n  1  b  a  bn  a = 1 m  Similarly Hn  H1  a an  b (interchange a and b) Hn  b Hence H  a  H  b 1 n  2n  1 b  a a= mn . mn ba ab 2nb  b  a  2na  a  b   2n ba  a2  a3  a 2 a3 a2  a3 Ex.32 If a a  a  a  3  a  a  then a1,a2 ,a3 ,a4 1 4 1 4  1 4  are in (1) A.P. (2) G.P. (3) H.P. (4) None of these Sol. [3] a1  a 4 a 2  a3 1 1 1 1     , so a1a4 a2a3 a 4 a1 a3 a2 or 1 1 1 1    a4 a3 a2 a1 Also 3 a2  a3  1 a 2 a3 1 ...... (1)  a1  a 4 ; a1a4 1 1 so 3  a  a   a  a  3 2  4 1 ...... (2) 1 1 1 1 1 1 Clearly, (1) and (2)  a  a  a  a  a  a ; 2 1 3 2 4 3 1 1 1 so a , a , a are in A.P.. 1 2 3 2 Ex.34 Insert 4 H.M between 3 and 13 . Sol. Let 'd' be the common difference of corresponding A.P.. 13 3  so d = 2 2 = 1. 5 1 3 5  H = +1= 2 2 1 2n  1 a  b  mn mn 1 (m  1) 1 – = n mn mn 1 mn T(m + n) = = a  (m  n  1) d 1 m  n  1 so 2 ab n  1 (m – n)d = 1 3 7 H2 = 2 + 2 = 2 1 3 9 H3 = 2 + 3 = 2 1 3 11 H4 = 2 + 4 = 2 2 or H1 = 5 or H2 = or or 2 7 2 H3 = 9 H4 = 2 . 11 Ex.35 If pth , q th, rth terms of an H.P. be a, b, c respectively, prove that (q – r)bc + (r – p) ac + (p – q) ab = 0 Sol. Let 'x' be the first term and 'd' be the common difference of the corresponding A.P.. 1 so a = x + (p – 1)d 1 b 1 c ...........(i) = x + (q – 1) d ..........(ii) = x + (r – 1) d ..........(iii) (i) - (ii)  ab(p – q)d = b – a ..........(iv) (ii) - (iii)  bc (q – r)d = c – b ..........(v) (iii) - (i)  ac (r – p) d = a – c .........(vi) (iv) + (v) + (vi) gives bc (q – r) + ac(r – p) + ab (p – q) = 0. 43 Sequence & Series Ex.38 If x,y,z are positive, then prove that (x + y + z) Relation between means : (i) If A, G, H are respectively A.M., G.M., H.M. between a & b both being positive, then G² = AH (i.e. A, G, H are in G.P.) and A  G  H. Ex.36 The A.M. of two numbers exceeds the G.M. by 6 3 and the G.M. exceeds the H.M. by 5 ; find the 2 numbers. Sol. Let the numbers be a and b, now using the relation G2 = AH  1 1 1      9 x y z Sol. Using the relation A.M.  H.M. 3 xyz  1 1 1 3   x y z  1 1 1  (x + y + z)  x  y  z   9   n Ex.39 If ai > 0  i  N such that a i  1 , then prove i 1 = 3  6  G   G   2  5   G=6 = G2 + 3 10 G– that (1 + a1) (1 + a2) (1 + a3) ........(1 + an)  2n 9 5 Sol. Using A.M.  G.M. 1 + a1  2 a1 i.e.ab = 36 also 1 + a2  2 a 2 a + b = 15  Hence the two numbers are 3 and 12. 1 + an  2 an A.M.  G.M.  H.M.  Let a1, a2, a3, .......an be n positive real numbers, then we define their a  a  a  .......  an A.M. = 1 2 3 , their n n 1 1   .......  a1 a 2 an . 1/ 2 As a1 a2 a3 ..... an = 1 Hence (1 + a1) (1 + a2) .......... (1 + an)  2n. 1  2  22  .......  2n1 > (1.2. 22. 23. ......... .2n–1)1/n n It can be shown that A.M.  G.M.  H.M. and equality holds at either places iff a1 = a2 = a3 = ..............= an a b c Ex.37 If a, b, c > 0, prove that b + c + a  3 Sol. Using the relation A.M.  G.M. we have 1 a b c   3 b c a   a . b . c  b c a 3  2n a1a 2a 3 ......a n Ex.40 If n > 0, prove that 2n > 1 + n 2n 1 Sol. Using the relation A.M.  G.M. on the numbers 1, 2, 22, 23,..........., 2n–1 , we have G.M. = (a1 a2 a3 .........an)1/n and their H.M. = 1 (1 + a1) (1 + a2) .........(1 + an)  a b c   3 b c a Equality does not hold as all the numbers are not equal. 1 n  (n 1) n  n 2 2       2 1 >n 2 1  2n – 1 > n. 2 (n 1) 2  2n > 1 + n. 2 ( n 1) 2 Ex.41 Find the greatest value of xyz for positive value of x, y, z subject to the condition xy + yz + zx = 12. Sol. Using the relation A.M.  G.M. xy  yz  zx  (x2 y2 z2)1/3  3  4  (x y z)2/3 xyz  8 44 Sequence & Series Ex. 42 If a, b, c are in H.P. and they are distinct and positive, then prove that an + cn > 2bn Arithmetic-Geometric Series The series whose each term is formed by multiplying Sol. Let an and cn be two numbers corresponding terms of an A.P. and G.P. is called an  c n > (an cn)1/2 2 then the Arithmetic–geometric series. an + cn > 2 (ac)n/2 ...........(i) For example – Also G.M. > H.M. i.e. ac > b, (ac) n/2 n >b * 1 + 2x + 4x2 + 6x3 + ....................... ...........(ii) * a + (a + d)r + (a + 2d)r2 + ............... hence from (i) and (ii), we get an + cn > 2bn Ex.43 If a, b and c are distinct positive real numbers and a 2  b2  c2  1 , then ab  bc  ca is 1. Summation of n terms of Arithmetic–Geometric Series : (1) less than 1 (2) equal to 1 Let S = a + (a + d)r + (a + 2d)r2 + ............. (3) greater than 1 (4) any real number (i) tn = [a + (n - 1)d].rn –1 Sol. [1] 2 Since a and b are unequal, (ii) Sn = a + (a + d)r + (a + 2d)r2 + ............. + 2 a b  a 2b 2 2 [a + (n – 1)d]rn – 1 (A.M. > G.M. for unequal numbers) Multiply by ‘r’ and rewrite the series in following  a2  b2  2ab way. Similarly b2  c 2  2bc and c 2  a2  2ca rSn = ar + (a + d)r2 + (a + 2d)r3 + ..............+ Hence 2 a2  b2  c 2  2 ab  bc  ca [a + (n – 2)d]rn – 1 + [a + (n – 1)d]rn  ab  bc  ca  1 On subtraction, Ex.44 a, b, c are three positive numbers and abc has 2 the greatest value 1 2 (1) a  b  ,c  1 64 . Then 1 4 [a + (n – 1)d]rn 1 4 (2) a  b  ,c  1 (3) a  b  c  3 1 2 (4) None of these or, Sn = 2. c c ab  2 2  4 a.b. c . c 4 2 2  44 4 abc 2  4 or ; or a b  c  4 4 abc 2 4 1 abc  abc 64 2 a + – .r n Summation of Infinite Terms Series : S = a + (a + d)r + (a + 2d)r2 + ......................  rS = a r + (a + d)r2 + ............................. 4  On subtraction the greatest value of Also for the greatest value of abc2 the numbers have to be equal, i.e., a  b  – [a + (n – 1)d].rn or, Sn(1 – r) = a + Sol. [2] a b c Sn(1 – r) = a + d (r + r2 + ................. + rn – 1) – c 2 Also, given the greatest value  S (1 – r) = a + d (r + r2 + r3 + ..................) S= 1 . 64 a + dr So a + b + c = 1 45 Sequence & Series Ex.47 Evaluate 1 + 2x + 3x + 4x + ......... upto infinity, where | x | < 1. 2 Ex.45 If r th term of a series is 2r  1 2r , then sum of its infinite terms is (1) 10 (2) 8 (3) 5 Sol. Let S = 1 + 2x + 3x2 + 4x3 + ..... .......(i) x + 2x2 + 3x3 + ....... ......(ii) xS = (4) 0 (i) - (ii)  (1 – x) S = 1 + x + x2 + x3 + .......... Sol. [3] Here Tr  2r  1 2 or Obviously the series in the bracket is ArithmeticGeometrical Series. Therefore by the formula S  7 4 1 5 S= 3n  2 5n rS = r + (1 + b) r2 + ......... 10 7 + 5 + 2 + 4 5 10 2 5 + + ...... + 3 7 5 3 4 5 3n  2 (1 – r)S = 1 + br + b2r2 + b3r3 + ......  S = (1  br )(1  r ) 1 5 + ....... + n 1 (1) 99.2100 (2) 100.2100 (3) 1  99.2100 (4) none of these Let S  1  2.2  3.22  4.23  ....  100.299 ....(i) 3n  5 5n1 .... (1)  2S  2  2.22  3.23  ....  99.299  100.2100 ... (2) Subtracting (1) from (2) we get + S  1  2  2 2  23  .....  299  100.2100 ..........(ii)  S  100.2100  2100  1 2 1  100.2100  2100  1 3 3 3 3 S = 1 + 5 + 2 + 3 + ....... + n1 – 5 5 5 S =1+ 3   1  1   5   5  1 1 5 3n  2 5    12n  7 7 – 4 4.5 n  n  1  99.2100 . Results : n 1  – 3n  2 (i) 5n 35  (ar ± br) =  ar ±  br.  (ii) 1 3n  2 3 3 = 1 + – × n1 – 5 5n 4 4 =  Sol. [3] (i) – (ii)  4 5 ......(ii) Ex.49 1  2.2  3.22  4.23  .....  100.299 equals Ex.46 Find the sum of the series 1 + 5 + 2 + 3 5 5 + ...... to n terms.  1   5 Sol. Let S = 1 + (1 + b)r + (1 + b + b2) r2 +... ....(i) 2   1  2    We have S  1  3   2  2   5 2 1  1  1  2  1      2   Sol. Let S = 1 + S (i) - (ii) a r  1  r 1 r 4 5 1 = (1  x )2 Ex.48 Evaluate : 1 + (1 + b) r + (1 + b + b2) r2 + ......... to infinite terms for | br | < 1. r 1 5 7  3   2  .....  2 2 2   Series is: 3 (12n  7) (iii)   k ar = k  ar.  S = 16 – 16 . 5n1 .    k = k + k + k +...............n times = nk;  where k is a constant. 46 (iv)   r = 1 + 2 + 3 +...........+ n =  (v)  r² = 12 + 22 + 32 +...........+ n2  = (vi)   r3 = 13 + 23 + 33 +...........+ n3 =  Sequence & Series Method of difference for finding nth term : Let u1, u2, u3 ........ be a sequence, such that u2 – u1, u3 – u2, ......... is either an A.P. or a G.P. then nth term un of this sequence is obtained as follows S = u1 + u2 + u3 + ........... + un ................(i) S = u1 + u2 + ........... + un–1 + un ................(ii) (i) – (ii)  un = u1 + (u2 – u1) + (u3 – u2) + ........... + (un – un–1) Where the series (u2 – u1) + (u3 – u2) + .......... + (un – un–1) is either in A.P. or in G.P. then we can find un. n Ex.50 Find the sum of the series to n terms whose general term is 2n + 1. Sol. Sn = Tn = (2n + 1) = 2n + 1 2(n  1) n + n = n2 + 2n 2 = n 2 k Ex.51 Tk = k + 2 , then find n Sol.  = k 1 k . k 1 n n Tk T  k2 + k 1 2 k k 1 2(2n  1) 2 1 = n (n  1) ( 2n  1) 6 + = n (n  1) ( 2n  1) 6 + 2n + 1 – 2. n i j Ex.52 Find the value of the expression  1 i  1 j  1k  1 n Sol. i n j  1 =  j u r r 1 Note : The above method can be generalised as follows : Let u1, u2, u3, ......... be a given sequence. The first differences are 1u1, 1u2, 1u3, ........ where 1u1 = u2 – u1, 1u2 = u3 – u2 etc. The second differences are 2u1, 2u2, 2u3, ......., where 2u1 = 1u2 – 1u1, 2u2 = 1u3 – 1u2 etc. This process is continued untill the kth differences ku1 , k u2 , ........ are obtained, where the kth differences are all equal or they form a GP with common ratio different from 1. Case - 1 : The kth differences are all equal. In this case the nth term, un is given by un = a0nk + a1nk–1 + .....+ ak , where a0, a1, ...., ak are calculated by using first 'k + 1' terms of the sequence. Case - 2 : The kth differences are in GP with common ratio r (r  1) The nth term is given by un =  rn – 1 + a0 nk–1 + a1 nk–2 + ..... + ak–1 i1 j1 i  1 j  1k  1 n i (i  1) 2 i 1 i So sum of series S = n 1  i2  2 i  1  =  = 1  n ( n  1) ( 2n  1)  n (n  1)   6 2  2  =   i i 1   n  n (n  1) (n  2) n (n  1) = [2n + 1 + 3] = . 6 12 Ex.53 Find the sum to n-terms 3 + 7 + 13 + 21 + ......... Sol. Let S = 3 + 7 + 13 + 21 + ......... + Tn ...........(i) S = 3 + 7 + 13 + ............ + Tn–1 + Tn ...........(ii) (i) – (ii)  Tn = 3 + 4 + 6 + 8 + .......... + (Tn – Tn–1) =3+ n 1 [8 + (n – 2)2] 2 = 3 + (n – 1) (n + 2) = n2 + n + 1 Hence S =  (n2 + n + 1) = n2 + n + 1 = n(n  1)( 2n  1) 6 + n n(n  1) + n = (n2 + 3n + 5) 3 2 47 Sequence & Series Ex.54 Find the sum to n-terms 1 + 4 + 10 + 22 + ........ Sol. Let S = 1 + 4 + 10 + 22 + .........+ Tn ........(i) S= 1 + 4 + 10 + ......... + Tn–1 + Tn ........(ii) (i) – (ii)  Tn = 1 + (3 + 6 + 12 + ......... + Tn – Tn–1 ) n 1   Putting n  10 S10   2 1 Tn = 1 + 3  2  1     Tn = 3 . 2n–1 – 2 S =  Tn = 3  2n–1 – 2 So  1  n n  1 2n  1 n n  1   4n 3 2 6 2   2n  1  = 3 .  2  1  – 2n = 3.2n – 2n – 3   1  10  11 21 10  11    40  2  2 2  1 1140 1155  55  40   570 2 2 Ex.56 Find the sum of n-terms of the series 1.2 + 2.3 + 3.4 + ............ Sol. Let Tr be the general term of the series So Tr Difference Method = r(r + 1). then To express tr = f(r) – f(r–1) multiply and divide tr by [(r + 2) – (r – 1)] (i) If (T2 – T1), (T3 – T2)............... (Tn – Tn – 1) are so Tr Let T1, T2, T3 ...... Tn are the terms of sequence, in A.P. then, the sum of the such series may be obtained by using summation formulae in nth term, = r 3 (r + 1) [(r + 2) – (r – 1)] = 1 3 [r (r + 1) (r + 2) – (r – 1) r (r + 1)]. 1 3 r (r + 1) (r + 2) (ii) If (T2 – T1), (T3 – T2) ................ (Tn – Tn – 1) are Let f(r) = found in G.P. then the sum of the such series may be so Tr = [f(r) – f(r – 1)]. n obtained by using summation formulae of a G.P. Now S = T r r 1 = T1 + T2 + T3 + .........+ Tn 1 Ex.55 Sum of the series 3  7  14  24  37  .....10 terms , is (1) 560 (2) 570 (3) 580 (4) None of these T1 = 3 [1 . 2 . 3 – 0] T2 = Sol. [2] 1 3 [2 . 3 . 4 – 1 . 2 . 3] 1 Here the given series is not A.P., G.P., or H.P. Let S  3  7  14  24  37  ....  Tn T3 = 3 [3 . 4 . 5 – 2 . 3 . 4]  S= Tn = after subtracting 1 3 [n(n+1) (n + 2) – (n – 1)n (n + 1)]  A.P.  Tn  3   Sn  n 1 2 1 2 2 4  n  2 3   3n  n  4 2 1 3n2  n  4n  2  1 S = 3 n (n + 1) (n + 2) Hence sum of series is f(n) – f(0). 48 1 Ex.57 Sum to n terms of the series (1  x )(1  2 x) 1 1 + (1  2 x )(1  3 x) + (1  3 x)(1  4x ) + ......... Sol. Let Tr be the general term of the series Tr = 1 (1  rx )(1  (r  1)x ) =  1 1  1    x 1  rx 1  (r  1) x  Tr = f(r) – f(r + 1)  S =  Tr = T1 + T2 + T3 + .......... + Tn =  1 1  1    x 1  x 1  (n  1)x  n = (1  x )[1  (n  1)x] 4 5 Ex.58 Sum to n terms of the series 1 . 2 . 3 + 2 . 3 . 4 6 + 3 . 4 . 5 + ......... r 3 = r(r  1)(r  2) Sol. Let Tr Tn = 2 + 10 + 24 + 44 + 70 + 102 + ....... + (Tn–1–Tn–2) + (Tn – Tn–1) ...........(iv) (iii) – (iv)  Tn – Tn–1 = 2 + 8 + 14 + 20 + 26 + ......... = 1  (1  (r  1) x )  (1  rx )    x  (1  rx )(1  (r  1)x )  So Tr = Sequence & Series (i) – (ii)  Tn = 2 + 10 + 24 + 44 + 70 + 102 + ............... + (Tn – Tn–1) ...........(iii) n [4 + (n – 1) 6] = n [3n – 1]  Tn – Tn–1 = 3n2 – n 2 general term of given series is  (Tn – Tn–1) =  (3n2 – n) = n3 + n2.  Hence sum of this series is S =  n3 +  n2 n(n  1)(2n  1) n2 (n  1)2 + 6 4 n (n  1) = (3n2 + 7n + 2) 12 1 = n (n + 1) (n + 2) (3n + 1) 12 = Ex.60 Find the general term and sum of n terms of the series 9 + 16 + 29 + 54 + 103 + ......... Sol. Let S = 9 + 16 + 29 + 54 + 103 + ................. + Tn ...........(i) S= 9 + 16 + 29 + 54 + 103 + ......... + Tn–1 +Tn ...........(ii) = (r  1)(r  2) + r(r  1)(r  2) (i) – (ii)  Tn = 9 + 7 + 13 + 25 + 49 + ................. + (Tn – Tn–1) ...........(iii) 1   1 = r  1  r  2   Tn = 9 + 7 + 13 + 25 + 49 + ........... + (Tn–1–Tn–2) + (Tn – Tn–1) ...........(iv) 1  = S= 3  1 3  1  +   r ( r  1 ) ( r  1 )( r  2 ) 2   1  1 2  n  2    1 3 1  +   2  2 (n  1)(n  2)   3  1 5 1 5 – – 2(n  1)(n  2) 1   = 4 n  2  2(n  1)  4 [2n + 5] Ex.59 Find the nth term and the sum of n term of the series 2 + 12 + 36 + 80 + 150 + 252 + .......... Sol. Let S = 2 + 12 + 36 + 80 + 150 + 252 + ................+Tn ...........(i) S = 2 + 12 + 36 + 80 + 150 + 252 + .........+ Tn–1 + Tn ...........(ii) (iii) – (iv)  T n – T n–1 = 9 + (–2) + n–2 n–2 6 12  24  ........     = 7 + 6 [2 – 1] = 6(2) + 1. (n  2 ) terms  General term is Tn = 6(2)n–1 + n + 2 Also sum S = Tn = 62n–1 + n + 2 =6. n (n  1) (2n  1) + + 2n 2 21 = 6(2n – 1) + n(n  5) 2 49