Sequence and Series PDF

Sequence and Series "1729 is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."...... S.Ramanujan Sequence : A sequence is a function whose domain is the set N of natural numbers. Since the domain for every sequence...

Sequence and Series "1729 is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."...... S.Ramanujan Sequence : A sequence is a function whose domain is the set N of natural numbers. Since the domain for every sequence is the set N of natural numbers, therefore a sequence is represented by its range. If f : N R, then f(n) = tn, n  N is called a sequence and is denoted by {f(1), f(2), f(3),...............} = {t1, t2, t3,...............} = {tn} 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,...................... Types of sequence : On the basis of the number of terms there are two types of sequence. (i) Finite sequences : A sequence is said to be finite if it has finite number of terms. (ii) Infinite sequences : A sequence is said to be infinite if it has infinitely many terms. Example # 1 : Write down the sequence whose n term is th  2 n  1n  2 Solution : Let tn =  2 n  1n  2 put n = 1, 2, 3, 4,.............. we get 4 16 t1 = –2, t2 = , t3 = –8, t4 = 3 3 16 so the sequence is –2, , –8, ,........ 3 Series : By adding or subtracting the terms of a sequence, we get an expression which is called a series. If a1, a2, a3,........an is a sequence, then the expression a1 + a2 + a3 +...... + an is a series. e.g. (i) 1 + 2 + 3 + 4 +............... + n (ii) 2 + 4 + 8 + 16 +............... (iii) – 1 + 3 – 9 + 27 –............... Progression : The word progression refers to sequence or series – finite or infinite Arithmetic progression (A.P.) : A.P. is a sequence whose successive terms are obtained by adding a fixed number 'd' to the preceding terms. This fixed number 'd' is called the common difference. If a is the first term & d the common difference, then A.P. can be written as a, a + d, a + 2 d,....... ,a + (n  1) d,........ e.g. – 4, – 1, 2, 5........... nth term of an A.P. : Let 'a' be the first term and 'd' be the common difference of an A.P., then tn = a + (n – 1) d, where d = tn – tn – 1 Example # 2 : Find the number of terms in the sequence 4, 7, 10, 13,........,82. Solution : Let a be the first term and d be the common difference a = 4, d = 3 so 82 = 4 + (n – 1)3  n = 27 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 1 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series 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 last term and t  n 1 is the middle term.) 2    2    2   Note : For any sequence {tn}, whose sum of first r terms is Sr, rth term, tr = Sr – Sr – 1. Example # 3 : If in an A.P., 3rd term is 18 and 7 term is 30, then find sum of its first 17 terms Solution : Let a be the first term and d be the common difference a + 2d = 18 a + 6d = 30 d = 3 , a = 12 17 s17 = [2 × 12 + 16 × 3] = 612 2 Example # 4 : Find the sum of all odd numbers between 1 and 1000 which are divisible by 3 Solution : Odd numbers between 1 and 1000 are 3, 5, 7, 9, 11, 13, ------ 993, 995, 997, 999. Those numbers which are divisible by 3 are 3, 9, 15, 21, ------- 993, 999 They form an A.P. of which a = 3 , d = 6,  = 999  n = 167 n S= [a + ] = 83667 2 Example # 5 : The ratio between the sum of n term of two A.P.’s is 3n + 8 : 7n + 15. Then find the ratio between their 12 th term Sn (n / 2)[2a  (n – 1)d] 3n  8 a  {(n – 1) / 2}d 3n  8 Solution :  = or  ----- (i) Sn ' (n / 2)[2a' (n – 1)d'] 7n  15 a' (n – 1) / 2d' 7n  15 T12 a  11d we have to find  T12 ' a' 11d' choosing (n – 1)/2 = 11 or n = 23 in (1), T a  11d 3(23)  8 77  we get 12  =   T12 ' a' 11d' (23)  7  15 176 16 Example # 6 : If sum of n terms of a sequence is given by Sn = 3n2 – 4n, find its 50th term. Solution : Let tn is nth term of the sequence so tn = Sn – Sn – 1. = 3n2 – 4n – 3(n – 1)2 + 4(n – 1) = 6n – 7 so t50 = 293. Self practice problems : (1) Which term of the sequence 2005, 2000, 1995, 1990, 1985,............. is the first negative term (2) For an A.P. show that tm + t2n + m = 2 tm + n (3) Find the maximum sum of the A.P. 40 + 38 + 36 + 34 + 32 +.............. (4) Find the sum of first 16 terms of an A.P. a1, a2, a3.......... If it is known that a1 + a4 + a7 + a10 + a13 + a16 = 147 Ans. (1) 403 (3) 420 (4) 392 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 2 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series Remarks : (i) The first term and common difference can be zero, positive or negative (or any complex number.) (ii) If a, b, c are in A.P.  2 b = a + c & if a, b, c, d are in A.P.  a + d = b + c. (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, a + d, a + 3d; five numbers in A.P. are a  2d, a  d, a, a + d, a + 2d ; six terms in A.P. are a  5d, a  3d, a  d, a + d, a + 3d, a + 5d etc. (iv) The sum of the terms of an A.P. equidistant from the beginning & end is constant and equal to the sum of first & last terms. (v) Any term of an A.P. (except the first) is equal to half the sum of terms which are equidistant from it. an = 1/2 (ank + an+k), k < n. For k = 1, an = (1/2) (an1+ an+1); For k = 2, an = (1/2) (an2+ an+2) and so on. (vi) If each term of an A.P. is increased, decreased, multiplied or divided by the same non-zero number, then the resulting sequence is also an AP. (vii) The sum and difference of two AP's is an AP. t2 Example # 7 : The numbers t (t2 + 1),  and 6 are three consecutive terms of an A.P. If t be real, then find 2 the the next two term of A.P. Solution : 2b = a + c   –t2 = t3 + t + 6 or t +t +t+6=0 3 2 or (t + 2) (t2 – t + 3) = 0  t2 – t + 3  0  t = –2 the given numbers are – 10, – 2, 6 which are in an A.P. with d = 8. The next two numbers are 14, 22 5 Example # 8 : If a1, a2, a3, a4, a5 are in A.P. with common difference  0, then find the value of  ai , when i 1 a3 = 2. Solution : As a1, a2, a3, a4, a5 are in A.P., we have a1 + a5 = a2 + a4 = 2a3. 5 Hence a i 1 i = 10. 1 1 1 Example # 9 : If a(b + c), b(c + a), c(a + b) are in A.P., prove that , , are also in A.P. a b c Solution : a(b + c), b(c + a), c(a + b) are in A.P.  subtract ab + bc + ca from each – bc, – ca, – ab are in A.P. divide by –abc 1 1 1 , , are in A.P. a b c ab bc 1 1 Example # 10 : If , b, are in A.P. then prove that ,b are in A.P. 1  ab 1  bc a c ab bc Solution :  , b, are in A.P. 1  ab 1  bc ab bc b– = –b 1  ab 1  bc  a b2  1  =  c 1  b2  1  ab 1  bc  –a + abc = c – abc a + c = 2abc divide by ac 1 1 1 1  = 2b  , b, are in A.P. c a a c Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 3 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series Arithmetic mean (mean or average) (A.M.) : If three terms are in A.P. then the middle term is called the A.M. between the other two, so if a, b, c are in A.P., b is A.M. of a & c. a  a  a .....  an A.M. for any n numbers a1, a2,..., an is; A = 1 2 3. n n - Arithmetic means between two numbers : If a, b are any two given numbers & a, A1, A2,...., An, b are in A.P., then A1, A2,... An are the n A.M.’s between a & b. ba 2 (b  a) n (b  a) A1 = a + ,A =a+ ,......, An = a + n1 2 n1 n1 Note : Sum of n A.M.’s inserted between a & b is equal to n times the single A.M. between a & b n ab i.e.  Ar = nA, where A is the single A.M. between a & b i.e. A = r 1 2 Example # 11 : If a, b, c,d,e, f are A. M’s between 2 and 12, then find a + b + c + d + e + f. 6(2  12) Solution : Sum of A.M.s = 6 single A.M. = = 42 2 Example # 12 : Insert 10 A.M. between 3 and 80. Solution : Here 3 is the first term and 80 is the 12th term of A.P. so 80 = 3 + (11)d  d=7 so the series is 3, 10, 17, 24,........, 73, 80  required means are 10, 17, 24,........, 73. Self practice problems : (5) There are n A.M.’s between 3 and 29 such that 6th mean : (n – 1)th mean : : 3 : 5 then find the value of n. an3  bn3 (6) For what value of n, n 2 , a  b is the A.M. of a and b. a  bn 2 Ans. (5) n = 12 (6) n = –2 Geometric progression (G.P.) : G.P. is a sequence of numbers whose first term is non zero & each of the succeeding terms is equal to the preceeding terms multiplied by a constant. Thus in a G.P. the ratio of successive terms is constant. This constant factor is called the common ratio of the series & is obtained by dividing any term by that which immediately preceeds it. Therefore a, ar, ar2, ar3, ar4,...... is a G.P. with 'a' as the first term & 'r' as common ratio. 1 1 1 1 e.g. (i) 2, 4, 8, 16,....... (ii) , , , ,....... 3 9 27 81 Results : (i) nth term of GP = a rn1 (ii) Sum of the first n terms of GP   a rn  1   , r 1 Sn =  r  1  na , r 1  (iii) Sum of an infinite terms of GP when r < 1. When n  rn 0 if r < 1 therefore, S = a 1 r  r  1 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 4 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series 1 Example # 13 : The nth term of the series 3, 3 , 1 , ------ is , then find n 243 n–1  1  1 Solution : 3.     n = 13  3 243 Example # 14 : The first term of an infinite G.P. is 1 and any term is equal to the sum of all the succeeding terms. Find the series. Solution : Let the G.P. be 1, r, r2, r3,......... r2 1 given condition   r=  r= , 1 r 2 1 1 1 Hence series is 1, , , ,.............. 2 4 8  Example # 15 : In a G.P.., T2 + T5 = 216 and T4 : T6 = 1 : 4 and all terms are integers, then find its first term : ar 3 1 Solution : ar (1 + r3) = 216 and  ar 5 4  r2 = 4  r = ± 2 when r = 2 then 2a(9) = 216  a = 12 when r = – 2, then – 2a (1–8) = 216 216 108  a= = , which is not an integer. 14 7 Self practice problems : (7) Find the G.P. if the common ratio of G.P. is 3, nth term is 486 and sum of first n terms is 728. (8) If x, 2y, 3z are in A.P. where the distinct numbers x, y, z are in G.P. Then find the common ratio of G.P. (9) A G.P. consist of 2n terms. If the sum of the terms occupying the odd places is S 1 and that of the terms occupying the even places is S2 , then find the common ratio of the progression. (10) If continued prodcut of three number in G.P. is 216 and sum of there product in pairs is 156. Find the numbers. 1 S Ans. (7) 2, 6, 18, 54, 162, 486 (8) (9) 2. 3 S1 (10) 2, 6, 18 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 =.......................... a (ii) Any three consecutive terms of a G.P. can be taken as , a , ar. r a a (iii) Any four consecutive terms of a G.P. can be taken as, 3 , ar, ar3. r r (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.. (v) If a1, a2, a3,........ and b1, b2, b3,......... are two G.P’s with common ratio r1 and r2 respectively, then the sequence a1b1, a2b2, a3b3,..... is also a G.P. with common ratio r1 r2. (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 is also true. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 5 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series Example # 16 : Three numbers form an increasing G.P. If the middle number is doubled, then the new numbers are in A.P. The common ratio of G.P. is : a Solution : Three number in G.P. are , a, ar r a then , 2a ar are in A.P. as given. r  1  2(2a) = a  r    r or r2 – 4r + 1 = 0 or r=2± 3 or r=2+ 3 as r > 1 for an increasing G.P. Example # 17 : The sum of an infinite geometric progression is 2 and the sum of the geometric progression made from the cubes of this infinite series is 24. Then find its first term and common ratio : Solution : Let a be the first term and r be the common ratio of G.P. a a3  2,  24 , –1 < r < 1 1– r 1– r 3 1 Solving we get a = 3, r = – 2 p Example # 18 : Express 0.423 in the form of , (where p, q , q  0) q 4 23 23 4 a 4 23 419 Solution : S=  3  5 +.......  =    = 10 10 10 10 1– r 10 990 990 Example # 19 : Evaluate 9 + 99 + 999 +........... upto n terms. Solution : Let S = 9 + 99 + 999 +..........upto n terms. = [9 + 99 + 999 +.......] = [(10 – 1) + (102 – 1) + (103 – 1) +........ + upto n terms]  10(10n  1)  = [10 + 102 + 103 +...........+ 10n – n] =   n  9  Geometric means (mean proportional) (G.M.): If a, b, c are in G.P., b is called as the G.M. of a & c. If a and c are both positive, then b = ac and if a and c are both negative, then b = – ac. b² = ac, therefore b = ac ; a > 0, c > 0. n-Geometric means between a, b : If a, b are two given numbers & a, G1, G2,....., Gn, b are in G.P.. Then G1, G2, G3,...., Gn are n G.M.s between a & b. G1 = a(b/a)1/n+1, G2 = a(b/a)2/n+1,......, Gn = a(b/a)n/n+1 Note : The product of n G.M.s between a & b is equal to the nth power of the single G.M. between a & b   n  n i.e. Gr = ab = Gn , where G is the single G.M. between a & b. r 1 Example # 20 : Between 4 and 2916 are inserted odd number (2n + 1) G.M’s. Then the (n + 1)th G.M. is Solution : 4, G1.G2,....... Gn+1,.... G2n, G2n+1, 2916 Gn+1 will be the middle mean of (2n +1) odd means and it will be equidistant from Ist and last term  4,Gn+1 , 2916 will also be in G.P.  Gn21 = 4 × 2916 = 4 × 9 × 324 = 4 × 9 × 4 × 81 Gn+1 = 2 × 3 × 2 × 9 = 108. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 6 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series Self practice problems : an 1  bn 1 (11) Find the value of n so that may be the G.M. between a and b. an  bn (12) If a = 111..........1, b = 1 + 10 + 102 + 103 + 104 and c = 1 + 105 + 1010 +..... + 1050, then prove 55 that (i) ‘a’ is a composite number (ii) a = bc. 1 Ans. (11) n=– 2 Harmonic progression (H.P.) A sequence is said to be in H.P if the reciprocals of its terms are in A.P.. If the sequence a1, a2, a3,...., an is in H.P. then 1/a1, 1/a2,...., 1/an is in A.P. Note : (i) Here we do not have the formula for the sum of the n terms of an H.P.. For H.P. whose first ab term is a and second term is b, the nth term is tn =. b  (n  1)(a  b) 2ac a ab (ii) If a, b, c are in H.P.  b = or =. ac c bc ab a (iii) If a, b, c are in A.P.  = bc a ab a (iv) If a, b, c are in G.P.  = bc b Harmonic mean (H.M.): 2ac If a, b, c are in H.P., b is called as the H.M. between a & c, then b = ac If a1, a2 ,........ an are ‘n’ non-zero numbers then H.M. 'H' of these numbers is given by 1 1 1 1 1 =   .......   H n  a1 a2 an  1 1 Example # 21 : The 7th term of a H.P. is and 12th term is , find the 20th term of H.P. 10 25 Solution : Let a be the first term and d be the common difference of corresponding A.P. a + 6d = 10 a + 11d = 25 5d = 15 d = 3, a = – 8 T20 = a + 19d = – 8 + 19 × 3 = 49 1 20 term of H.P. = 49 3 3 Example # 22 : Insert 4 H.M between and. 4 19 Solution : Let 'd' be the common difference of corresponding A.P.. 19 4  so d = 3 3 = 1. 5 1 4 7 3   = +1= or H1 = H1 3 3 7 1 4 10 3 = +2= or H2 = H2 3 3 10 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 7 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029  Sequence and Series 1 4 13 3 = +3= or H3 = H3 3 3 13 1 4 16 3 = +4= or H4 =. H4 3 3 16 2 12 Example # 23 : Find the largest positive term of the H.P., whose first two term areand. 5 23 5 23 30 23 16 9 2 –5 Solution : The corresponding A.P. is ,........ or , , , , ,....... 2 12 12 12 12 12 12 12 12 12 12 12 12 12 The H.P. is , , , , , – ,....... 30 23 16 9 2 5 12 Largest positive term = =6 2 Self practice problems : (13) If a, b, c, d, e are five numbers such that a, b, c are in A.P., b, c, d are in G.P. and c, d, e are in H.P. prove that a, c, e are in G.P. (14) If the ratio of H.M. between two positive numbers 'a' and 'b' (a > b) is to their G.M. as 12 to 13, prove that a : b is 9 : 4. ba bc (15) a, b, c are in H.P. then prove that + =2 ba bc (16) If a, b, c, d are in H.P., then show that ab + bc + cd = 3ad Arithmetico-geometric series : A series, each term of which is formed by multiplying the corresponding terms of an A.P. & G.P. is called the Arithmetico-Geometric Series. e.g. 1 + 3x + 5x2 + 7x3 +..... Here 1, 3, 5,.... are in A.P. & 1, x, x2, x3..... are in G.P.. Sum of n terms of an arithmetico-geometric series: Let Sn = a + (a + d) r + (a + 2 d) r² +..... + [a + (n  1)d] rn1, then Sn = a   dr 1 r n 1 a  (n 1)dr n , r  1. 1 r 1 r  1 r 2 Sum to infinity: If r < 1 & n  , then Limit rn = 0 and Limit n.rn = 0  n n a dr   S = . 1 r 1 r 2 2 3  4n  1   4n  1   4n  1  Example # 24 : The sum to n terms of the series 1 + 5   +9   + 13   +....... is.  4n – 3   4n – 3   4n – 3  4n  1 Solution : Let x= , then 4n – 3 –4 1 (4n – 3) 1–x= , =– 4n – 3 1– x 4 x (4n  1) =– 1– x 4 S = 1 + 5x + 9x2 +....... + (4n – 3)xn–1 Sx = x + 5x2 +........ (4n – 3)xn S – Sx = 1 + 4x + 4x2 +......... + 4xn–1 – (4n – 3)xn. 4x S(1 – x) = 1 + [1 –xn–1] – (4n – 3)xn 1– x 1  4x 4xn  (4n – 3) S=  1  – – (4n – 3)xn  = – [1– (4n  1)  (4n – 3)xn – (4n – 3)xn ] = n (4n – 3). 1– x  1– x 1– x  4 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 8 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029  Sequence and Series Example # 25 : Find sum to infinite terms of the series 1 + 2x + 3x2 + 4x3 +....., –1 < x < 1 Solution : let S = 1 + 2x + 3x2 + 4x3 +.............(i) xS = x + 2x2 + 3x3 +.................(ii) (i) - (ii)  (1 – x) S = 1 + x + x + x +.......... 2 3 1 or S= (1  x)2 Example # 26 : Evaluate : 12 + 22x + 32x2 + 42x3....... upto infinite terms for | x | < 1. Solution : Let s = 12 + 22x + 32x2 + 42x3............ ...(i) xs = 12x + 22x2 + 32x3............... ...(ii) (i) – (ii) (1 – x) s = 1 + 3x + 5x2 + 7x3 +........ 1 2x (1 – x) s = + 1 x 1  x  2 1 2x s= + 1  x  1  x  2 3 1  x  2x s= 1  x  3 1 x s= 1  x  3 Self practice problems : 4d 4  2d (17) If 4 + +......... = 1, then find d. 5 52 (18) Evaluate : 1 + 3x + 6x2 + 10x3 +...... upto infinite term, where | x | < 1. 2  1  1 (19) Sum to n terms of the series : 1 + 2  1   + 3  1   +......  n  n 64 Ans. (17) – 5 1 (18) (1  x)3 (19) n2 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. 8 Example # 27 : The A.M. of two numbers exceeds the G.M. by 2 and the G.M. exceeds the H.M. by ; find the 5 numbers. Solution : Let the numbers be a and b, now using the relation  8 G2 = AH = (G + 2)  G   G = 8 ; A = 10  5 i.e. ab = 64 also a + b = 20 Hence the two numbers are 4 and 16. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 9 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029  Sequence and Series A.M.  G.M.  H.M. Let a1, a2, a3,.......an be n positive real numbers, then we define their a  a2  a3 .......  an A.M. = 1 , their n G.M. = (a1 a2 a3.........an)1/n and their n H.M. =. 1 1 1  .......  a1 a2 an It can be shown that A.M.  G.M.  H.M. and equality holds at either places iff a1 = a2 = a3 =.........= an ab bc ca Example # 28 : If a, b, c > 0, prove that 2 3 + 2 + c a b2 Solution : Using the relation A.M.  G.M. we have ab bc ca   1 c 2 a2 b2   ab. bc. ca  3  ab bc ca + 2+ 2 3  2 2 2 3 c a b  c 2 a b 1 1 1 1 Example # 29 : If ai > 0  i = 1, 2, 3,....... prove that (a1 + a2 + a3.... + an)   .....    n2  1 a a 2 a 3 a n  Solution : Using the relation A.M.  H.M. a1  a2  a3.......  an n   n 1 1 1 1  .....  a1 a2 a3 an 1 1 1 1    (a1 + a2 + a3.... + an)   .....    n2  1 a a 2 a 3 a n   1 1  1 1   1 1  Example # 30 : If x, y, z are positive then prove that (x + y)(y + z)(z + x)          64  x y  y z   z x  Solution : Using the relation A.M.  H.M. xy 2  1 1   (x + y)     4.....(i) 2 1 1  x y  x y  1 1 similarly (y + z)     4.....(ii) y z  1 1 (z + x)  4   .....(iii) z x  1 1  1 1   1 1  by (i), (ii) & (iii) (x + y)(y + z)(z + x)          64  x y  y z   z x  Example # 31 : If n > 0, prove that 2n > 1 + n 2n1 Solution : Using the relation A.M.  G.M. on the numbers 1, 2, 22, 23,..........., 2n–1 , we have 1  2  22 .......  2n1 > (1.2. 22. 23...........2n–1)1/n n Equality does not hold as all the numbers are not equal. 1 2n  1  (n 1) n  n (n 1)  > n 2 2   2n – 1 > n 2 2  2 1   (n 1)    2n > 1 + n 2 2. Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 10 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series Example # 32 : If x, y, z are positive and x + y + z = 7 then find greatest value of x 2 y3 z2. Solution : Using the relation A.M.  G.M. x x y y y z z       1 2 2 3 3 3 2 2   x 2 y 3 z  2 7 ..  7  4 27 4  1  x 2 y3 z2  7  1  ..   432  x2 y3 z2  4 27 4  Self practice problems : (20) If a, b, c are real and distinct, then show that a2 (1 + b2) + b2 (1 + c2) + c2 (1 + a2) > 6abc (21) Prove that 2.4.6.8.......2n < (n + 1)n. (n  N) bcd cda dab abc (22) If a, b, c, d are positive real numbers prove that  2  2  2 >a+b+c+d a2 b c d (23) If x6 – 12x5 + ax4 + bx3 + cx2 + dx + 64 = 0 has positive roots then find a, b, c, d, (24) If a, b > 0, prove that [(1 + a) (1 + b)]3 > 33 a2 b2 Ans. (23) a = 60, b = –160, c = 240, d = –192 Results : n n n n n (i)  r 1 (ar ± br) =  r 1 ar ±  r 1 br. (ii)  r 1 k ar =  r 1 k ar. n (iii)  r 1 k = k + k + k +...............n times = nk; where k is a constant. n n (n  1) (iv)  r 1 r = 1 + 2 + 3 +...........+ n = 2 n n (n  1) (2n  1) (v)  r 1 r² = 12 + 22 + 32 +...........+ n2 = 6 n n2 (n  1)2 (vi)  r 1 r3 = 13 + 23 + 33 +...........+ n3 = 4 Example # 33 : Find the sum of the series to n terms whose nth term is 3n + 2. 3(n  1) n n Solution : Sn = Tn = (3n + 2) = 3n + 2 = + 2n = (3n + 7) 2 2 n Example # 34 : Tk = k3 + 3k , then find T k 1 k. n n 2 2  n(n  1)  3(3n  1)  n(n  1)    n 3 Solution :  Tk = k 1 k3 + 3k =   2   + 3 1 =   2   + 2 (3n –1) k 1 k 1 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 11 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and 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 n either in A.P. or in G.P. then we can find un. So sum of series S =  r 1 ur 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 k th differences ku1 , ku2 ,........ are obtained, where the k th 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 Example # 35 : Find the nth term of the series 1, 3, 8, 16, 27, 41,.......... Solution : s = 1 + 3 + 8 + 16 + 27 + 41 +...... T n.....(i) s= 1 + 3 + 8 + 16 + 27.......Tn–1 + Tn.....(ii) (i) – (ii) Tn = 1 + 2 + 5 + 8 + 11 +....... (T n – Tn – 1)  n  1 1 Tn = 1 +   [2 × 2 + (n – 2)3] = [3n2 – 5n + 4]  2  2 Example # 36 : Find the sum to n terms of the series 5, 7, 13, 31, 85 +...... Solution : Successive difference of terms are in G.P. with common ratio 3. Tn = a(3)n –1 + b a+b=5 3a + b = 7  a = 1, b = 4 Tn = 3n – 1 + 4 Sn = Tn = (3n – 1 + 4) = (1 + 3 + 32 +...... + 3n – 1) + 4n 1 n [3 + 8n – 1] 2 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 12 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series Method of difference for finding sn : If possible express rth term as difference of two terms as tr = ± (f(r) – f(r ± 1)). This can be explained with the help of examples given below. t1 = f(1) – f(0), t2 = f(2) – f(1), tn = f(n) – f(n-1)  Sn = f(n) – f(0) Example # 37 : Find the sum of n-terms of the series 2.5 + 5.8 + 8.11 +........... Solution : Tr = (3r – 1) (3r + 2) = 9r2 + 3r – 2 n n n n Sn =  r 1 Tr = 9  r 1 Tr + 3  2 r 1 r – r 1  n  n  1 2n  1   n  n  1  =9  + 3  – 2n  6   2      = 3n(n + 1)2 – 2n 1 1 1 Example # 38 : Sum to n terms of the series + + +......... (1  x)(1  3x) (1  3x)(1  5x) (1  5x)(1  7x) Solution : Let Tr be the general term of the series 1 Tr = 1   2r  1 x  1  (2r  1)x  1  1  (2r  1)x   1  (2r  1)x    1 1  So Tr =   =    2x  1  (2r  1)x  (1  (2r  1)x)   1  (2r  1)x  1  (2r  1)x    Sn =  Tr = T1 + T2 + T3 +.......... + Tn 1  1 1  n =   = 2x  1  x 1  (2n  1)x   (1  x)[1  (2n  1)x] 1 1 1 Example # 39 : Sum to n terms of the series   +............ 1.4.7 4.7.10 7.10.13 1 1  1 1  Solution : Tn = =     3n  2 3n  1 3n  4  6   3n  2 3n  1 3n  13n  4   1  1 1   1 1  1 1  =       ......    6  1.4 4.7   4.7 7.10  3n  23n  1 3n  13n  4   1 1 1  =    6  4  3n  1 3n  4   Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 13 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029 Sequence and Series Example # 40 : Find the general term and sum of n terms of the series 1 + 5 + 19 + 49 + 101 + 181 + 295 +......... Solution : The sequence of difference between successive term 4, 14, 30, 52, 80..... The sequence of the second order difference is 10, 16, 22, 28,...... clearly it is an A.P> so let nth term Tn = an3 + bn2 + cn + d a+b+c+d =1....(i) 8a + 4b + 2c + d = 5....(ii) 27a + 9b + 3c + d = 19....(iii) 64a + 16b + 4c + d = 49....(iv) from (i), (ii), (iii) & (iv) a = 1, b = –1, c = 0, d = 1  Tn = n3 – n2 + 1  n  n  1  sn = (n – n + 1 ) =  – n  n  1 2n  1 2 n n2  1  3n  2    3  2  +n= +n  2  6 12 Self practice problems : (25) Sum to n terms the following series 3 5 7 (i) 2 2  2 2  2 2 +........... 1.2 2.3 3.4 (ii) 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4)........ (iii) 4 + 14 + 30 + 52 + 82 + 114 +.......... n n 1 (26) If  T = (n + 1)(n + 2)(n + 3) then find  T r 1 r r 1 r 2n  n2 n n  1 n  2  n Ans. (25) (i) (ii) (iii) n(n + 1)2 (26) n  1 2 6 6 n  2 Reg. & Corp. Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] ADVSS- 14 Toll Free : 1800 258 5555 | CIN : U80302RJ2007PLC024029