PROGRESSIONS AND SERIES

PROGRESSIONS AND SERIES A sequece is lso clled progressio. We ow study three importt types of sequeces: () The Arithmetic Progressio, () The Geometric Progressio, () The Hrmoic Progressio. Arithmetic Progressio. Def. (A.P) A sequece is clled rithmetic progressio (Abbrev. A.P.), if its terms cotiully icrese or decrese by the sme umber. The fixed umber by which they icrese or decrese is clled the Commo differece. The followig re exmple of sequeces i A.P. Sequeces Commo differece, 6, 0, 4. 4 0, 5, 0, 5, 0,. 5, + d, d, + d.. d [Test. Subtrct from ech term, the precedig term, If the differece is equl, the progressio is rithmetic] Thus three qutities,, b, c will be i A.P. if b = c b i.e. b = + c Aid to memory: Three qutities re i A.P. If twice the middle = sum of the extremes. The lst sequece i these exmples is stdrd A.P. th term of A.P. Let us cosider the rithmetic progressio, + d, + d, + d,.. Wht is the coefficiet of d i the 0th term, the 5th term, the 8rd term? It is obvious tht the coefficiet of d i y term is oe less th the umber of the term, d so the th term of A.P. is = ( ) d. If the sme is represeted by l, we hve T = l = + ( ) d. Aid to memory: Lst term = [First term + (umber of the terms ) commo differece]. Note.. The formul l = + ( ) d cotis for qutities l,,, d. Three qutities beig give, the fourth c foud out. If oly two qutities re give, two coditios of the problem should be give. Note.. d = T T = T T =.. = T T. Illustrtio : The th term of A.P. is 4. Write dow the first 4 terms d the 8th term of the A.P. Solutio: T = 4 Puttig =,,, 4 d 8, we get T = 4 x =, T = 4 x = 7, T = 4 x =. T 4 = 4 x 4 = 5, T 8 = 4 x 8 = 7. Hece, the firs four term of the A.P. re, 7,, 5 d the 8th term is 7. Illustrtio : Fid the 9th d pth term of the A.P., 5, 8. Solutio: Here =, d = 9th term = + (9 ) X = 6 pth term = + (p ) X =p +. http://kishoremths.i/

Illustrtio : Which term of the series + 9 + 7 +. is? Solutio: Here l is give d is to be determied; =, d = = + ( ) ( ) = + = 0 = 5 Hece is the 5th term. Illustrtio 4: Solutio: The 8th term of series i A.P. is d the 0th term is 05. Fid the series. Let be the first term d d be the commo differece. The, T 8 = + 7d = T 0 = + 0 d = 05 Solvig () d (), =, d = Hece, the required series is + 5 + 8 + +. Illustrtio 5: Solutio: I certi A.P. the 4th term is twice the 0th term. Prove tht the 7d term is twice the 4th term. Let be the first term d d the commo differece. The, By the give coditio, + d =( + 9d) = 5d Now, 7d term = + 7d = 5d + 7d = 76d. [From ()] 4th term = + d = 5d + d = 8d [From ()] Obviously, () is twice (). Hece proved. Illustrtio 6: Solutio: If p times the pth term of A.P. is equl to q times the qth term, prove tht (p + q) th term is zero. Let be the first term d d the commo differece. It is give tht p.t p = q.t q. i.e., p [ + (p ) d] =q [+ (q ) d] (p q) = (q q p + p) d (q p) = (q p) [(q + p) ] d = [(q + p) ] d + [(q + p) ] d = 0 t p+q = 0 Exercise : (i) Fid the vlue of k so tht 8k + 4, 6k, d k + 7 will form A.P. (ii) Fid, b such tht 7,, b, re i A.P. Exercise : Determie d term d rth term of A.P. whose 6th term is d 8th term is. Exercise : If 7 times the 7th term of A.P. is equl to times its th term, show tht the 8th term of A.P. is zero. Sum of terms of A.P. Let the A.P. be, + d, + d,..let l be the lst term, d S the required sum. The, S = + ( + d) + ( + d) + + (l d) + l Reversig the right hd terms S = l + (l d) + (l d) + + ( + d) +. Addig, S = ( + l) + ( +l) + to terms = ( + ) S ( l) If we substitute the vlue of l viz., l = + ( ) d, i this formul, we get S [ + + ( ) d] i.e., S [ + ( ) d] http://kishoremths.i/

Note.. Ech of these formule cotis four qutities. Three beig kow, the fourth c be foud out. Note.. If the sum of terms be fuctio of, the sum of ( ) terms c be foud o puttig ( ) for. Thus if S = +, S = ( ) + ( ). Illustrtio 7: Fid the sum of (i) 0 terms, (ii) terms of the progressio,, 5, 7, 9. Solutio: Here, =, d = (i) S 0 = 0 { x + (0 ) x } = 0 x 40 = 400. (ii) S = {+ + ( ) d} = { x + ( ) } =. Illustrtio 8: Solutio: The sum of terms of A.P. is 4 + 5. Fid the series. S = 4 x + 5 x = 9, S = 4 x + 5 x = 6, S = 4 x + 5 x = 5,.. T = S = 9, T = S S = 6 9 = 7, T = S S = 5 6 = 5 Hece the series is 9 + 7 + 5 +. Altertively, T = S S = (4 + 5) [4( ) + 5 ( )] = 8 +. Now, put =,,, Exercise 4: I childre s potto rce, pottoes re plce metre prt i stright lie. A competitor strts from poit i this lie which is 5 metres from the erest potto. Fid expressio for the totl distce ru i collectig the pottoes, oe t time, d brigig them bck oe t time to the strtig poit. Clculte the vlue of, if the totl distce ru is 6 metres. Exercise 5: The third term of rithmeticl progressio is 7, d the seveth term is more th times the third term. Fid the first term, the commo differece d the sum of the first 0 terms. Exercise 6: The iterior gles of polygo re i rithmetic progressio. The smllest gle is 5 d the commo differece is 8. Fid the umber of sides of the polygo. Arithmetic Me If three terms re i A.P., the the middle term is clled the rithmetic me (A.M.) betwee the other two i.e. If d b re the give umbers d x is their rithmetic me the, x, b re i A.P. b x = b x x = + b x Illustrtio 9: Isert 4 rithmetic mes betwee d. Solutio: Let A, A, A, A 4 be the 4 rithmetic mes The,, A, A, A, A 4, is A.P., cotiig 6 terms. = 6th term = + 5d d = 4 A = + 4 = 7, A = + X 4 =, A = + X 4 = 5. A 4 = + 4 x 4 = 9 GEOMETRIC SEQUENCES Wht is Geometric Sequece? Cosider the followig sequeces (), 4, 8, 6 (b), 6, 8, 54, 6 http://kishoremths.i/

(c) 0, 0, 40, 80, 60,.. (d) 9, 6, 4, 8, 6 9, These re ot rithmetic sequeces. Wht is the specilty of sequeces? Do you see tht i ech of them, the rtio of the secod term to the first is equl to the rtio of the third to its predecessor, the rtio of the fourth term to its predecessor d so o. Such sequece is clled geometric sequece. Defiitio: Geometric Sequece (G.S.) is oe i which the rtio of y term to its predecessor is lwys the sme umber. This rtio is clled the commo rtio. A geometric sequece is lso clled geometric progressio (G.P.). The followig re lso exmples of geometric sequeces. Sequece Commo rtio,, 9, 7, 8,. r = 6/7, 8/9, 4/,. r = / x, x, x, x 4 r = x [Test: Divide ech term by precedig term, if the quotiets re equl the sequece is geometric progressio.] If deotes the first term d r be the commo rtio i G.P., them the stdrd G.P. is,, r, r, Illustrtio 0: Solutio: The first three terms of geometric sequece re 48, 4, d. Wht is the commo rtio d fourth term of this sequece? Step.. To fid the commo rtio, divide y term by its predecessor. 4 48 = or 4 = the commo rtio is Step.. To fid the fourth term, Multiply the third term by the commo rtio. x the fourth term is 6. The th Term The followig tble suggests expressios for t i terms of, r, d st term d term rd term 4th term... th term t t t t 4 t r r r r - Observig tht the expoet of r i ech term is oe less th the umber of the term, you c mke the followig cojecture. The th term of geometric sequece whose first term is d commo rtio is o zero umber r is t = r -. Illustrtio : The th term of geometric sequece is (.5) for ll vlues of. Write dow the vlue of (the first term) d the rtio (r). Solutio: t = (.5) = t =. (.5) = (Puttig = ) t = (.5) = Commo rtio r = =.5. Geometric Mes Whe three umbers form geometric sequece, the middle oe is clled the geometric me of the umbers. For exmple, 5, 5, 5 form G.S., therefore 5 is the Geometric me of 5 d 5. http://kishoremths.i/

Let, G, b form geometric sequece, the G = b G G = b G = b, Where, b re positive rel umbers. The positive geometric me betwee two umbers is the me proportiol betwee them d is G = b, this shows tht we hve uique G.M. betwee two umbers. Note. This lso shows tht if there re umbers,, b, c which form G.S., the b = c i.e., (Squre of the middle term) = Product of the extremes. Note. If there re positive itegers,,, the their geometric me is defied to be equl to (.. ) /. Illustrtio : Fid the geometric me of 6 d 4. Solutio: Let G be the geometric me, the G = 6 x 4 = 44 G = The geometric me is. Just s you my hve more th oe rithmetic me, you my lso hve more th oe geometric me. The terms betwee y two give terms of geometric sequece re geometric mes. Thus, i the G.S., 4, 8, 6.The terms 4 d 8 re the geometric mes betwee d 6. To isert geometric me betwee oly for G.P. s of positive umbers. Illustrtio : Isert rel geometric mes betwee d 6. Solutio: Let, G, G, G be the required mes. The,, G, G, G, from G.S. Let r be the commo rtio. 6 The = 5th term = r4 6 r 4 = 6 r4 6 = 0 r + 4 r 4 =0 or, r4 = 6 r 4 = 0 r + 4 0 for y r R r = 4 r = r 4 = G = = or G = =, r = G = = 4 or G = = 4, G = = 8 or G = =. 8 /, /4, /8 re the required geometric mes betwee d /6, s geometric mes re defied oly for GP s of positive umbers. 4 Illustrtio 4: Fid geometric mes betwee two give umbers d b d prove tht their products equl to the th power of sigle geometric me betwee the give umbers. Solutio: Let d b be the give umbers d G, G,.G their geometric mes. The, G, G,.. G, b is G.P. cotiig ( + ) terms. http://kishoremths.i/

If the commo rtio is r, the ( + )th term = r + = r + b = b r = b G = r = (ii) Now, G, G, G, G b = b., G = r b = b. b ( )... sum of A.P. b (). = b =.b b..... G = x r b = = b... Hece proved /() Exercise 7: Fid the vlue of x for which x + 9, x 6, 4 re the first three terms of geometricl progressio d clculte the fourth term of progressio i ech cse. Exercise 8: If 5, x, y, z, 405 re the first five terms of geometric progressio, fid the vlues of x, y, d z. Exercise 9: Isert () geometric mes betwee 6 d 56 ; (b) 5 geometric mes betwee d 4. Sum of Terms of Geometric Series Cosider the series + r +.. + r Let S deote the sum of terms of this series. Write the series d subtrct from it, term by term, the product of r d the series s show below. S = + r + + r + r Sr = r +.. + r + r + r Subtrctig. S Sr = r S ( r) = ( r ) S = ( r ) (r ). r We c write d use the bove formul i the form S = lr lr Or, S = (or). if l is the lst term. r r Where l = t = r so tht lr = r. (r ) ;(r ). r Illustrtio 5: Fid the sum of ech of the followig series: (i) 5 0 + 0.. to 6 terms. (ii) 4 + + + + 4 +.. to 0 terms (iii) 4 + 4 + 4 + to terms. ( r ) Solutio: (i) S = ( r ). Hece = 5, r =, d = 6 r http://kishoremths.i/

S 6 = 6 5[ ( ) ] 5 5x64 5 05. ( ) (ii) = 4, r = ½ d = 0 0 4 4 04 0 s 0 8X 8( Approx ). 04 4 4 r ( ) (iii) 4, r ( whichis ) s 4 r 4 4 4 4. 6 4 6 X (4 ). 4 Illustrtio 6: The ivetor of the chess bord suggested rewrd of oe gri of whet for the first squre, gris for the secod d so o, doublig the mout of the gris for subsequet squres. How my gris would hve to be give to the ivetor? (There re 64 squres i the chess bord). Solutio: The required umber of gris = + + + to 64 terms = 64 x( ) 64. Illustrtio 7: Solutio: A isect strts from poit d trvels i stright pth oe mm i the first secod d hlf of the distce covered i the previous secod i the succeedig secod. I how much time would it rech poit mm wy from its strtig prt? Let the required time be secods. x The... upto terms, or. There is o vlue of for which. Hece, the isect would ever rech poit mm wy from its strtig poit. Exercise 0: Fid the sum of geometric series i which = 6, r = 4, l = Exercise : Fid the sum of the series 8 7 + 9... 7 64 Exercise : The first two terms of geometric progressio re d b respectively. Fid the sum of the first 0 terms. http://kishoremths.i/

HARMONIC PROGRESSION Defiitio A series is sid to be i Hrmoic Progressio whe the reciprocls of the terms form rithmeticl progressio. For short it is deoted by H.P. just s rithmeticl progressio d geometric progressio re deoted by A.P. d G.P. respectively. The followig series re some of the exmples of H.P. ()...; ()...; ()... 4 5 7 9 d d d The bove re i H.P. becuse the series formed by the reciprocls of the terms of these series, viz., (i) + + +4 +..; (ii) 5 7 9...; (iii) + ( +d ) + ( + d) + ( + d)+.. re i A.P. If follows, therefore, tht if series be i A.P., the reciprocls of the terms will form H.P. Note. The most geerl or stdrd H.P. is,,,,... d d d Illustrtio 8: If, b, c be i H.P., show tht b. b c c Solutio: Sice, b, c, re i H.P., therefore,, re i A.P., i.e. b c b c b b b c b b b or or i.e.,. b bc b c bc c b c c Workig the sme process bckwrds, it c be show tht if b, the, b, c re i H.P. b c c Hece, H.P. my lso be defied s series i which every three cosecutive term stisfy this reltio. Illustrtio 9: Fid the first term of H.P whose secod d third terms re 5,. Solutio: Let be the first term. The,, 5 Usig the reltio, we get 5 5 re i H.P. =, or 5 5 or 5 =, i.e., 4 =, or = 4 To fid the th term of H.P. Evidetly, the th term of H.P is the reciprocl of th term of the A.P. formed by the reciprocls of the H.P. Thus the th term of the H.P.,,, is d d d ( )d Illustrtio 0: Fid the 8th term of the series,,,... 6 http://kishoremths.i/

6 Solutio: The reciprocls of of the terms re,,,... This is A.P., the commo differece beig. The 8th term of this series is + (8 ) = The 8th term of the give H.P. is. Hrmoic Me Defiitio: If, c, b re three qutities i H.P., the c is sid to be the Hrmoic Me betwee d b rithmetic me betwee two umbers or qutities is geerlly deoted by H just s the rithmetic me is deoted by A d the geometric me by G. If, H, H. H, b re i H.P. the H, H,. H re defied to be Hrmoic Mes betwee d b. Let, b be give umbers d let H be the Hrmoic Me betwee them. The, sice, H, b re i H.P., Therefore.,, re i A.P. H b or b,or,or H b. b H b H b Thus, the H.M. betwee two qutities d b is b b. Illustrtio : (i) Fid the hrmoic me betwee d. (ii) Isert three hrmoic mes betwee 5 d 6. Solutio: (i) Hrmoic me betwee d b is b b Here =, b =, H.M. = x x. 5 (ii) Let us isert rithmetic mes betwee d d the tke their reciprocls. 5 6 Let A, A, A, be the rithmetic mes i betwee d 5 6 The 5, A,, A, A, re i A.P. 6 Let d be the commo differece. T 5 =,or (5 )d d 6 5 6 0 A = d ; 5 5 0 0 A = d ; 5 5 60 60 A = d 7 ; 5 5 40 40 The required hrmoic mes re 0 60 40,,. 7 http://kishoremths.i/

Exercise : The sum of the reciprocls of three umbers i H.P. is d the product of the umbers is. 48 Fid the umbers. Exercise 4: The sum of three umbers i H.P. is 7 60 d the first umber is. Fid the umbers 4 Exercise 5: If, b, c re i A.P., d x, y d z be i H.P. show tht c x z by xz ANSWERS TO EXERCISES. Exercise : (i) 7, (ii) = 49, b = Exercise : 8, 5r 8 Exercise 4: + 9, = 9 Exercise 5:, 4, 740 Exercise 6: Exercise 7: x = 0, 6;, 5 Exercise 8: 5, 45, 5 Exercise 9: (), 64, 8 (b),, 9, 7, 8 Exercise 0: 64 Exercise : Exercise : Exercise : Exercise4: 0 60 7 b 0 0 8 (b ),, 4 6,, 4 5 6 http://kishoremths.i/

ASSIGNMENTS SUBJECTIVE LEVEL I. Write the 5th d 8th terms of AP whose 0th term is 4 d commo differece is 4.. The sum of AP is 6. The commo differece 4 d the lst term, fid.. = ;d.fidt dt. 4. Determie the th term of G.P. whose 8th term is 9 d commo rtio is. 5. I H.P., 4,,,... fid the 4th d 7th terms. 7 6. The rd term of AP is 7 d its 7th term is more th thrice of its third term. Fid the first term, commo differece d sum of its first 0 terms. 7. Give GP with = 79 d 7th term 64, determie S 7 8. I H.P the third term is 7 d 7th term is. Show tht the 5th term is. 5 9. Fid the sum lf ll three digit umbers which leve the remider ; whe divided by 4. 0. Sum 7,, 7.. to terms LEVEL II. Fid the sum of term of the progressio 7, 77, 777, 7777. How my terms of GP,,... re eeded to give sum 069? 4 5. Fid the sum to terms of the series d hece to 0 terms of the series. + ( + ) + ( + + ) + ( + + + 4 ) 4. Express s frctio 0.7 5. Fid two umbers betwee /6 d /6. Such tht the first three my be i G.P d the lst three i H.P. http://kishoremths.i/

OBJECTIVE LEVEL I. If the first term of G.P. is ; d 6th term is 96; the the c.r. of the G.P. is (A) (B) (C) (D). The A.M d G.M of two umbers re 0 d 8; the two umbers re (A) 5, 5 (B) 6, 4 (C) 4, 6 (D), 8. If x, y, z re i G.P.; the /x, /y, /z re i (A) H.P (B) A.P (C) G.P (D) Noe of A, B, C 4. Commo rtio of the series,,,,... is (A) (B) (C) / (D) 5. The G.M. of the two positive umbers is 4; d H.M. is ; the A.M is (A) 6 (B) (C) 6 (D) 8 6. The hrmoic me of /4 d /8 is (A) /5 (B) /6 (C) / (D) /7 7. /x = b /y = c /z ; if, b, c re i G.P.; the x, y, z will be i (A) A.P (B) G.P (C) H.P (D) Noe of A, B, C 8. If five GM s re iserted betwee 8 d 58, the fifth term of the geometric series is (A) 648 (B) 8 (C) 68 (D) 944 9. If x, x +, x + re i G.P., the the fourth term is (A) 7 (B) 7 (C).5 (D).5 0. The miimum umber of terms of + + 5 + 7 +... tht dd up to umber exceedig 57 is (A) 5 (B) 7 (C) 5 (D) 7 http://kishoremths.i/

LEVEL II More th oe swer. Give the series + + ½ + ¼ +... (A) The sum icreses without limit. (B) The sum decreses without limit. (C) The sum pproches limit (D) The differece betwee the sum d 4 c be mde less th y positive qutity o mtter how smll.. If the first term of A.P is d commo differece is, the (A) t 7 = 9 (B) s 0 = 45 (C) t = (D) s 5 = 50. If 4 A.M s, A, A, A, A 4 re iserted betwee 4 d 6, the (A) t = 6.4 (B) t 4 =. (C) A =. (D) A 4 =.6 4. If the product of first three terms of G.P is 8, d the 8 th term is 87, the (A) = (B) r = (C) t = 8 (D) t 7 = 79 5. If, x +, 8 re i G.P., the (A) x = (B) x = (C) x = 7 (D) x = 6 6. If ( b) + (b c) + (c ) = 0, the, b, c re i (A) A.P (B) H.P (C) G.P (D) oe of these 7. If, b, c re i A.P, the (A) k b, k, c k re i A.P (B) k, k b, k c re i G.P m m m k k k (C),, re i H.P (D) oe of these b c 8. If x, y, z re i H.P, the (A) y x x (B) xy + yz zx = 0 (C) x y z z y z y z x (D) xy + yz + zx = 0 9. If S =...to terms the s is equl to (A) ( + ) (B) (C) + (D)...upto terms 0. Let S deotes the sum of the cubes of the first turl umbers d s deotes the sum of the first turl umbers. The (A) ( )( ) 6 r (B) S s r r ( ) is equl to (C) 6 (D) ( ) http://kishoremths.i/

ANSWERS SUBJECTIVE LEVEL I., 5. = 8. + 4. 07 5., 4 6. =, d = 4, s 0 = 740 7. 059 9.,, 55 / 7 / 0. LEVEL II. 7 [0(0 ) 9]. = 0 8. 7 [( )( ) ], 0 4. 999 5., or, 9 4 OBJECTIVE LEVEL I. D. C. C 4. A 5. D 6. B 7. A 8. D 9. D 0. B LEVEL II. C, D. A, B, C. A, B, C, D 4. A, B, C, D 5. A, C 6. A, B, C 7. A, B, C 8. A, B 9. C, D 0. A, D http://kishoremths.i/