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If-a-b-c-are-in-AP-b-c-d-are-in-GP-and-c-d-e-are-in-HP-then-a-c-e-are-in-




Question Number 7524 by Little last updated on 02/Sep/16
If  a, b, c  are in AP;  b, c, d  are in GP and  c, d, e  are in HP, then a, c, e are in
Ifa,b,careinAP;b,c,dareinGPandc,d,eareinHP,thena,c,earein
Commented by Rasheed Soomro last updated on 02/Sep/16
Complete Answer  If  a, b, c  are in AP;  b, c, d  are in GP and  c, d, e  are in HP, then a, c, e are in  −−−−−−−−−−−−−−−−−−−   a,b,c are in AP :  c=a+2d ′=a+2(b−a)=2b−a [d ′ is common difference]  b,c,d are in GP :  d=br^2 =b(c/b)^2 =c^2 /b =(((2b−a)^2 )/b) [r is common ratio]  c,d,e are in HP ⇒d=((2ce)/(c+e))⇒e=(dc/(2c−d))  e=(dc/(2c−d))=(((((2b−a)^2 )/b)×(2b−a))/(2(2b−a)−(((2b−a)^2 )/b)))=((((2b−a)^3 )/b)/((2b(2b−a)−(2b−a)^2 )/b))                                                            =(((2b−a)^3 )/((2b−a)[2b−2b+a]))=(((2b−a)^2 )/a)  a,c,e  a,2b−a, (((2b−a)^2 )/a)  Let′s  assume that this is GP  Testing for common ratio  ((2b−a)/a)=^(?) ((((2b−a)^2 )/a)/(2b−a))  ((2b−a)/a) =^(?) (((2b−a)^2 )/a)×(1/(2b−a))  ((2b−a)/a) =((2b−a)/a)  Hence a,2b−a, (((2b−a)^2 )/a)   is  a GP,whose common ratio  is ((2b−a)/a)  Or    a ,  c  ,  e   are in GP
CompleteAnswerIfa,b,careinAP;b,c,dareinGPandc,d,eareinHP,thena,c,eareina,b,careinAP:c=a+2d=a+2(ba)=2ba[discommondifference]b,c,dareinGP:d=br2=b(c/b)2=c2/b=(2ba)2b[riscommonratio]c,d,eareinHPd=2cec+ee=dc2cde=dc2cd=(2ba)2b×(2ba)2(2ba)(2ba)2b=(2ba)3b2b(2ba)(2ba)2b=(2ba)3(2ba)[2b2b+a]=(2ba)2aa,c,ea,2ba,(2ba)2aLetsassumethatthisisGPTestingforcommonratio2baa=?(2ba)2a2ba2baa=?(2ba)2a×12ba2baa=2baaHencea,2ba,(2ba)2aisaGP,whosecommonratiois2baaOra,c,eareinGP
Answered by Rasheed Soomro last updated on 07/Sep/16
Answer is in comment.
Answerisincomment.

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