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Question Number 7526 by Little last updated on 02/Sep/16
If a, b, c are in AP then (a/(bc)) , (1/c) , (2/b) are in
Ifa,b,careinAPthenabc,1c,2barein
Answered by Rasheed Soomro last updated on 04/Sep/16
a,b,c are in AP⇒b=((a+c)/2) (a/(bc)) , (1/c) , (2/b) (a/((((a+c)/2))c)) , (1/c) , (2/((a+c)/2)) ((2a)/(c(a+c))) , (1/c) , (4/(a+c)) (1/c)−((2a)/(c(a+c)))=^(?) (4/(a+c))−(1/c) ((a+c−2a)/(c(a+c)))=^(?) ((4c−a−c)/(c(a+c))) ((c−a)/(c(a+c)))≠((3c−a)/(c(a+c))) (a/(bc)) , (1/c) , (2/b) are not in AP (1/c)/((2a)/(c(a+c)))=^(?) (4/(a+c))/(1/c) (1/c)×((c(a+c))/(2a))=^(?) ((4c)/(a+c)) ((a+c)/(2a))≠((4c)/(a+c)) (a/(bc)) , (1/c) , (2/b) are not in GP ((2a)/(c(a+c))) , (1/c) , (4/(a+c)) 1/c is HM if above are in GP. So, 1/c =^(?) ((2(((2a)/(c(a+c))))((4/(a+c))))/(((2a)/(c(a+c)))+(4/(a+c)))) =^(?) (((16a)/(c(a+c)^2 ))/((2a+4c)/(c(a+c))))=(((8a)/(a+c))/(a+2c))=((8a)/((a+c)(a+2c))) ≠((8a)/((a+c)(a+2c))) Hence (a/(bc)) , (1/c) , (2/b) are not in HP.
a,b,careinAP⇒b=a+c2abc,1c,2ba(a+c2)c,1c,2a+c22ac(a+c),1c,4a+c1c−2ac(a+c)=?4a+c−1ca+c−2ac(a+c)=?4c−a−cc(a+c)c−ac(a+c)≠3c−ac(a+c)abc,1c,2barenotinAP1c/2ac(a+c)=?4a+c/1c1c×c(a+c)2a=?4ca+ca+c2a≠4ca+cabc,1c,2barenotinGP2ac(a+c),1c,4a+c1/cisHMifaboveareinGP.So,1/c=?2(2ac(a+c))(4a+c)2ac(a+c)+4a+c=?16ac(a+c)22a+4cc(a+c)=8aa+ca+2c=8a(a+c)(a+2c)≠8a(a+c)(a+2c)Henceabc,1c,2barenotinHP.

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