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Question Number 82628 by jagoll last updated on 23/Feb/20
If a,b, c are in Harmonic progression find the value of ((a+b)/(b−a)) + ((b+c)/(b−c)) . ?
Ifa,b,careinHarmonicprogressionfindthevalueofa+bb−a+b+cb−c.?
Commented by $@ty@m123 last updated on 23/Feb/20
(1/a), (1/b),(1/c) are in AP ⇒(1/b)−(1/a)=(1/c)−(1/b)=d, say ((a+b)/(b−a)) + ((b+c)/(b−c)) =((a+b)/(ab((1/a)−(1/b))))+((b+c)/(bc((1/c)−(1/b)))) =−((a+b)/(abd))+((b+c)/(bcd)) =((−c(a+b)+a(b+c))/(abcd)) =((b(a−c))/(abcd)) =(((a−c))/(acd)) =(1/d)((1/c)−(1/a)) =(1/d)×(d+(1/b)+d−(1/b)) =((2d)/d) =2
1a,1b,1careinAP⇒1b−1a=1c−1b=d,saya+bb−a+b+cb−c=a+bab(1a−1b)+b+cbc(1c−1b)=−a+babd+b+cbcd=−c(a+b)+a(b+c)abcd=b(a−c)abcd=(a−c)acd=1d(1c−1a)=1d×(d+1b+d−1b)=2dd=2
Answered by TANMAY PANACEA last updated on 23/Feb/20
(1/a),(1/b),(1/c) are in A.P (2/b)=(1/a)+(1/c) →2=(b/a)+(b/c) ((a+b)/(b−a))+((b+c)/(b−c)) ((1+(b/a))/((b/a)−1))+(((b/c)+1)/((b/c)−1)) ((1+(b/a))/((b/a)−1))+((2−(b/a)+1)/(2−(b/a)−1)) (((b/a)+1)/((b/a)−1))+((3−(b/a))/(1−(b/a))) (((b/a)+1−3+(b/a))/((b/a)−1)) →((2((b/a)−1))/((b/a)−1))=2 answer
1a,1b,1careinA.P2b=1a+1c→2=ba+bca+bb−a+b+cb−c1+baba−1+bc+1bc−11+baba−1+2−ba+12−ba−1ba+1ba−1+3−ba1−baba+1−3+baba−1→2(ba−1)ba−1=2answer
Commented by jagoll last updated on 23/Feb/20
thank you mister
thankyoumister

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