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Question Number 171403 by mnjuly1970 last updated on 14/Jun/22
a , b , c ∈R and a≠b≠c and (a/(b+c)) , (b/(a+c)) , (c/(a+b)) are three consecutive terms of AP (Arithmetic progression ) find −: ((b^( 2) −a^( 2) )/(c^( 2) −b^( 2) ))
a,b,c∈Randa≠b≠candab+c,ba+c,ca+barethreeconsecutivetermsofAP(Arithmeticprogression)find−:b2−a2c2−b2
Answered by Rasheed.Sindhi last updated on 15/Jun/22
(b/(a+c))− (a/(b+c))=(c/(a+b))−(b/(a+c)) ((b(b+c)−a(a+c))/((a+c)(b+c)))=((c(a+c)−b(a+b))/((a+b)(a+c))) ((b^2 +bc−a^2 −ac))/((a+c)(b+c)))=((ac+c^2 −ab−b^2 )/((a+b)(a+c))) (((b−a)(b+a)+c(b−a))/((a+c)(b+c)))=(((c−b)(c+b)+a(c−b))/((a+b)(a+c))) (((a−b)(a+b+c))/((a+c)(b+c)))−(((b−c)(a+b+c))/((a+b)(a+c)))=0 (a+b+c)∙(((a−b)/((a+c)(b+c)))−((b−c)/((a+b)(a+c))))=0 { ((a+b+c=0)),(( OR)),((((a−b)/(b+c))=((b−c)/(a+b))⇒a^2 −b^2 =b^2 −c^2 )) :} ((a^2 −b^2 )/(b^2 −c^2 ))=1⇒((b^( 2) −a^( 2) )/(c^( 2) −b^( 2) ))=1
ba+c−ab+c=ca+b−ba+cb(b+c)−a(a+c)(a+c)(b+c)=c(a+c)−b(a+b)(a+b)(a+c)b2+bc−a2−ac)(a+c)(b+c)=ac+c2−ab−b2(a+b)(a+c)(b−a)(b+a)+c(b−a)(a+c)(b+c)=(c−b)(c+b)+a(c−b)(a+b)(a+c)(a−b)(a+b+c)(a+c)(b+c)−(b−c)(a+b+c)(a+b)(a+c)=0(a+b+c)⋅(a−b(a+c)(b+c)−b−c(a+b)(a+c))=0{a+b+c=0ORa−bb+c=b−ca+b⇒a2−b2=b2−c2a2−b2b2−c2=1⇒b2−a2c2−b2=1
Commented by Rasheed.Sindhi last updated on 15/Jun/22
The condition should be a+b+c≠0 instead of a≠b≠c
Theconditionshouldbea+b+c≠0insteadofa≠b≠c
Commented by Rasheed.Sindhi last updated on 15/Jun/22
Thanks sir, you′re right.
Thankssir,you′reright.

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