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If-a-b-c-are-3-positive-numbers-in-an-A-P-and-T-a-8b-2b-a-8b-c-2b-c-Then-the-value-of-T-2-is-Ans-given-is-361-




Question Number 32532 by rahul 19 last updated on 28/Mar/18
If a,b,c are 3 positive numbers in an  A.P and   T= ((a+8b)/(2b−a))+((8b+c)/(2b−c)).  Then the value of T^(  2 )  is ?  Ans. given is 361.
Ifa,b,care3positivenumbersinanA.PandT=a+8b2ba+8b+c2bc.ThenthevalueofT2is?Ans.givenis361.
Commented by Rasheed.Sindhi last updated on 28/Mar/18
b=a+d ,c=a+2d  T= ((a+8b)/(2b−a))+((8b+c)/(2b−c))     = ((a+8(a+d))/(2(a+d)−a))+((8(a+d)+(a+2d))/(2(a+d)−(a+2d)))     = ((9a+8d)/(a+2d))+((9a+10d)/a)     =((9a^2 +8ad+9a^2 +10ad+18ad+20d^2 )/(a(a+2d)))     =((18a^2 +36ad+20d^2 )/(a(a+2d)))     =((2(9a^2 +18ad+10d^2 ))/(a(a+2d)))  T can′t be free of a & d  Also T is even(Its square should also even)  ∴ T^( 2) ≠ 361
b=a+d,c=a+2dT=a+8b2ba+8b+c2bc=a+8(a+d)2(a+d)a+8(a+d)+(a+2d)2(a+d)(a+2d)=9a+8da+2d+9a+10da=9a2+8ad+9a2+10ad+18ad+20d2a(a+2d)=18a2+36ad+20d2a(a+2d)=2(9a2+18ad+10d2)a(a+2d)Tcantbefreeofa&dAlsoTiseven(Itssquareshouldalsoeven)T2361
Commented by Tinkutara last updated on 28/Mar/18
When (a,b,c)=(3,5,7)  T=((43)/7)+((47)/3)  T^2 ≠361
When(a,b,c)=(3,5,7)T=437+473T2361
Commented by rahul 19 last updated on 28/Mar/18
Ok, thank you both !
Ok,thankyouboth!

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