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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-

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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.
$$\boldsymbol{{I}}{f}\:{a},{b},{c}\:{are}\:\mathrm{3}\:{positive}\:{numbers}\:{in}\:{an} \\ $$$$\boldsymbol{{A}}.\boldsymbol{{P}}\:{and}\: \\ $$$${T}=\:\frac{{a}+\mathrm{8}{b}}{\mathrm{2}{b}−{a}}+\frac{\mathrm{8}{b}+{c}}{\mathrm{2}{b}−{c}}. \\ $$$${Then}\:{the}\:{value}\:{of}\:{T}^{\:\:\mathrm{2}\:} \:{is}\:? \\ $$$${Ans}.\:{given}\:{is}\:\mathrm{361}. \\ $$
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
$${b}={a}+{d}\:,{c}={a}+\mathrm{2}{d} \\ $$$${T}=\:\frac{{a}+\mathrm{8}{b}}{\mathrm{2}{b}−{a}}+\frac{\mathrm{8}{b}+{c}}{\mathrm{2}{b}−{c}} \\ $$$$\:\:\:=\:\frac{{a}+\mathrm{8}\left({a}+{d}\right)}{\mathrm{2}\left({a}+{d}\right)−{a}}+\frac{\mathrm{8}\left({a}+{d}\right)+\left({a}+\mathrm{2}{d}\right)}{\mathrm{2}\left({a}+{d}\right)−\left({a}+\mathrm{2}{d}\right)} \\ $$$$\:\:\:=\:\frac{\mathrm{9}{a}+\mathrm{8}{d}}{{a}+\mathrm{2}{d}}+\frac{\mathrm{9}{a}+\mathrm{10}{d}}{{a}} \\ $$$$\:\:\:=\frac{\mathrm{9}{a}^{\mathrm{2}} +\mathrm{8}{ad}+\mathrm{9}{a}^{\mathrm{2}} +\mathrm{10}{ad}+\mathrm{18}{ad}+\mathrm{20}{d}^{\mathrm{2}} }{{a}\left({a}+\mathrm{2}{d}\right)} \\ $$$$\:\:\:=\frac{\mathrm{18}{a}^{\mathrm{2}} +\mathrm{36}{ad}+\mathrm{20}{d}^{\mathrm{2}} }{{a}\left({a}+\mathrm{2}{d}\right)} \\ $$$$\:\:\:=\frac{\mathrm{2}\left(\mathrm{9}{a}^{\mathrm{2}} +\mathrm{18}{ad}+\mathrm{10}{d}^{\mathrm{2}} \right)}{{a}\left({a}+\mathrm{2}{d}\right)} \\ $$$${T}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{free}\:\mathrm{of}\:{a}\:\&\:{d} \\ $$$$\mathrm{Also}\:{T}\:\mathrm{is}\:\mathrm{even}\left(\mathrm{Its}\:\mathrm{square}\:\mathrm{should}\:\mathrm{also}\:\mathrm{even}\right) \\ $$$$\therefore\:{T}^{\:\mathrm{2}} \neq\:\mathrm{361} \\ $$
Commented by Tinkutara last updated on 28/Mar/18
When (a,b,c)=(3,5,7) T=((43)/7)+((47)/3) T^2 ≠361
$${When}\:\left({a},{b},{c}\right)=\left(\mathrm{3},\mathrm{5},\mathrm{7}\right) \\ $$$${T}=\frac{\mathrm{43}}{\mathrm{7}}+\frac{\mathrm{47}}{\mathrm{3}} \\ $$$${T}^{\mathrm{2}} \neq\mathrm{361} \\ $$
Commented by rahul 19 last updated on 28/Mar/18
Ok, thank you both !
$${Ok},\:{thank}\:{you}\:{both}\:! \\ $$

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