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Question Number 32532 by rahul 19 last updated on 28/Mar/18

$$\mathit{I}fa,b,care3positivenumbersinan$$$$\mathit{A}.\mathit{P}and$$$$T=\frac{a+8b}{2b-a}+\frac{8b+c}{2b-c}.$$$$Thenthevalueof{T}^{2}is?$$$$Ans.givenis361.$$

Commented by Rasheed.Sindhi last updated on 28/Mar/18

$$b=a+d,c=a+2d$$$$T=\frac{a+8b}{2b-a}+\frac{8b+c}{2b-c}$$$$=\frac{a+8(a+d)}{2(a+d)-a}+\frac{8(a+d)+(a+2d)}{2(a+d)-(a+2d)}$$$$=\frac{9a+8d}{a+2d}+\frac{9a+10d}{a}$$$$=\frac{9a^2+8ad+9a^2+10ad+18ad+20d^2}{a(a+2d)}$$$$=\frac{18a^2+36ad+20d^2}{a(a+2d)}$$$$=\frac{2(9a^2+18ad+10d^2)}{a(a+2d)}$$$$T{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{free}\mathrm{of}a\mathrm{\&}d$$$$\mathrm{Also}T\mathrm{is}\mathrm{even}\left(\mathrm{Its}\mathrm{square}\mathrm{should}\mathrm{also}\mathrm{even}\right)$$$$\therefore T^2\ne 361$$

Commented by Tinkutara last updated on 28/Mar/18

$$When(a,b,c)=(3,5,7)$$$$T=\frac{43}{7}+\frac{47}{3}$$$$T^2\ne 361$$

Commented by rahul 19 last updated on 28/Mar/18

$$Ok,thankyouboth!$$

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