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Question Number 24858 by Tinkutara last updated on 27/Nov/17

$$\mathrm{If}\mathrm{three}\mathrm{positive}\mathrm{numbers}a,b,c\mathrm{are}\mathrm{in}$$$$\mathrm{A}.\mathrm{P}.\mathrm{and}\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}\mathrm{also}\mathrm{in}\mathrm{A}.\mathrm{P}.,\mathrm{then}$$$$\left(1\right)a=b=c$$$$\left(2\right)2b=3a+c$$$$\left(3\right)b^2=\frac{ac}{8}$$$$\left(4\right)2c=2b+a$$

Answered by ajfour last updated on 28/Nov/17

$$leta=b-d$$$$andc=b+d$$$$\frac{2}{b^2}=\frac{a^2+c^2}{a^2{c}^2}=\frac{2(b^2+d^2)}{{(b^2-d^2)}^2}$$$$letd=kb$$$$\Rightarrow \frac{1}{b^2}=\frac{(1+k^2)b^2}{{(1-k^2)}^2{b}^4}$$$$or{(1-k^2)}^2=1+k^2$$$$let1-k^2=t$$$$\Rightarrow t^2=2-t$$$$\Rightarrow {(t+\frac{1}{2})}^2=\frac{9}{4}$$$$ort=1$$$$\Rightarrow 1-k^2=1\Rightarrow k=0$$$$soa=b=c.$$

Commented by prakash jain last updated on 30/Nov/17

$$t=-2$$$$k=\sqrt{3}$$$$b=b$$$$a=b(1-\sqrt{3})$$$$c=b(1+\sqrt{3})$$$$\mathrm{That}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{provided}$$$$\mathrm{by}\mathrm{jota}+.$$

Commented by Tinkutara last updated on 28/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}!$$

Commented by prakash jain last updated on 30/Nov/17

$$\mathrm{What}\mathrm{about}\mathrm{the}\mathrm{solution}\mathrm{posted}$$$$\mathrm{below}\mathrm{by}\mathrm{jota}+.\mathrm{I}\mathrm{have}\mathrm{put}$$$$\mathrm{my}\mathrm{calculation}\mathrm{as}\mathrm{a}\mathrm{comment}\mathrm{there}.$$

Answered by jota+ last updated on 29/Nov/17

$$onesolution$$$$a=1-\sqrt{3}b=1c=1+\sqrt{3}$$

Commented by prakash jain last updated on 30/Nov/17

$$ok.$$

Commented by prakash jain last updated on 30/Nov/17

$$\frac{1}{a^2}=\frac{1}{4-2\sqrt{3}}=\frac{1}{2(2-\sqrt{3})}$$$$=\frac{2+\sqrt{3}}{2}=1+\frac{\sqrt{3}}{2}$$$$\frac{1}{b^2}=1$$$$\frac{1}{c^2}=\frac{1}{4+2\sqrt{3}}=\frac{1}{2(2+\sqrt{3})}=1-\frac{\sqrt{3}}{2}$$

Commented by Tinkutara last updated on 30/Nov/17

$$\mathrm{But}\mathrm{my}\mathrm{ques}.\mathrm{says}a,b,c\mathrm{to}\mathrm{be}+\mathrm{ve}.$$

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