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Question Number 30780 by mondodotto@gmail.com last updated on 25/Feb/18

Answered by Rasheed.Sindhi last updated on 25/Feb/18

$$(4\mathrm{x},4\mathrm{y},4\mathrm{z}\mathrm{are}\mathrm{in}\mathrm{GP}\Rightarrow \mathrm{x},\mathrm{y},\mathrm{z}\mathrm{are}\mathrm{in}\mathrm{GP}$$$$\mathrm{because}\frac{4\mathrm{y}}{4\mathrm{x}}=\frac{4\mathrm{z}}{4\mathrm{y}}\Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{z}}{\mathrm{y}})$$$$\mathrm{y}=\mathrm{x}+\mathrm{d},\mathrm{z}=\mathrm{x}+2\mathrm{d}$$$$\mathrm{x}+(\mathrm{x}+\mathrm{d})+(\mathrm{x}+2\mathrm{d})=70$$$$3\mathrm{x}+3\mathrm{d}=70\Rightarrow \mathrm{x}+\mathrm{d}=\frac{70}{3}\Rightarrow \mathrm{x}=\frac{70-3\mathrm{d}}{3}$$$$\frac{\mathrm{x}+\mathrm{d}}{\mathrm{x}}=\frac{\mathrm{x}+2\mathrm{d}}{\mathrm{x}+\mathrm{d}}$$$${(\mathrm{x}+\mathrm{d})}^2=\mathrm{x}(\mathrm{x}+2\mathrm{d})$$$${\left(\frac{70}{3}\right)}^2=\left(\frac{70-3\mathrm{d}}{3}\right)(\frac{70-3\mathrm{d}}{3}+2\mathrm{d})$$$${\left(\frac{70}{3}\right)}^2=\left(\frac{70-3\mathrm{d}}{3}\right)\left(\frac{70+3\mathrm{d}}{3}\right)$$$${70}^2-{\left(3\mathrm{d}\right)}^2={70}^2$$$$\mathrm{d}=0$$$$\mathrm{x}=\frac{70-3\left(0\right)}{3}=\frac{70}{3}$$$$\mathrm{x}=\mathrm{y}=\mathrm{z}=\frac{70}{3}$$$$\mathrm{common}\mathrm{difference}=0$$$$\mathrm{common}\mathrm{ratio}=1$$$$$$

Commented by mondodotto@gmail.com last updated on 25/Feb/18

$$\mathrm{sir}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{are}\mathrm{in}\mathrm{AP}$$

Commented by Rasheed.Sindhi last updated on 25/Feb/18

$$\mathrm{Yes}\mathrm{sir},{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{why}\mathrm{I}\mathrm{take}\mathrm{y}=\mathrm{x}+\mathrm{d}\mathrm{\&}\mathrm{z}=\mathrm{x}+2\mathrm{d}$$

Commented by mondodotto@gmail.com last updated on 25/Feb/18

$$\mathrm{thanx}$$

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