Three numbers are in an increasing geometric | vedclass

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Question

Let numbers be $\frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}, \mathrm{ar} \rightarrow$ $G.P.$

$\mathrm{a}_{\mathrm{r}}, 2 \mathrm{a}, \mathrm{ar} \rig

Vedclass · Accepted Answer

$7+\sqrt{3}$

Vedclass · Answer

Let numbers be $\frac{\mathrm{a}}{\mathrm{r}}, \mathrm{a}, \mathrm{ar} ightarrow$ $G.P.$

$\mathrm{a}_{\mathrm{r}}, 2 \mathrm{a}, \mathrm{ar} ightarrow \mathrm{A} \cdot \mathrm{P} \Rightarrow 4 \mathrm{a}=\frac{\mathrm{a}}{\mathrm{r}}+\mathrm{ar} \Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}=4$

$\mathrm{r}=2 \pm \sqrt{3}$

$4^{	ext {th }}$ form of G.P $=3 \mathrm{r}^{2} \Rightarrow \mathrm{ar}^{2}=3 \mathrm{r}^{2} \Rightarrow \mathrm{a}=3$

$\mathrm{r}=2+\sqrt{3}, \mathrm{a}=3, \mathrm{~d}=2 \mathrm{a}-\frac{\mathrm{a}}{\mathrm{r}}=3 \sqrt{3}$

$\mathrm{r}^{2}-\mathrm{d}=(2+\sqrt{3})^{2}-3 \sqrt{3}$

$=7+4 \sqrt{3}-3 \sqrt{3}$

$=7+\sqrt{3}$

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