Three numbers are in an increasing geometric progression

Question:

Three numbers are in an increasing geometric progression with common ratio $\mathrm{r}$. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference $d$. If the fourth term of GP is $3 r^{2}$, then $r^{2}-d$ is equal to :

  1. $7-7 \sqrt{3}$

  2. $7+\sqrt{3}$

  3. $7-\sqrt{3}$

  4. $7+3 \sqrt{3}$


Correct Option: , 2

Solution:

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

$\frac{\mathrm{a}}{\mathrm{r}}, 2 \mathrm{a}$, ar $\rightarrow$ A.P $\Rightarrow 4 \mathrm{a}=\frac{\mathrm{a}}{\mathrm{r}}+\mathrm{ar} \Rightarrow \mathrm{r}+\frac{1}{\mathrm{r}}=4$

$r=2 \pm \sqrt{3}$

$4^{\text {th }}$ form of G.P $=3 r^{2} \Rightarrow a r^{2}=3 r^{2} \Rightarrow a=3$

$r=2+\sqrt{3}, a=3, d=2 a-\frac{a}{r}=3 \sqrt{3}$

$r^{2}-d=(2+\sqrt{3})^{2}-3 \sqrt{3}$

$=7+4 \sqrt{3}-3 \sqrt{3}$

$=7+\sqrt{3}$

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