Question

Three numbers are in an increasing geometric progression with common ratio $r$. 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 $3r^2$, then $r^2-d$ is equal to:

• A. $7-7\sqrt{3}$
• B. $7+\sqrt{3}$
• C. $7-\sqrt{3}$
• D. $7+3\sqrt{3}$

Answer 1

Answer: (B)

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