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 $3 r^{2}$, 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 $\frac{a}{r}, a, a r \rightarrow$ G.P
$\frac{ a }{ r }, 2 a , ar \rightarrow A . P $
$\Rightarrow 4 a =\frac{ a }{ r }+ ar $
$\Rightarrow r +\frac{1}{ 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, $
$d=2 a-\frac{a}{r}=3 \sqrt{3}$
$r ^{2}- d =(2+\sqrt{3})^{2}-3 \sqrt{3}$
$=7+4 \sqrt{3}-3 \sqrt{3}$
$=7+\sqrt{3}$