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

Answer 1

The correct option is A $7+\sqrt{3}$

Let the number be $\frac{a}{r},a,ar$
As it is increasing G.P., so $r>1$
Fourt term of G.P. is $a{r}^2$
$a{r}^2=3r^2\Rightarrow a=3$
So, the G.P. becomes
$\frac{3}{r},3,3r$
Also, $\frac{3}{r},6,3r$ are in A.P., so
$\frac{3}{r}+3r=12\Rightarrow r^2-4r+1=0\cdots \left(1\right)\Rightarrow r=\frac{4\pm \sqrt{16-4}}{2}\Rightarrow r=2+\sqrt{3}[\because r>1]$
Now,
$r^2-d=r^2-(3r-6)=r^2-3r+6=r^2-4r+1+r+5=r+5\left[\text{From equation}\right(1\left)\right]=7+\sqrt{3}$