- ✓$9: 6: 4$
- B$16: 4: 1$
- C$25: 10: 4$
- D$6: 3: 2$
$\therefore \mathrm{b}^2=\mathrm{ac}$
Also
$\log _{\circ}\left(\frac{a}{2 b}\right), \log _{\circ}\left(\frac{2 b}{3 c}\right), \log _{\circ}\left(\frac{3 c}{a}\right)$ are in $A.P.$
$\left(\frac{2 b}{3 \mathrm{c}}\right)^2=\frac{\mathrm{a}}{2 \mathrm{~b}} \times \frac{3 \mathrm{c}}{\mathrm{a}} $
$ \frac{\mathrm{b}}{\mathrm{c}}=\frac{3}{2}$
Putting in eq. $(i)$ $b^2=a \times \frac{2 b}{3}$
$ \frac{a}{b}=\frac{3}{2}$
$ a: b: c=9: 6: 4$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$R=\left\{(x, y): \max \left\{0, \log _{e} x\right\} \leq y \leq 2^{x}, \frac{1}{2} \leq x \leq 2\right\}$
is, $\alpha\left(\log _{e} 2\right)^{-1}+\beta\left(\log _{e} 2\right)+\gamma$, then the value of $(\alpha+\beta-2 \gamma)^{2}$ is equal to: