JEE 2024 — Mathematics PYQ

log_ea, log_eb, log_ec are in AP

⇒ 2 log_eb = log_ea + log_ec

⇒ b² = ac

 log_ea – log_e2b, log_e2b – log_e3c

${\log }_{e}3c-{\log }_{e}a\text{ are in AP}$

$\Rightarrow {\log }_e\frac{a}{2b},{\log }_e\frac{2b}{3c},{\log }_e\frac{3c}{a}\text{ are in AP}$

$\Rightarrow 2{\log }_e\frac{2b}{3c}={\log }_e\frac{a}{2b}+{\log }_e\frac{3c}{a}$

$\Rightarrow {\left(\frac{2b}{3c}\right)}^2=\frac{a}{2b}\times \frac{3c}{a}\Rightarrow \frac{4b^2}{9c^2}=\frac{3c}{2b}$

$\Rightarrow 8b^3=27c^3\Rightarrow 2b=3c$

$\because b^2=ac\Rightarrow {\left(\frac{3c}{2}\right)}^2=ac\Rightarrow 9c=4a$

$\therefore a:b:c=\frac{9c}{4}:\frac{3c}{2}:c=9:6:4$