JEE Main 2024 — Mathematics Algebra

Given: $\mathrm{log}a,\mathrm{log}b,\mathrm{log}c$loga,logb,logc are in $A.P.$A.P.

$$\Rightarrow 2\mathrm{log}b=\mathrm{log}a+\mathrm{log}c$$⇒2logb=loga+logc

$$\Rightarrow {b}^{2}=ac...(1)$$⇒b2=ac...(1)

And $\mathrm{log}a-\mathrm{log}2b,\mathrm{log}2b-\mathrm{log}3c,\mathrm{log}3c-\mathrm{log}a$loga−log2b,log2b−log3c,log3c−loga are in $A.P$A.P

$$\Rightarrow 2\mathrm{log}\frac{2b}{3c}=\mathrm{log}\frac{a}{2b}+\mathrm{log}\frac{3c}{a}$$⇒2log3c2b​=log2ba​+loga3c​

$$\Rightarrow {(\frac{2b}{3c})}^{2}=\frac{a}{2b}\times \frac{3c}{a}$$⇒(3c2b​)2=2ba​×a3c​

$$\Rightarrow {(\frac{2b}{3c})}^{2}=\frac{3c}{2b}$$⇒(3c2b​)2=2b3c​

$$\Rightarrow 2b=3c$$⇒2b=3c

\Rightarrow 4{b}^{2}=9{c}^{2}&6b=9c...(2)

Now solving the equation (1)&(2) we get,

$$4ac=9{c}^{2}$$4ac=9c2

$$\Rightarrow 4a=9c...(3)$$⇒4a=9c...(3)

Then from equation (2)&(3) we get,

$$4a=6b=9c=k$$4a=6b=9c=k

\Rightarrow a=\frac{k}{4},b=\frac{k}{6}&c=\frac{k}{9}

$$\Rightarrow a:b:c=\frac{1}{4}:\frac{1}{6}:\frac{1}{9}=9:6:4$$⇒a:b:c=41​:61​:91​=9:6:4