Question

If $a,b,c,d$ are positive real numbers such that $\frac{a}{3}=\frac{a+b}{4}=\frac{a+b+c}{5}=\frac{a+b+c+d}{6}$, then $\frac{a}{b+2c+3d}$ is equal to $?$

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. Cannot be determined

Answer 1

The correct option is A $\frac{1}{2}$

Given that, $\frac{a}{3}=\frac{a+b}{4}=\frac{a+b+c}{5}=\frac{a+b+c+d}{6}$

So, we have $\frac{a}{3}=\frac{a+b}{4}$

$\Rightarrow a=3b$

We have, $\frac{a}{3}=\frac{a+b+c}{5}$

$\Rightarrow a=3c$

We have, $\frac{a}{3}=\frac{a+b+c+d}{6}$

$\Rightarrow a=3d$

So, the value of $\frac{a}{b+2c+3d}=\frac{a}{\frac{a}{3}+\frac{2a}{3}+a}=\frac{a}{2a}=\frac{1}{2}$

Therefore, the correct option is $A$.