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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 $\frac{a}{b+2 c+3 d}$ is -
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$1 / 2$
$\begin{array}{l}
\frac{a}{3}=\frac{a+b}{4}=\frac{a+b+c}{5}=\frac{a+b+c+d}{6}=k \\
a=3 k, b=k, c=k, d=k \\
\text { Hence:- } \frac{a}{b+2 c+3 d}=\frac{3 k}{k+2 k+3 k}=\frac{1}{2}
\end{array}$
\frac{a}{3}=\frac{a+b}{4}=\frac{a+b+c}{5}=\frac{a+b+c+d}{6}=k \\
a=3 k, b=k, c=k, d=k \\
\text { Hence:- } \frac{a}{b+2 c+3 d}=\frac{3 k}{k+2 k+3 k}=\frac{1}{2}
\end{array}$
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