If a, b, c, d are positive real numbers such | vedclass

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Question

(A) Let $\frac{a}{3} = \frac{a+b}{4} = \frac{a+b+c}{5} = \frac{a+b+c+d}{6} = k$.From $\frac{a}{3} = k$,we get $a = 3k$.From $\frac{a+b}{4} = k$,we hav

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$\frac{1}{2}$

Vedclass · Answer

(A) Let $\frac{a}{3} = \frac{a+b}{4} = \frac{a+b+c}{5} = \frac{a+b+c+d}{6} = k$.From $\frac{a}{3} = k$,we get $a = 3k$.From $\frac{a+b}{4} = k$,we have $a+b = 4k$. Substituting $a=3k$,we get $3k+b = 4k$,so $b = k$.From $\frac{a+b+c}{5} = k$,we have $a+b+c = 5k$. Substituting $a+b=4k$,we get $4k+c = 5k$,so $c = k$.From $\frac{a+b+c+d}{6} = k$,we have $a+b+c+d = 6k$. Substituting $a+b+c=5k$,we get $5k+d = 6k$,so $d = k$.Now,substitute these values into the expression $\frac{a}{b+2c+3d}$:$\frac{3k}{k + 2(k) + 3(k)} = \frac{3k}{k + 2k + 3k} = \frac{3k}{6k} = \frac{1}{2}$.Thus,the correct option is $A$.

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 the value of $\frac{a}{b+2c+3d}$ is

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