Question

If the median of $\frac{a}{3},\frac{a}{2},\frac{a}{4},\frac{2a}{5},\frac{a}{6}$ is $12$ then find the value of $a\left(a>0\right)$.

Answer 1

We have,

$\frac{a}{3},\frac{a}{2},\frac{a}{4},\frac{2a}{5},\frac{a}{6}$

On taking LCM of denominators,

$\frac{a}{3}=\frac{20a}{60}$

$\frac{a}{2}=\frac{30a}{60}$

$\frac{a}{4}=\frac{15a}{60}$

$\frac{2a}{5}=\frac{24a}{60}$

$\frac{a}{6}=\frac{10a}{60}$

Now, arranging in asscending order

$\frac{a}{6},\frac{a}{4},\frac{a}{3},\frac{2a}{5},\frac{a}{2}$

So, the median term is $3^{rd}$, we get

$\frac{a}{3}$

Therefore,

$\frac{a}{3}=12$

$a=36$

Hence, this is the answer.