Question

Let the harmonic mean of two positive real number $a$ and $b$ be $4$. If $q$ is a positive real number such that a, $5$, $q$, $b$ is an arithmetic progression, then the value(s) of $|q-a|$ is (are)

Answer 1

$\begin{array}{c}\frac{1}{a}+\frac{1}{b}=\frac{2}{4}(1+M)\\ a,5,q,b\to AP\\ q-a=2d\\ b-a=3d\\ 5-a=d\\ \therefore \frac{1}{a}+\frac{1}{a+3d}=\frac{1}{2}.....\left(1\right)\\ a+d=\mathrm{5.....}\left(2\right)\\ d=5-a\\ \frac{1}{a}+\frac{1}{a+15-3a}=\frac{1}{2}\\ \frac{15-2a+a}{\left(15-2a\right)a}=\frac{1}{2}\\ 30-2a=15a-2a^2\\ 2a^2-17a+30=0\\ \Rightarrow a=6,2.5\\ \Rightarrow d=-1,2.5\\ for,a=6,d=-1\\ q=4\\ \therefore |q-a|=2\\ for,a=2.5,d=2.5\\ |q-a|=5\\ \therefore 2or5\end{array}$