Let the harmonic mean of two positive real numbers  $a$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 $\left|q-a\right|$ is (are)

It is given that the harmonic mean of $a$ and $b$ is 4.

$\therefore \frac{2ab}{a+b}=4\Rightarrow ab=2(a+b)$                                ..(i)

It is given that $a, 5, q, b$ are in A.P. with common difference $d$(say). then,

$\begin{array}{l}5=a+d\text{ and }b=a+3d\\ \Rightarrow a=5-d\text{ and }b=5-d+3d\\ \Rightarrow a=5-d\text{ and }b=5+2d\end{array}$

Substituting the values of $a$and $b$in (i), we obtain

$\begin{array}{l}(5-d)(5+2d)=2(5-d+5+2d)\\ \Rightarrow 25+5d-2d^2=20+2d\\ \Rightarrow 2d^2-3d-5=0\Rightarrow (2d-5)(d+1)=0\Rightarrow d=-1,5/2\\ \therefore q=a+2d\Rightarrow |q-a|=\left|2d\right|=2,5\end{array}$