Speed of an object is defined as the distance travelled by the object per unit time.

Let's assume that, $AB= d$. So, $BC= d, AD= 3d$. Also, assume that the time required by the object to travel $AB, BC, CD$ are $t_1, t_2, t_3$ respectively.

Now,

$\begin{array}{rcl}AD& =& 3AB\\ AB+BC+CD& =& 3AB\\ AB+AB+CD& =& 3AB\\ CD& =& AB\end{array}$

Hence,

$\begin{array}{rcl}{t}_1& =& \frac{AB}{v_1}\\ & =& \frac{d}{v_1}\\ t_2& =& \frac{d}{v_2}\\ t_3& =& \frac{d}{v_3}\end{array}$

The formula to calculate the average speed $\left(v_a\right)$ of the object is given by

$v_a= \frac{AD}{t_1+t_2+t_3}........................\left(1\right)$

Substitute the expressions for the parameters into equation (1) and simplify to obtain the required average speed of the object.

$\begin{array}{rcl}{v}_a& =& \frac{3d}{{\displaystyle \frac{d}{v_1}}+{\displaystyle \frac{d}{v_2}}+{\displaystyle \frac{d}{v_3}}}\\ & =& \frac{3v_1{v}_2{v}_3}{v_1{v}_2+v_2{v}_3+v_3{v}_1}\end{array}$