We have to consider sign convention as only magnitude of focal length is given.

If $f_1$  is focal length of convex (converging) lens and  $f_2$  is focal length of concave (diverging) lens. Then their equivalent focal length F would be

 $\begin{array}{l}\frac{1}{F}=\frac{1}{f_1}+\frac{1}{-f_2}=\frac{f_2-f_1}{f_1{f}_2}[\because \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}]\\ \therefore F=\frac{f_1{f}_2}{f_2-f_1}\end{array}$