If a parallel beam of light is incident on a combination of a convex lens of focal length f and a concave lens of focal length $f/2$ , the image formed is

The equivalent focal length of the combination 
$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f}+\frac{1}{\left(-\left(f/2\right)\right)}$ but $\frac{1}{\upsilon }-\frac{1}{u}=\frac{1}{f_{eq}}$
$=\frac{1}{f}-\frac{2}{f}=-\frac{1}{f}\text{or\hspace{0.17em}}{f}_{eq}=-f$ 
For a parallel bean of rays, $u=\infty$
$\frac{1}{\upsilon }=\frac{1}{-f_{eq}}\Rightarrow \upsilon =-f$ (virtual image at a distance $\left|\upsilon \right|=\left|f\right|$)