What is the position and nature of image formed by lens combination shown in figure? (where,  $f_1$ and $f_2$ are focal lengths)

For lens A, object distance  $u=-20cm$
Focal length,  $f=+5cm$
From lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

We have,  $\frac{1}{v}=\frac{1}{5}-\frac{1}{20}$
 $\frac{1}{v}=\frac{20-5}{20\times 5}=\frac{15}{100}$
$v=\frac{20}{3}cm$ 
For lens B, image of A is object for B
 $\therefore u=\frac{20}{3}-2=+\frac{14}{3}cm$
 $f=-5cm$
Now, from lens formula we have
 $$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{-5}+\frac{3}{14} \\ \frac{1}{v}=\frac{15-14}{5\times 14} \\ v=70cm \end{gathered}$$
Hence, image is on right of lens B and is real in nature.