Find the position of the image formed by the lens combination given in the figure.

(N/A) Step $1$: Image formed by the first lens $(f_1 = +10 \, cm)$:
Using the lens formula $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$,where $u_1 = -30 \, cm$:
$\frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$
So,$v_1 = 15 \, cm$ to the right of the first lens.
Step $2$: Image formed by the second lens $(f_2 = -10 \, cm)$:
The image formed by the first lens acts as an object for the second lens. The distance between the first and second lens is $5 \, cm$. Thus,the object distance for the second lens is $u_2 = +(15 - 5) \, cm = +10 \, cm$ (virtual object).
Using $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:
$\frac{1}{v_2} - \frac{1}{10} = \frac{1}{-10} \implies \frac{1}{v_2} = 0 \implies v_2 = \infty$.
Step $3$: Image formed by the third lens $(f_3 = +30 \, cm)$:
The rays emerging from the second lens are parallel,so they act as an object at infinity for the third lens $(u_3 = \infty)$.
Using $\frac{1}{v_3} - \frac{1}{u_3} = \frac{1}{f_3}$:
$\frac{1}{v_3} - \frac{1}{\infty} = \frac{1}{30} \implies v_3 = 30 \, cm$.
The final image is formed $30 \, cm$ to the right of the third lens.