Sign convention in combination of lenses The problem statement:
Two convex lenses, each of focal length 10cm, are placed at a separation of 15cm with their principal axes coinciding.


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*Show that a light beam coming parallel to the principal axis diverges as it comes out of the lens system.


My approach:
I've applied the focal length equivalent formula,
 $$\frac{1}{f_{eq}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1}*f_{2}}$$
$$f_{eq}=+20 cm$$
Upon looking at the focal length I'm confused as my answer's focal length is positive(converging) but the nature of the given lens is diverging in the first subpart.
Is my approch right or I'm using wrong sign convention?
 A: I believe your calculations are correct you just didn't complete it. The formula you used are to find a lens whose focal length would produce an image at the same position when replacing the two lenses, but where is it going to be placed ?!
The formula that is used to find at what distance should this equivalent lens be placed from the second lens is
$$ x = \dfrac{d ~ f_{eq}}{f_1}$$
It is better to use the lens maker's equation twice, which is given by 
$$ \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$$
Where, $u$ is the distance between the object and the lens, and $v$ is the distance between the image and the lens, and if you showed that the final image is formed in front of the second lens then you would have proved that this system is actually a diverging system "for objects that is infinitely distant from the second lens".
Knowing that the light incident on the first lens is parallel to the principal-axis, this means that the object is infinitely distant from the lens, and substituting in the lens maker's equation, it is obvious that the image would be formed at the focal point of the first lens ``the image would be real and formed behind the lens''.
Thus, the image would be formed in front the focal point of the next lens, which is separated a distance 15 cm from the first lens, which means that the image which now serves as a real object for the second lens, is at a distance of 5 cm from the second lens.
A convex lens, is capable of producing both real and virtual images "it can either converge or diverge the light incident on it", depending on the position of the object with the respect to its focal point.
And since the image is in front the focal point of the second lens, then the second lens would diverge the image formed by the first lens "which is the object for the second lens", and if you want to find where the image is formed, then you can use the lens maker's equation once again, substituting the focal length value and the distance $u$ to be 5 cm, then we can find $v$ which would have a negative sign of 10 cm.
However, the final image of this optical system is a diverging, where the final image is formed in front the second lens.
