If the sides of a triangle are in A.P, the perimeter of the triangle is $30$ cm. The difference between the longer and shorter side is $4$ cm. Then find the all sides of the triangles.

Let the sides of triangle be $a,b,c$ such that $a<b<c$

Given:

$2b=a+c$ ....... $(1)$

$a+b+c=30cm$ ....... $(2)$

$c-a=4cm$ ....... $(3)$

From $(1)$ and $(2)$, we get

$2b+b=30\Rightarrow 3b=30\Rightarrow b=10$ 

$\Rightarrow a+c=20$ ....... $(4)$

Solving $(3)$ and $(4)$, we get

$c=12\Rightarrow a=8$

Hence, $a=8,b=10,c=12$