Sides of a Progressive Triangle

Let the sides of the triangle be $(a-d),a,(a+d)$ since they are in arithmetic progression.
The perimeter of the triangle $=$ the perimeter of a equilateral triangle $= a-d+a+a+d = 3a.$ Therefore the sides of the equilateral triangle are $a,a,a.$
The area of the equilateral triangle $= \dfrac{a^2\sqrt3}{4}$ by formula.
The area of the original triangle $= \dfrac{3}{5}\times$ area of the equilateral triangle $=\dfrac{3a^2\sqrt3}{20}.- - - - - 1$
We find the area of the original triangle by heron's formula : [Semi-Perimeter(s) $=\frac{3a}{2}$ ]
$=\sqrt{s(s-x)(s-y)(s-z)}$ where $x,y,z$ are the sides of the triangle.
$=\sqrt{(\frac{3a}{2})(\frac{a}{2}+d)(\frac{a}{2})(\frac{a}{2}-d)}$
$=\sqrt{(\frac{3a^2}{4})(\frac{a^2-4d^2}{4})} - - - - - - 2$
Since $1=2, \Rightarrow \dfrac{3a^{\cancel{2}}\cancel{\sqrt3}}{\cancel{20}\small{5}}=\dfrac{\cancel{a\sqrt3}\sqrt{a^2-4d^2}}{\cancel{4}}\Rightarrow d=\dfrac{2a}{5}.$
Therefore the sides of the triangle are $(a-d)=a-\dfrac{2a}{5}=\dfrac{3a}{5} ; a ; (a+d)=a+\dfrac{2a}{5}=\dfrac{7a}{5}.$
Hence the ratio of the sides of the triangle $= \dfrac{3a}{5} : a : \dfrac{7a}{5} = 3:5:7.$
$a=3 , b=5 , c=7.$ So $a+b+c=3+5+7=\boxed{15}.$