Question

Find the lengths of the sides of a triangle if they are expressed by integers, form an arithmetic progression, and the perimeter of the triangle is equal to 15.

Answer 1

Let the common difference be $d$.

And smallest side ${}^{\prime }{a}^{\prime }$.

Therefore, the sides will be $a,a+d,a+2d$.

$Perimeter=15=3a+3d$

$a+d=5$

Now longest side of triangle should have length less than sum of lengths of remaining sides.

Hence $a+2d<2a+d$

$d<a$.

As $a$ and $d$ should be integers,

$a,d$ can be $4,1or3,2$.

Hence sides can be $4,5,6$ or $3,5,7$.