Question

the sides of a right angled triangle are in a.p.,prove that they are in the ratio of 3,4,5.

Answer 1

sides must be a, a+d, and a+2d
a^2 + (a+d)^2 = (a+2d)^2
a^2 + a^2 + 2ad + d^2 = a^2 + 4ad + 4d^2
a^2 - 2ad - 3d^2 = 0
(a - 3d)(a+d) = 0
a = 3d or a = -d

case 1: a = 3d
let d = 1
then a = 3 , and the sides are 3, 4, and 5
let d = 2
then a = 6, and the sides are 6, 8, and 10
which is just a multiple of 3,4,5, thus a similar triangle.
6:8:10 = 3:4:5

case2: a = -d , d could only be a negative
if d = -1, a = 1, the other sides are 0, and -1 ,
silly
if d = -5, a = 5, then the other sides are 4 and 3
we already had that
if d = -6, a = 6, and the other sides are 5 and 4
but 6^2 ≠ 5^2 + 4^2

go with case 1