Total areas of all right triangles whose sides are in AP
A is the set of areas of all right triangles whose sides (sides of the right triangle) are in an Arithmetic progression. It is given that the elements of A are sorted by ascending order of areas of each of these right triangles. It is also given that the lengths of the sides are positive integers and that the common difference of the respective arithmetic progressions are a positive integers.
What is the sum or total of the first 100 elements of the set A ?
The answer is 2030100.
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1 solution
As the sides of the triangle are in an AP, let the sides of the right triangle be (a-d), a and (a+d) where d is the common difference and a is the first term of an AP.
Using the property of right triangles we know that (a-d)^2 + a^2 = ( a+d)^2
From the above equation we infer that a = 4d.
So the three sides of all right triangles whose sides are in AP are 3d, 4d and 5d . The area of any right triangle with the properties mentioned in the question is hence 6*d^2.
Hence the sum of the areas of the first 100 elements of the set mentioned in the question is 6(1 + 2^2 + 3^2 +...100^2) or is equal to 100 * 101 * 201 = 2030100