Question 1150514
a) This problem involves an arithmetic sequence with common difference of -1, and the last term equal to 1.
The nth term of an arithmetic sequence is a_n = a_1 + (n-1)d
So we have 1 = a_1 + (1-n)
Thus a_1 = n
The sum of an arithmetic sequence with n terms is 
S_n = (n/2)(a_1 + a_n)
6480 = a_1^2 + a_1
The solutions are a_1 = 80, -81
Take the positive solution, a_1 = 80
b) The above equation can be written 2S = n((n + 1) -> n^2 + n - 2S = 0
c) n^2 + n - 2*2100 = 0
n^2 + n - 4200 = 0
There are no integer solutions, thus this number of cans cannot be made into a triangular pile