Posted by jazzmin on Sunday, June 5, 2011 at 10:08pm.
The cost is the arithmetic progression,
A(n)=A(1)+(n-1)d
where A(1)=100, d=25
and the sum to the nth term is
S(n)=(n/2)(A(1)+A(n)
For n=9 (90')
S(n)=(9/2)(100 + 100+(9-1)*25)
=?
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