Penalty for delay of work
A contractor has agreed to pay a penalty if he uses more than a specified length of time to finish a certain job. The penalties for excess time are for the first day and, thereafter, more for each day than the preceding day. If he pays a total penalty of , how many excess days did he need to finish the work?
The answer is 36.
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1 solution
The penalties are $ 2 5 , $ 3 0 , $ 3 5 , . . . , which form an A.P. where a 1 = 2 5 , d = 5 , and s = 4 0 5 0 . We wish to find the number of terms, n .
From, s = 2 n [ 2 a 1 + ( n − 1 ) d ] ,
4 0 5 0 = 2 n [ 5 0 + 5 ( n − 1 ) ]
Multiply by 2 on both sides , and simplify:
n 2 + 9 n − 1 6 2 0 = 0
By use of factoring or the quadratic formula, we find n = 3 6 and n = − 4 5 . Hence, there are 3 6 excess days.