Question

A contract on construction job specifies a penalty for delay in completion of the work beyond acertain date is as follows: Rs. 200 for the first day, Rs.250 for the second day, Rs. 300 for the third day etc., the penalty for each succeeding day being 50 more than that of the preceding day. How much penalty should the contractor pay if he delays the work by 10 days?

• A. Rs. 4950
• B. Rs. 4250
• C. Rs. 3600
• D. Rs. 650

Answer 1

The correct option is B Rs. 4250

The penalty for delay on first day is 200, for second day is 250, for third day is 300 and so on.
The penalties form an arithmetic progression (AP) with a common difference of 50 and first element
as 200.

The sum of penalties for 10 days is thus same as the sum of numbers in the A.P. with n as 10.
Sum of AP (formula) $=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{10}{2}[2\times 200+(10-1)\times 50]=4250.$