A contract on construction job specifies a penalty for delay in completion of the work beyond a certain 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 Rs 50 more than that of the preceding day. How much penalty should the contractor pay if he delays the work by 10 days?

Correct Option (B)

The problem describes a penalty structure that forms an arithmetic progression (AP).

• The first day's penalty (first term, a) = Rs 200.
• The increase in penalty for each succeeding day (common difference, d) = Rs 50.
• The total number of days of delay (number of terms, n) = 10.

The total penalty is the sum of this arithmetic progression for 10 terms. The formula for the sum of an AP is:

$S_n=\frac{n}{2}[2a+(n-1)d]$

Substituting the given values:

$S_{10}=\frac{10}{2}[(2\times 200)+(10-1)50]$

$S_{10}=5[400+(9)50]$

$S_{10}=5[400+450]$

$S_{10}=5(850)$

$S_{10}=4250$

Therefore, the contractor should pay a total penalty of Rs 4250.

Incorrect Options:

Options (A) Rs 4950, (C) Rs 3600, and (D) Rs 650 are incorrect as they do not correspond to the sum of the arithmetic progression calculated based on the specified penalty structure and delay period.