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,…

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:

$S_n = \frac{n}{2}\left\lbrack 2a + (n - 1)d \right\rbrack$Sn​=2n​[2a+(n−1)d]

Substituting the given values:

$S_{10} = \frac{10}{2}\left\lbrack (2 \times 200) + (10 - 1)50 \right\rbrack$S10​=210​[(2×200)+(10−1)50]

$S_{10} = 5\left\lbrack 400 + (9)50 \right\rbrack$S10​=5[400+(9)50]

$S_{10} = 5\left\lbrack 400 + 450 \right\rbrack$S10​=5[400+450]

$S_{10} = 5(850)$S10​=5(850)

$S_{10} = 4250$S10​=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.