Question

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: 200 for the first day, 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay the penalty if he has delayed the work by 30 days?

Answer 1

It is given that the penalty for each succeeding day is Rs. 50 more than the preceding day, so the amount to be paid in each penalty are in AP with common difference Rs. 50

Number of days in the delay of the work = 30

The amount to be paid in each penalty is Rs. 200, Rs. 250, Rs. 300, Rs. 350, .... Up to 30 terms.

Therefore, total amount of money paid by the contractor as penalty, $S_{30}$= Rs. 200 + Rs. 250 +Rs. 300 + Rs. 350+ ... Up to 30

Here, a = Rs. 200, d = Rs. 50 and n = 30

Using the formula, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
we get

$S_{30}=\frac{30}{2}[2\times 200+(30–1)\times 50]$

$=15\times (400+1450)$

$=15\times 1850$
$=27750$

Hence, the contractor has to pay Rs. 27,750 as the penalty.