A contract on construction job specifies a penalty for delay of completion beyond a a certain date as follows:
A contract on construction job specifies a penalty for delay of completion beyond a a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he delayed the work by 30 days?
It is given that the penalty for each succeeding day is ₹50 more than for the preceding day, so the amount of penalties are in AP with common difference ₹50.
Number of days in the delay of the work = 30
The amount of penalties are ₹200, ₹250, ₹300,... up to 30 terms.
∴ Total amount of money paid by the contractor as penalty, S30 = ₹200 + ₹250 + ₹300 + ... up to 30 terms
Here, a = ₹200, d = ₹50 and n = 30
Using the formula, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$, we get
$S_{30}=\frac{30}{2}[2 \times 200+(30-1) \times 50]$
$=15(400+1450)$
$=15 \times 1850$
$=27750$
Hence, the contractor has to pay ₹27,750 as penalty.