A contract on construction job specifies a penalty for delay of completion beyond a a certain date as follows:

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.