Question

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?

Answer 1

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

$$\begin{gathered} S_{30}=\frac{30}{2}\left[2\times 200+\left(30-1\right)\times 50\right] \\ =15\left(400+1450\right) \\ =15\times 1850 \\ =27750 \end{gathered}$$

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