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, ₹ 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 has delayed the work by 30 days?

Answer 1

Penalty for first day = ₹ 200

Penalty for second day = ₹ 250

Penalty for third day = ₹ 300

It is given that penalty for each succeeding day is ₹ 50 more than the preceding day.

This makes it an arithmetic progression because the difference between consecutive terms is constant.

We want to know how much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

So, we have an AP of the form 200, 250, 300, 350....30 terms

First term $=a=200$

Common difference $=d=50$

$n=30$

Applying formula, $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$ to find sum of n terms of AP , we get

$S_n=\frac{30}{2}(400+(30-1\left)50\right)$

$=15(400+29\times 50)$

$=15(400+1450)$

$=27750$

Therefore, penalty for 30 days is ₹ 27,750.