A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 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 has the contractor to pay as penalty if he has delayed the work by 30 days?
Rs \(27750\)
Step 1: Recognise the pattern as an AP.
The penalties form an arithmetic progression (AP):
First day: Rs 200
Second day: Rs 250
Third day: Rs 300
...
Each day the penalty increases by Rs 50.
So, first term \(a = 200\) and common difference \(d = 50\).
The contractor is late by 30 days, so the number of terms \(n = 30\).
Step 2: Recall the sum formula for an AP.
The sum of the first \(n\) terms of an AP is:
\[S_n = \frac{n}{2}\big[2a + (n - 1)d\big]\]
Here, we need \(S_{30}\).
Step 3: Substitute the known values.
\[S_{30} = \frac{30}{2}\big[2 \cdot 200 + (30 - 1) \cdot 50\big]\]
First simplify each part:
\[\frac{30}{2} = 15\]
\[2 \cdot 200 = 400\]
\[(30 - 1) = 29\]
\[29 \cdot 50 = 1450\]
Now add inside the bracket:
\[2a + (n - 1)d = 400 + 1450 = 1850\]
So,
\[S_{30} = 15 \times 1850\]
Step 4: Multiply to get the total penalty.
\[15 \times 1850 = 27750\]
Conclusion: The contractor has to pay a total penalty of Rs 27,750.