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?
The contractor has to pay ₹ 27,750 as penalty.
Explanation
This question is from the Arithmetic Progression chapter. The penalties for each day form an AP with first term ₹ 200 and common difference ₹ 50. The total penalty is the sum of first 30 terms of this AP, which can be calculated using the sum formula S_n = n/2 × [2a + (n-1)d].
Solution Steps
Identify the AP parameters: The penalty for first day = ₹ 200, so first term a = 200. Each succeeding day penalty is ₹ 50 more than preceding day, so common difference d = 50. Number of days delayed n = 30.
Apply the sum formula: The total penalty is the sum of first 30 terms of this AP. Using formula: S_n = n/2 × [2a + (n-1)d]
Substitute values: S_30 = 30/2 × [2(200) + (30-1)(50)]
Calculate: S_30 = 15 × [400 + ] = 15 × [400 + 1450] = = ₹ 27,750