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Question
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 does the contractor have to pay as a penalty if he has delayed the work by 30 days?
Sum
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Solution
Here, penalty for delay on
1th day = 200
2nd day = 250
3rd day = 300
Now, 200, 250, 300, etc. are in AP such that a = 200,
d = 250 - 200 = 50
S30 is given by
S30 = `30/2 [2 (200) + (30 - 1)xx50]` ..[using, `S_n = n/2 [2a + (n -1)]d`]
= 15 [400 + 29 × 50]
= 15 [400 + 1450]
= 15 × 1850
= 27,750
Thus, a penalty for the delay for 30 days is < 27,750.
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