What is the best way to download the problem of arithmetic progressions of ncert of exercise 5.3 of class 10th math, how i solve this easy method A contract on 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 the contractor has to pay as penalty, if he has delayed the work by 30 days.
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A contract on 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 the contractor has to pay as penalty, if he has delayed the work by 30 days. Q.15
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We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.
Therefore, a = 200 and d = 50
Penalty that has to be paid if contractor has delayed the work by 30 days = S30
By the formula of sum of nth term, we know,
Sn = n/2[2a+(n -1)d]
Therefore,
S30= 30/2[2(200)+(30 – 1)50]
= 15[400+1450]
= 15(1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.