## Friday, 21 October 2011

### Progressions-3 (CSAT-2011)

A contract on construction job specifies a penalty for delay in completion of work beyond a certain date is 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 is Rs 50 more than the preceding day. How much penalty should the contractor pay if he delays the work by 10 days?
(A) Rs 4950
(B) Rs 4250
(C) Rs 3600
(D) Rs 650
Solution follows here:
Solution:
Penalty is ----- Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day...
Penalty per day follows an arithmetic progression: 200,250,300,....
With initial term = a = 200 and common difference = d = 50.
Here we have two formulae, one is for nth term in A.P, tn = a+(n-1)d
and the second one is Sum of n-terms in A.P, Sn = n/2 * {2a+(n-1)d}
The only chance of mistake in this problem is here: whether to find penalty on 10th day or penalty for 10 days. It is asked that “How much penalty should the contractor pay if he delays the work by 10 days”
=> it is asked -penalty for 10 days
=> Sum to n-terms formula should be used
=>   Sn = 10/2 * (2*200 + 9*50) = 5*850 = 4250
If we go on the wrong way and find out tn, we get 650, which is also available in the answer options. So, here we should be careful.