 Hello and welcome to the session. Let us understand the following question today. A contract on construction jobs 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? Now let us write the solution. It is given to us that penalty for first day is equal to Rs. 200, then penalty for second day is equal to Rs. 250 and penalty for third day is equal to Rs. 300 and so on we can get the penalty for each successive day by adding Rs. 350 to it. Now thus AP formed is Rs. 50, Rs. 300, Rs. 350 and so on. Here Rs. 200 is equal to Rs. 250 minus Rs. 200 which is equal to Rs. 50. N is equal to, since it is given the question that worker has delayed the work by 30 days, so N is equal to 30 and paid by the worker will be the sum of the charges up to 30 days. So we have to find the sum and we know SN is equal to N by 2 2A plus N minus 1D. Now substituting the values and we have to find S30, so 30 by 2 multiplied by minus 1 multiplied by 50 which is equal to 30 gets cancelled by 2 and we get here 15. So 15 multiplied by 400 plus 29 into 50 which is equal to 15 multiplied by 41450 which is equal to 15 multiplied by 1850 which is equal to 27,750. So by the worker 7,750. I hope you have answered the question. Bye and have a nice day.