Question

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs.200 for the first day, Rs250 for the second day, Rs for the third day, etc., 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 word by 30 days.

Answer 1

The given question is in the form of an Arithmetic progression and we will solve it using the formula of sum of n terms of an AP which is given as, ${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$, where a is the first term, n number of terms and d is the common difference.

Complete step by step answer:
According to the given condition in the question, the penalty for the first day = 200, the penalty for the second day = 250 and the penalty for the third day= 300.
Hence the series is obtained as 200, 250, 300……….
Since difference in between the terms if the series is same, it is an AP and the common difference = d = 50
We need to find the total penalty if work is delayed by 30 days, i.e. we need to find S30(Sum of the terms till 30 terms).
We know that the sum of n terms of an AP is given by,
${{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right)$, where a is the first term, n number of terms and d is the common difference.
Here, n = 30, a = 200, d = 50.
Whereas n is the time that is 30 days, a is the minimum penalty i.e. Rs.200, d is the change in penalty i.e. Rs.50.
Putting the values in the formula we get,

& {{S}_{n}}=\dfrac{n}{2}\left( 2a+(n-1)d \right) \\\ & \Rightarrow {{S}_{n}}=\dfrac{30}{2}\left( 2\left( 200 \right)+(30-1)50 \right) \\\ & \Rightarrow {{S}_{n}}=15((400+(29)50) \\\ & \Rightarrow {{S}_{n}}=15((400+1450) \\\ & \Rightarrow {{S}_{n}}=15(1850) \\\ & \Rightarrow {{S}_{n}}=27750 \\\ \end{aligned}$$ **Hence, the total penalty if work is delayed by 30 days is Rs 27750.** **Note:** The possibility of error in this question can be found in the 30th term of the given AP, but it would not be the solution of the question. Because we had to find the total penalty amount till 30 days, so we need to find the sum of n terms till 30.