Question

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for 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?

Answer 1

Hint: If you will notice, you will find that the sequence of penalties is forming an A.P. with a common difference of Rs. 50 and the first term of the A.P. is Rs. 200. So, just apply the formula of sum of first 30 terms of an A.P. to get the required answer.
Complete step-by-step answer:
Before starting with the solution, let us discuss what an AP is. AP stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by ${{T}_{r}}$ and sum till r terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)=\dfrac{r}{2}\left( a+l \right)$
Now moving to the solution to the above question. If you will notice, you will find that the sequence of penalties is forming an A.P. with a common difference of Rs. 50 and the first term of the A.P. is Rs. 200.
Therefore, the penalty that would be charged for a delay of 30 days is the sum of the first 30 terms of the given A.P.
${{S}_{30}}=\dfrac{30}{2}\left( 2\times 200+\left( 30-1 \right)50 \right)=15(400+1450)=27750.$
So, the contractor has to pay a penalty of Rs. 27,750 for a delay of work by 30 days.

Note: Always, while dealing with an arithmetic progression, try to find the first term and the common difference of the sequence, as once you have these quantities, you can easily solve the questions related to the given arithmetic progressions.