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# A contractor 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 the preceding day. How much money the contractor has to pay as penalty, if he delayed the work by 30 days.

Last updated date: 20th Jun 2024
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Hint: The problem is of arithmetic progression as on each day after the cutoff date 50 Rs. is being added as the penalty.

Complete step by step solution: It is given that for the first day after the cutoff date the penalty is 200 Rs.
On the second day the penalty is 250 Rs.
On the third day the penalty is 300 Rs.
This continues for 30 days like this after the cutoff day and 50 Rs. is being added on each day as the penalty.
This follows a unique series, which is better known as arithmetic progression series. The first term of the series is 200, second is 250, and the third is 300.
The total penalty is to be calculated at the end of 30th day. Therefore, the number of terms of an AP is 30.
The N-th term of an arithmetic progression is given by,
${a_n} = a + \left( {n - 1} \right)d \cdots (1)$
Where,
$a =$ First term
$n =$ Number of terms.
$d =$ Common difference.
Here, $a = 200$ , $n = 30$ and $d = 50$.
Substituting the values in equation (1),
${a_n} = 200 + (30 - 1)50$
$\begin{gathered} {a_n} = 200 + \left( {29} \right)50 \\ {a_n} = 200 + 1450 \\ {a_n} = 1650 \\ \end{gathered}$
Therefore, the contractor has to pay Rs.1650 as penalty at the end of the 30-th day of the work.

Note: This is a generalized real life problem which is formed into an arithmetic progression.
The important thing is to realize that the successive terms of the sequence differ by a common number which is 50 Rs. in this case. The formulas to be remembered for the arithmetic progression
N-th term, ${a_n} = a + (n - 1)d$
Sum of N-th term, ${S_n} = \dfrac{a}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$