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Q15:
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 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?

Solution :

Given:

The penalty for the first day ($a_1$) = ₹ $200$.

The penalty for the second day ($a_2$) = ₹ $250$.

The penalty for the third day ($a_3$) = ₹ $300$.

The common difference ($d$) between consecutive days = $250 - 200 = 50$ and $300 - 250 = 50$.

The total number of days of delay ($n$) = $30$.

To Find:

The total penalty amount to be paid for $30$ days, which is the sum of the first $30$ terms of the arithmetic progression ($S_{30}$).

Step 1: Identifying the Progression

The sequence of penalties forms an Arithmetic Progression (AP) because the difference between consecutive terms is constant.

The sequence is: $200, 250, 300, \dots$

Here, the first term $a = 200$.

The common difference $d = 50$.

The number of terms $n = 30$.

Step 2: Selecting the Formula

To find the sum of the first $n$ terms of an arithmetic progression, we use the formula:

$S_n = \frac{n}{2} [2a + (n - 1)d]$

[Where $S_n$ is the sum of $n$ terms, $a$ is the first term, $n$ is the number of terms, and $d$ is the common difference.]

Step 3: Substituting the Values

Substitute $n = 30$, $a = 200$, and $d = 50$ into the formula:

$S_{30} = \frac{30}{2} [2(200) + (30 - 1)50]$

Step 4: Performing the Calculations

First, simplify the fraction outside the brackets:

$S_{30} = 15 [2(200) + (29)50]$

Next, calculate the values inside the brackets:

$S_{30} = 15 [400 + 1450]$

[Since $2 \times 200 = 400$ and $29 \times 50 = 1450$]

Add the values inside the brackets:

$S_{30} = 15 [1850]$

Finally, multiply the result by $15$:

$S_{30} = 27750$

Conclusion:

The total penalty for a delay of $30$ days is calculated by summing the arithmetic series of the daily penalties.

Final Answer: The contractor has to pay a total penalty of ₹ 27,750.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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