Consider the following data for the next three (03) items that follow : The marks obtained by 51 students in a class are in AP with its first term 4 and common difference 3.
What is the mean of the marks ?
79
Understanding the Problem: Calculating Mean of Marks in AP
The problem provides data about the marks obtained by 51 students in a class. These marks form an Arithmetic Progression (AP). We are given the first term and the common difference of this AP. Our goal is to find the mean (average) of these marks.
Breaking Down the Arithmetic Progression (AP) Data
Let's list the information given about the AP of student marks:
- Number of students (and terms in AP), $n = 51$
- First term of the AP, $a_1$ or $a = 4$
- Common difference of the AP, $d = 3$
The marks of the students can be represented as the sequence $4, 4+3, 4+2(3), ..., 4+(51-1)3$.
Calculating the Sum of Marks ($S_n$)
To find the mean of the marks, we first need to calculate the sum of all the marks. The sum of an arithmetic progression ($S_n$) can be found using the formula:
\( S_n = \frac{n}{2}[2a + (n-1)d] \)
Let's substitute the given values into this formula:
\( S_{51} = \frac{51}{2}[2(4) + (51-1)3] \)
\( S_{51} = \frac{51}{2}[8 + (50)3] \)
\( S_{51} = \frac{51}{2}[8 + 150] \)
\( S_{51} = \frac{51}{2}[158] \)
\( S_{51} = 51 \times \frac{158}{2} \)
\( S_{51} = 51 \times 79 \)
Now, let's calculate the product:
\( 51 \times 79 = 4029 \)
So, the total sum of the marks obtained by the 51 students is 4029.
Determining the Mean of the Marks
The mean (average) of a set of numbers is calculated by dividing the sum of the numbers by the count of the numbers. In this case, the mean of the marks is the total sum of marks divided by the number of students.
\( \text{Mean} = \frac{\text{Sum of marks}}{\text{Number of students}} \)
\( \text{Mean} = \frac{S_{51}}{n} \)
Substitute the values we calculated:
\( \text{Mean} = \frac{4029}{51} \)
Let's perform the division:
\( \frac{4029}{51} = 79 \)
Alternatively, using the intermediate step from the sum calculation:
\( \text{Mean} = \frac{51 \times 79}{51} \)
The 51 in the numerator and denominator cancel out, leaving:
\( \text{Mean} = 79 \)
The mean of the marks obtained by the 51 students is 79.
Revision Table: Key AP Concepts for Marks Calculation
| Concept | Formula | Description |
|---|---|---|
| Arithmetic Progression (AP) | $a, a+d, a+2d, ...$ | A sequence where the difference between consecutive terms is constant (common difference $d$). |
| $n$-th term of AP ($a_n$) | $a_n = a + (n-1)d$ | The value of the term at the $n$-th position. |
| Sum of $n$ terms of AP ($S_n$) | $S_n = \frac{n}{2}[2a + (n-1)d]$ OR $S_n = \frac{n}{2}[a + a_n]$ |
The total sum of the first $n$ terms of the sequence. |
| Mean of an AP | $\text{Mean} = \frac{S_n}{n}$ | The average value of the terms in the AP. For an AP, the mean is also the average of the first and last term: $\frac{a_1 + a_n}{2}$. |
Additional Information: Properties of Mean and AP
The mean of an arithmetic progression has some interesting properties:
- For an odd number of terms, the mean is equal to the middle term. In this problem, $n=51$, which is odd. The middle term is the $\frac{51+1}{2} = 26$-th term. Let's find the 26th term: $a_{26} = a + (26-1)d = 4 + 25(3) = 4 + 75 = 79$. This matches the mean we calculated.
- The mean is always the average of the first and the last term: $\frac{a_1 + a_n}{2}$. The first term $a_1=4$. The last term is the 51st term, $a_{51} = a + (51-1)d = 4 + 50(3) = 4 + 150 = 154$. The average of the first and last term is $\frac{4 + 154}{2} = \frac{158}{2} = 79$. This also confirms our result.
- The concept of mean is widely used in statistics to represent the central value of a dataset. When the data is in an AP, the mean calculation simplifies nicely.
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