Finding the Median of Marks in an Arithmetic Progression

The problem provides data about the marks obtained by 51 students in a class. These marks form an Arithmetic Progression (AP) with a given first term and common difference. We need to find the median of these marks.

Understanding Arithmetic Progression (AP)

An Arithmetic Progression is a sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference.

In this problem:

• The first term ($a_1$) is 4.
• The common difference ($d$) is 3.
• The number of terms ($n$), which is the number of students, is 51.

The terms of the AP can be represented as $a_1, a_1+d, a_1+2d, \dots, a_1+(n-1)d$.

Calculating the Median

The median is the middle value in a dataset that is ordered from least to greatest. When the number of observations ($n$) is odd, the median is the value of the term at the position $\frac{n+1}{2}$.

In this case, the number of students ($n$) is 51, which is an odd number. The marks are in an AP, which means they are already ordered.

The position of the median term is:

\begin{math}\text{Median position} = \frac{n+1}{2} = \frac{51+1}{2} = \frac{52}{2} = 26\end{math}

So, the median of the marks is the value of the 26th term in the arithmetic progression.

Finding the 26th Term of the AP

The formula for the $k$-th term ($a_k$) of an Arithmetic Progression is:

\begin{math}a_k = a_1 + (k-1)d\end{math}

Here, we want to find the 26th term ($k=26$).

• $a_1 = 4$
• $d = 3$
• $k = 26$

Substitute these values into the formula:

\begin{math}a_{26} = 4 + (26-1) \times 3\end{math}

\begin{math}a_{26} = 4 + (25) \times 3\end{math}

\begin{math}a_{26} = 4 + 75\end{math}

\begin{math}a_{26} = 79\end{math}

The 26th term of the AP is 79. Therefore, the median of the marks obtained by the 51 students is 79.

Let's verify the first few terms and the last term:

• 1st term: $a_1 = 4$
• 2nd term: $a_2 = 4 + 3 = 7$
• 3rd term: $a_3 = 4 + 2(3) = 10$
• ...
• 26th term: $a_{26} = 79$
• ...
• 51st term: $a_{51} = 4 + (51-1) \times 3 = 4 + 50 \times 3 = 4 + 150 = 154$

The sorted list of marks starts with 4 and ends with 154. The 26th term, 79, is indeed the middle value with 25 terms before it and 25 terms after it.

The median of the marks is 79.

Summary of Calculation

Revision Table: Key Concepts

Additional Information: Median in Data Analysis

The median is a measure of central tendency. It is often used instead of the mean (average) when the data might contain outliers, as the median is less affected by extreme values.

• Median for Even 'n': If the number of observations 'n' is even, the median is typically calculated as the average of the two middle terms. These are the $(n/2)$-th term and the $(n/2 + 1)$-th term.
• Mean: The sum of all values divided by the number of values.
• Mode: The value that appears most frequently in the dataset.

In this problem, since the data is in an AP and the number of terms is odd, finding the middle term directly provides the median.