If the average marks of 17 students in a class is A. The marks of the students when arranged in either an ascending or a descending order was found to be in arithmetic progression. The class teacher found that the students who were ranked 3rd, 7th, 8th, 11th, 15th had copied in the exam and hence got all of them rusticated. The average of the remainder of the class was B. Then

Options

A. A = B
B. `A gt B`
C. `A lt B`
D. Data Insufficient

Answer

To solve the problem step by step, we will analyze the information given and derive the relationship between the average marks of the class before and after the rustication of certain students.

### Step 1: Understand the Average Marks
The average marks of 17 students in a class is given as A. This means that the total marks of all students can be calculated as:
\[ \text{Total Marks} = \text{Average} \times \text{Number of Students} = A \times 17 \]

**Hint:** Remember that the average is calculated by dividing the total marks by the number of students.

### Step 2: Identify the Students Affected
The students who were rusticated are ranked 3rd, 7th, 8th, 11th, and 15th. This means that 5 students are removed from the class, leaving us with:
\[ \text{Remaining Students} = 17 - 5 = 12 \]

**Hint:** Keep track of how many students are left after removing those who were rusticated.

### Step 3: Analyze the Marks in Arithmetic Progression
Since the marks of the students are in arithmetic progression (AP), the marks of the students can be represented as:
- 1st student: \( A - 8d \)
- 2nd student: \( A - 6d \)
- 3rd student: \( A - 4d \)
- 4th student: \( A - 2d \)
- 5th student: \( A \)
- 6th student: \( A + 2d \)
- 7th student: \( A + 4d \)
- 8th student: \( A + 6d \)
- 9th student: \( A + 8d \)
- 10th student: \( A + 10d \)
- 11th student: \( A + 12d \)
- 12th student: \( A + 14d \)
- 13th student: \( A + 16d \)
- 14th student: \( A + 18d \)
- 15th student: \( A + 20d \)
- 16th student: \( A + 22d \)
- 17th student: \( A + 24d \)

**Hint:** Understand that in an arithmetic progression, the middle value represents the average.

### Step 4: Calculate the New Average After Rustication
When the students ranked 3rd, 7th, 8th, 11th, and 15th are removed, we need to find the total marks of the remaining students. The marks of the rusticated students are:
- 3rd: \( A - 4d \)
- 7th: \( A + 4d \)
- 8th: \( A + 6d \)
- 11th: \( A + 12d \)
- 15th: \( A + 20d \)

Now, we can calculate the total marks of these students:
\[ \text{Total Marks of Rusticated Students} = (A - 4d) + (A + 4d) + (A + 6d) + (A + 12d) + (A + 20d) = 5A + 38d \]

The total marks of the remaining students is:
\[ \text{Total Marks of Remaining Students} = (A \times 17) - (5A + 38d) = 17A - 5A - 38d = 12A - 38d \]

**Hint:** When calculating the new average, subtract the total marks of the rusticated students from the original total.

### Step 5: Calculate the New Average
The new average \( B \) of the remaining 12 students can be calculated as:
\[ B = \frac{\text{Total Marks of Remaining Students}}{\text{Remaining Students}} = \frac{12A - 38d}{12} \]

**Hint:** Remember to divide the total marks of the remaining students by the number of students left to find the new average.

### Step 6: Compare A and B
To compare the averages \( A \) and \( B \):
- Since the marks of the student ranked 8th (which is \( A + 6d \)) is greater than the average \( A \), removing this student will decrease the average. Thus, \( B < A \).

**Conclusion:** Therefore, the correct answer is that \( A \) is greater than \( B \).

### Final Answer
**A is greater than B.**

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