In a school, the number of students in each class, from Class I to X, in that order, are in an arithmetic progression. The total number of students from Class I to V is twice the total number of students from Class VI to X.
If the total number of students from Class I to IV is 462, how many students are there in Class VI?

Question

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Answer 1

Let a represent the number of students in Class I, and let d be the common difference of the arithmetic progression.

Step 1: Formulate equations for the total number of students. The total number of students from Class I to IV is:

S₄ = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Given S₄ = 462:

4a + 6d = 462

Simplifying the equation yields:

2a + 3d = 231 (1)

The total number of students from Class I to V is:

S₅ = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d

The total number of students from Class VI to X is:

S_5-10 = (a + 5d) + (a + 6d) + (a + 7d) + (a + 8d) + (a + 9d) = 5a + 35d

Given the condition S₅ = 2S_5-10:

5a + 10d = 2(5a + 35d)

Simplifying this equation:

5a + 10d = 10a + 70d

Rearranging terms, we get:

5a = −60d, which simplifies to a = −12d (2)

Step 2: Solve the system of equations. Substitute the expression for a from equation (2) into equation (1):

2(−12d) + 3d = 231

−24d + 3d = 231

−21d = 231

Solving for d:

d = −11

Substitute the value of d = −11 back into the equation a = −12d:

a = −12(−11) = 132

Step 3: Determine the number of students in Class VI. The number of students in Class VI is given by the formula a + 5d:

a + 5d = 132 + 5(−11) = 132 − 55 = 88

Answer: 88

The Correct Option is D