Students Progression in Classes I to X

Let the number of students in classes I through X be $a, a+d, a+2d, \dots, a+9d$.

Sum of Class I to IV = $4a + 6d = 462 \implies 2a + 3d = 231$ ... (1)

Sum of Class I to V = $5a + 10d$ Sum of Class VI to X = $5a + 35d$ Given: $5a + 10d = 2(5a + 35d) \implies 5a + 10d = 10a + 70d \implies 5a + 60d = 0 \implies a + 12d = 0$.

From $a = -12d$, substitute into (1): $2(-12d) + 3d = 231 \implies -21d = 231 \implies d = -11$. Then $a = -12(-11) = 132$.

Class VI has $a + 5d = 132 + 5(-11) = 132 - 55 = 77$.