March 19, 2025

Finding the Final Grade in a Recurring Exam System

A student, whenever they take an exam, scores the arithmetic mean of the grades obtained in the previous two exams (except for the first two exams, where they received grades $a_1$ and $a_2$ ). We define $a_n$ as the grade of the $n$ -th exam. If we assume the student is taking an infinitely long degree program and their final grade is given by $\lim_{n \to \infty} a_n$ , determine their final grade.

Step 1: Characteristic Equation

The recurrence equation can be written as:

$2a_n - a_{n-1} - a_{n-2} = 0.$

The associated characteristic polynomial is:

$2x^2 - x - 1 = 0.$

Solving this quadratic equation:

$x = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}.$

We obtain two roots:

$x_1 = 1, \quad x_2 = -\frac{1}{2}.$

Thus, the general solution of the sequence is:

$a_n = A(1)^n + B\left(-\frac{1}{2}\right)^n.$

Which simplifies to:

$a_n = A + B \left(-\frac{1}{2}\right)^n.$

Step 2: Determining the Limit

Since $\left(-\frac{1}{2}\right)^n$ tends to $0$ as $n \to \infty$ , it follows that:

$\lim_{n \to \infty} a_n = A.$

To find $A$ and $B$ , we impose the initial conditions:

$a_1 = A + B\left(-\frac{1}{2}\right),$

$a_2 = A + B\left(-\frac{1}{2}\right)^2.$

Solving the system of equations:

$a_1 = A - \frac{B}{2},$

$a_2 = A + \frac{B}{4}.$

Summing both equations:

$a_1 + a_2 = 2A - \frac{B}{2} + \frac{B}{4}.$

Factoring $B$ :

$a_1 + a_2 = 2A - B\left(\frac{1}{4}\right).$

Solving for $A$ :

$A = \frac{a_1 + 2a_2}{3}.$

Conclusion

The final grade of the student, after an infinite number of exams, is:

$\lim_{n \to \infty} a_n = \frac{a_1 + 2a_2}{3}.$