Key Concepts and Formulas

• Arithmetic Progression (A.P.): A sequence of numbers such that the difference between the consecutive terms is constant. This constant difference is called the common difference, denoted by $d$.
• N-th Term Formula of an A.P.: The $n$-th term ($a_n$) of an A.P. is given by $a_n = a_1 + (n-1)d$, where $a_1$ is the first term and $d$ is the common difference.
• Divisors of an Integer: An integer $a$ is a divisor of an integer $b$ if $b$ can be written as $b = ka$ for some integer $k$.

Step-by-Step Solution

Step 1: Setting up the Fundamental Equation

We are given an Arithmetic Progression with the first term $a_1 = 100$ and the last term $a_n = 199$. The formula for the $n$-th term of an A.P. is $a_n = a_1 + (n-1)d$. Substituting the given values, we get: $199 = 100 + (n-1)d$ Subtracting 100 from both sides to isolate the term involving $n$ and $d$: $199 - 100 = (n-1)d$ $99 = (n-1)d$ This equation establishes a relationship between the number of terms ($n$) and the common difference ($d$).

Step 2: Analyzing the Constraints on $n$ and $d$

The problem states the following constraints:

1. The common difference, $d$, must be an integer ($d \in \mathbb{Z}$).
2. The number of terms, $n$, must be at least 3 and at most 33 ($3 \le n \le 33$).

From the constraint on $n$, we can determine the possible range for $(n-1)$: Since $n \ge 3$, then $n-1 \ge 3-1$, which means $n-1 \ge 2$. Since $n \le 33$, then $n-1 \le 33-1$, which means $n-1 \le 32$. Therefore, the possible range for $(n-1)$ is $2 \le n-1 \le 32$.

From the equation $99 = (n-1)d$, and knowing that $d$ must be an integer, it implies that $(n-1)$ must be an integer divisor of 99. Also, since $a_1 = 100$ and $a_n = 199$, and $n \ge 3$, the terms are increasing, so $d$ must be positive. Consequently, $(n-1)$ must also be positive.

Step 3: Identifying Possible Values for $(n-1)$

We need to find the positive integer divisors of 99. First, find the prime factorization of 99: $99 = 9 \times 11 = 3^2 \times 11^1$. The positive divisors of 99 are: $3^0 \times 11^0 = 1$ $3^1 \times 11^0 = 3$ $3^2 \times 11^0 = 9$ $3^0 \times 11^1 = 11$ $3^1 \times 11^1 = 33$ $3^2 \times 11^1 = 99$ So, the set of positive integer divisors of 99 is $\{1, 3, 9, 11, 33, 99\}$.

Now, we must filter these divisors based on the constraint $2 \le n-1 \le 32$:

• If $n-1 = 1$: This implies $n=2$. This violates the condition $n \ge 3$. So, this is not a valid case.
• If $n-1 = 3$: This satisfies $2 \le 3 \le 32$. This is a valid case.
• If $n-1 = 9$: This satisfies $2 \le 9 \le 32$. This is a valid case.
• If $n-1 = 11$: This satisfies $2 \le 11 \le 32$. This is a valid case.
• If $n-1 = 33$: This implies $n=34$. This violates the condition $n \le 33$. So, this is not a valid case.
• If $n-1 = 99$: This implies $n=100$. This violates the condition $n \le 33$. So, this is not a valid case.

Thus, the only possible values for $(n-1)$ that satisfy all conditions are $3, 9, 11$.

Step 4: Calculating the Corresponding Common Differences ($d$)

For each valid value of $(n-1)$, we find the corresponding common difference $d$ using the relation $d = \frac{99}{n-1}$:

1. When $n-1 = 3$: $d = \frac{99}{3} = 33$. For this case, $n=4$, which is within the allowed range.

2. When $n-1 = 9$: $d = \frac{99}{9} = 11$. For this case, $n=10$, which is within the allowed range.

3. When $n-1 = 11$: $d = \frac{99}{11} = 9$. For this case, $n=12$, which is within the allowed range.

These are the only possible integral common differences for A.P.'s that meet the given criteria.

Step 5: Summing the Common Differences

The problem asks for the sum of all such common differences. Sum $= 33 + 11 + 9 = 53$.

Common Mistakes & Tips

• Forgetting the integer constraint on $d$: If $d$ were not required to be an integer, there would be infinitely many possibilities for $n-1$ and $d$.
• Not checking the range of $n$: It is crucial to verify that the calculated values of $n$ (derived from $n-1$) fall within the specified range of $3 \le n \le 33$. Forgetting this can lead to including invalid common differences.
• Missing divisors: Systematically finding all divisors of 99 using prime factorization prevents errors and ensures all potential cases are considered.

Summary

The problem involves finding Arithmetic Progressions with a fixed first and last term, and integral common differences, subject to constraints on the number of terms. By using the $n$-th term formula, we derived the relationship $99 = (n-1)d$. The integer constraint on $d$ means $(n-1)$ must be a divisor of 99. We then applied the constraint on the number of terms ($3 \le n \le 33$) to narrow down the possible values of $(n-1)$ to $\{3, 9, 11\}$. Calculating the corresponding common differences ($33, 11, 9$) and summing them yielded the final answer.

The final answer is $\boxed{53}$.