Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant. The $n^{\text{th}}$ term is given by $T_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference.
• Common Terms of Two APs: The common terms of two APs also form an AP. The first term of this new AP is the first common term of the original APs, and its common difference is the Least Common Multiple (LCM) of the common differences of the original APs.
• Finding the Last Term of an AP: The $n^{\text{th}}$ term formula, $T_n = a + (n-1)d$, can be used to find the last term if the number of terms is known.

Step-by-Step Solution

Step 1: Analyze the first Arithmetic Progression (AP1). We are given the progression $4, 9, 14, 19, \ldots$. The first term, $a_1$, is $4$. The common difference, $d_1$, is $9 - 4 = 5$. The number of terms is $n_1 = 25$. We need to find the $25^{\text{th}}$ term of this AP to determine the range of its terms. Using the formula $T_n = a + (n-1)d$: $T_{25} = 4 + (25-1) \times 5 = 4 + 24 \times 5 = 4 + 120 = 124$. So, the first AP is $4, 9, 14, \ldots, 124$.

Step 2: Analyze the second Arithmetic Progression (AP2). We are given the progression $3, 6, 9, 12, \ldots$. The first term, $a_2$, is $3$. The common difference, $d_2$, is $6 - 3 = 3$. The number of terms is $n_2 = 37$. We need to find the $37^{\text{th}}$ term of this AP to determine the range of its terms. Using the formula $T_n = a + (n-1)d$: $T_{37} = 3 + (37-1) \times 3 = 3 + 36 \times 3 = 3 + 108 = 111$. So, the second AP is $3, 6, 9, \ldots, 111$.

Step 3: Find the first common term. We need to find the smallest number that appears in both sequences. AP1: $4, 9, 14, 19, 24, 29, 34, \ldots$ AP2: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, \ldots$ By inspection, the first common term is $9$.

Step 4: Find the common difference of the common terms. The common terms of two APs form a new AP. The common difference of this new AP is the LCM of the common differences of the original APs. $d_1 = 5$ and $d_2 = 3$. LCM$(5, 3) = 15$. So, the common difference of the AP formed by the common terms is $15$.

Step 5: Determine the range of the common terms. The common terms must be present in both APs. The terms of AP1 range from $4$ to $124$. The terms of AP2 range from $3$ to $111$. Therefore, the common terms must be less than or equal to the minimum of the last terms of the two APs, which is $\min(124, 111) = 111$. The common terms form an AP with first term $9$ and common difference $15$. Let the $k^{\text{th}}$ common term be $C_k$. $C_k = 9 + (k-1)15$. We need to find the largest value of $k$ such that $C_k \le 111$. $9 + (k-1)15 \le 111$ $(k-1)15 \le 111 - 9$ $(k-1)15 \le 102$ $k-1 \le \frac{102}{15}$ $k-1 \le 6.8$ Since $k-1$ must be an integer, the largest integer value for $k-1$ is $6$. $k-1 = 6 \implies k = 7$.

Step 6: Verify the common terms. The common terms are: $C_1 = 9 + (1-1)15 = 9$ $C_2 = 9 + (2-1)15 = 9 + 15 = 24$ $C_3 = 9 + (3-1)15 = 9 + 30 = 39$ $C_4 = 9 + (4-1)15 = 9 + 45 = 54$ $C_5 = 9 + (5-1)15 = 9 + 60 = 69$ $C_6 = 9 + (6-1)15 = 9 + 75 = 84$ $C_7 = 9 + (7-1)15 = 9 + 90 = 99$ Let's check if these terms are within the bounds of both APs. The last term of AP1 is $124$. All these terms are $\le 124$. The last term of AP2 is $111$. All these terms are $\le 111$. The next common term would be $99 + 15 = 114$, which is greater than the last term of AP2 ($111$), so it's not a common term. Thus, there are $7$ common terms.

Common Mistakes & Tips

• Incorrectly identifying the first common term: Carefully list out initial terms of both sequences to avoid errors.
• Forgetting to consider the upper bounds of both sequences: The common terms must exist within the range of both original sequences. The last common term cannot exceed the smaller of the two last terms.
• Confusing LCM with GCD: The common difference of the common terms is the LCM of the individual common differences, not the GCD.

Summary

To find the number of common terms in two arithmetic progressions, we first determine the individual sequences, including their first terms, common differences, and the number of terms. We then find the last term of each sequence to establish their respective ranges. The common terms themselves form an arithmetic progression whose first term is the smallest common term of the original sequences, and whose common difference is the LCM of the original common differences. By setting the general term of this common AP to be less than or equal to the minimum of the last terms of the original APs, we can determine the total number of common terms. In this case, the first AP goes up to $124$ and the second up to $111$. The common terms start at $9$ and have a common difference of $15$. The common terms are $9, 24, 39, 54, 69, 84, 99$. The next term would be $114$, which is beyond the limit of the second AP. Therefore, there are $7$ common terms.

The final answer is \boxed{7}.