Animated Solution for Mathematics - Sequence and Series: The number of common terms in the progressions $4,9,14,19,\dots$, up to $25^{th}$ term and $3,6,9,12,\dots$, up to $37^{th}$ term is :

The first progression, $A{P}_1$, is defined by its first term $a_1=4$ and a common difference $d_1=5$. It runs for $n_1=25$ terms.

To find its final destination, we use the general term formula $T_n=a+(n-1)d$. Plugging in our values:

The first runner stops at $124$.

Now, consider $A{P}_2$. It starts at $a_2=3$ with a common difference $d_2=3$, running for $n_2=37$ terms. Its final term is:

The crucial realization is that if we are looking for common terms, we are restricted by the shorter path. Even though the first progression continues until $124$, the second one has already finished its journey at $111$.

Therefore, any common term must be less than or equal to $\min (124,111)=111$. This value serves as our search boundary.

Now, let us find the first point of contact. Scanning the early terms, we see $A{P}_1$ contains $4,9,14,\dots$ and $A{P}_2$ contains $3,6,9,12,\dots$.

The number $9$ appears in both sequences. This is our starting point, $a=9$.

The common terms form their own arithmetic progression. The common difference $D$ of this new sequence is the Least Common Multiple of the original differences, $d_1=5$ and $d_2=3$.

Since $5$ and $3$ are coprime, $D=\text{LCM}(5,3)=15$. Our common terms will appear at $9,24,39,\dots$ and so on.

We have the first term $a=9$ and the common difference $D=15$. We can express any common term as $T_n=9+(n-1)15$.

We know this term cannot exceed our boundary of $111$. So, we set up the inequality:

Subtracting $9$ from both sides, we get $(n-1)15\le 102$. Dividing by $15$, we find:

Adding $1$ to both sides, we arrive at $n\le 7.8$. Since $n$ must be a positive integer representing the count of terms, the largest possible value is $n=7$.

We have successfully navigated the paths and found that there are exactly 7 common terms.