Animated Solution for Mathematics - Sequence and Series: The sum of the common terms of the following three arithmetic progressions. $3,7,11,15,\dots ,399$, $2,5,8,11,\dots ,359$ and $2,7,12,17,\dots ,197$, is equal to

Imagine you are standing at the starting line of a grand mathematical race. You have three runners—three arithmetic progressions—each moving at their own pace, with their own unique stride length.

The first runner, $A{P}_1$, takes steps of $4$. The second, $A{P}_2$, takes steps of $3$. And the third, $A{P}_3$, takes steps of $5$.

They are all running along the same path, and you are tasked with finding the exact moments where all three runners land on the same spot simultaneously. This is not just a problem of arithmetic; it is a problem of finding harmony in a system of different rhythms.

When we look for common terms across multiple arithmetic progressions, we are essentially looking for the points of intersection in their periodic behavior. If $A{P}_1$ has a common difference $d_1=4$, it hits numbers like $3,7,11,15,\dots$. If $A{P}_2$ has $d_2=3$, it hits $2,5,8,11,\dots$.

The common terms must satisfy the step-size requirements of all three sequences. This means the new sequence of common terms must have a common difference $D$ that is a multiple of $d_1$, $d_2$, and $d_3$.

To find the most efficient, or the 'fundamental' rhythm, we calculate the Least Common Multiple:

Since $4$, $3$, and $5$ are pairwise coprime—meaning they share no common factors other than $1$—their LCM is simply their product:

This is a powerful realization. It tells us that once we find the first common term, the next one will appear exactly $60$ units later. The chaos of the individual sequences collapses into a predictable, elegant pattern.

Now, how do we find that elusive first common term, $a$? A novice might try to list every term of every sequence, but that is a trap. We are smarter than that.

We use the 'Largest Stride' strategy. We look at the sequence with the largest common difference, which is $A{P}_3$ with $d_3=5$. Because it covers the ground fastest, testing its terms ($2,7,12,17,22,27,32,37,42,47,\dots$) allows us to check fewer candidates.

Let us test $47$:

1. For $A{P}_1$: $47=3+(n-1)4\Rightarrow 44=(n-1)4\Rightarrow n-1=11\Rightarrow n=12$. This fits. 2. For $A{P}_2$: $47=2+(m-1)3\Rightarrow 45=(m-1)3\Rightarrow m-1=15\Rightarrow m=16$. This also fits.

We have found our anchor: $a=47$.

In any physical system, there are constraints. Here, the constraint is the length of the tracks. The sequences do not go on forever.

$A{P}_1$ ends at $399$, $A{P}_2$ at $359$, and $A{P}_3$ at $197$. If we try to find a common term beyond $197$, we are running on empty track.

The common terms must exist within the intersection of all three domains. Therefore, our upper limit $L$ is defined by the shortest sequence:

This is our 'wall.' Any term we generate must be less than or equal to $197$.

With our first term $a=47$, our common difference $D=60$, and our upper limit $L=197$, the sequence of common terms is simple to construct:

If we add another $60$, we get $227$, which is greater than $197$. We must stop. The common terms are $47,107,$ and $167$.

The final step is to sum them up:

There it is. The beauty of the solution lies not in the arithmetic, but in the logic of the structure. We identified the rhythm, found the anchor, respected the boundaries, and arrived at the final answer of 321.