Animated Solution for Mathematics - Sequence and Series: Let $3,7,11,15,\dots ,403$ and $2,5,8,11,\dots ,404$ be two arithmetic progressions. Then the sum, of the common terms in them, is equal to

The first train, $A{P}_1$, follows the sequence $3,7,11,15,\dots ,403$. This sequence has a first term $a_1=3$ and a common difference $d_1=4$.

The second train, $A{P}_2$, follows the sequence $2,5,8,11,\dots ,404$. This sequence has a first term $a_2=2$ and a common difference $d_2=3$.

To find the intersection, we identify the first term common to both sequences. By observing the terms, we see that $11$ appears in both $A{P}_1$ and $A{P}_2$.

Thus, the first common term is $a=11$. This serves as the anchor for our new sequence of intersections.

The common terms of two arithmetic progressions form a new arithmetic progression. The common difference $D$ of this new sequence is the Least Common Multiple of the individual differences.

The sequence of common stations is therefore $11,23,35,\dots$ with $a=11$ and $D=12$.

A station is common only if it exists on both tracks. Since $A{P}_1$ ends at $403$ and $A{P}_2$ ends at $404$, the common terms must satisfy $T_n\le 403$.

Using the general term formula $T_n=a+(n-1)D$, we set up the following inequality:

Subtracting $11$ from both sides yields $(n-1)12\le 392$. Dividing by $12$, we obtain:

Since $n$ must be an integer, we take $n-1=32$, which results in $n=33$ total common terms.

Summation

We calculate the sum of these $33$ common terms using the arithmetic series formula:

Substituting $n=33$, $a=11$, and $D=12$:

Simplifying the expression inside the brackets:

The sum of all stations where both trains stop simultaneously is 6699.