Animated Solution for Mathematics - Sequence and Series: The $8^{th}$ common term of the series $S_1=3+7+11+15+19+\dots$, $S_2=1+6+11+16+21+\dots$ is

Imagine two runners on a track. Runner One starts at the $3$-meter mark and leaps forward in consistent $4$-meter bounds. Runner Two starts at the $1$-meter mark and takes larger, $5$-meter leaps.

Our goal is to find the eighth time they land on the exact same spot. This is the essence of finding common terms in arithmetic progressions.

Let us first analyze our runners. Runner One follows the series $S_1=3,7,11,15,19,\dots$. Here, the first term $a_1=3$ and the common difference $d_1=4$.

Runner Two follows $S_2=1,6,11,16,21,\dots$. Here, the first term $a_2=1$ and the common difference $d_2=5$. They are moving at different speeds, starting from different positions.

If you scan the sequences, you will notice something fascinating. In $S_1$, we have $3,7,11,15,19$. In $S_2$, we have $1,6,11,16,21$.

There it is! At the number $11$, they collide. This is our first common term, $a=11$. This is the anchor point for our new sequence.

The sequence of common terms is not random; it is itself an Arithmetic Progression. For the runners to meet again, the distance traveled must be a multiple of both $4$ and $5$.

This is the definition of the Least Common Multiple, or LCM. Since $4$ and $5$ are co-prime, their LCM is simply their product:

This means that after they meet at $11$, they will meet again exactly $20$ units later. Our new series is $11,31,51,71,\dots$.

We have reduced a complex problem of two series into a single, elegant AP. We need the $8^{th}$ term. We use our master key, the formula for the $n^{th}$ term:

With $n=8$, $a=11$, and $d=20$, the calculation becomes a simple, satisfying exercise:

Simplifying the bracket gives us $T_8=11+7\times 20$. Multiplying $7$ by $20$ yields $140$, and adding $11$ brings us to our destination: