Let $a_n = 46 + 8n$ and $b_n = 98 + 4n$ be two sequences for natural numbers $n \le 100$. Then, the sum of all terms common to both the sequences is

We need to find which numbers appear in both sequences, then add them up.

---

For a term to be common to both sequences:

$46 + 8n_1 = 98 + 4n_2$

Rearranging: $8n_1 - 4n_2 = 52$

Dividing by 4: $2n_1 - n_2 = 13$

Therefore: $n_2 = 2n_1 - 13$

---

Since both $n_1$ and $n_2$ must be natural numbers between 1 and 100:

For $n_2 \geq 1$: $2n_1 - 13 \geq 1$, so $n_1 \geq 7$

For $n_2 \leq 100$: $2n_1 - 13 \leq 100$, so $n_1 \leq 56$

Therefore: $n_1 \in \{7, 8, 9, ..., 56\}$

---

Let's verify with the first common term:

When $n_1 = 7$: $a_7 = 46 + 8(7) = 102$ and $n_2 = 2(7) - 13 = 1$, so $b_1 = 98 + 4(1) = 102$ ✓

---

The common terms are: $a_{n_1} = 46 + 8n_1$ where $n_1 = 7, 8, 9, ..., 56$

We need:

$\sum_{n_1=7}^{56} (46 + 8n_1) = \sum_{n_1=7}^{56} 46 + 8\sum_{n_1=7}^{56} n_1$

Number of terms: $56 - 7 + 1 = 50$ terms

$\sum_{n_1=7}^{56} 46 = 46 \times 50 = 2300$

For $\sum_{n_1=7}^{56} n_1$:

$\sum_{n_1=7}^{56} n_1 = \sum_{n_1=1}^{56} n_1 - \sum_{n_1=1}^{6} n_1 = \frac{56 \times 57}{2} - \frac{6 \times 7}{2} = 1596 - 21 = 1575$

Therefore: $8 \times 1575 = 12600$

Final Answer: $2300 + 12600 = 14900$