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

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1. **Find the Common Terms**:
   - The sequences are $$ a_n = 46 + 8n $$ and $$ b_m = 98 + 4m $$.
   - To find common terms, equate $$ a_n = b_m $$:
     $$
     46 + 8n = 98 + 4m
     $$
     Simplify:
     $$
     8n - 4m = 52 \quad \Rightarrow \quad 2n - m = 13
     $$
   - From this, $$ m = 2n - 13 $$.

2. **Smallest Values of $$ n $$ and $$ m $$**:
   - For $$ n = 7 $$, $$ m = 2(7) - 13 = 14 - 13 = 1 $$.
   - The first common term is $$ a_7 = b_1 = 102 $$.

3. **General Form of Common Terms**:
   - The common terms will be in the form:
     $$
     a_n = b_m = 102 + 8k, \quad k \geq 0.
     $$

4. **Last Common Term**:
   - The largest $$ n \leq 100 $$:
     $$
     a_{100} = 46 + (8)(100) = 846
     $$
   - The largest $$ m \leq 100 $$:
     $$
     b_{100} = 98 + (4)(100) = 498
     $$
   - The largest term less than or equal to both values is $$ a_{62} = b_{49} = 494 $$.

5. **Number of Terms**:
   - The sequence of common terms is an arithmetic progression (AP): $$ T_k = 102, 110, ..., 494 $$.
   - First term ($$ T_1$$) = $$102$$, last term ($$ T_{50}$$) = $$494$$, common difference ($$ d$$) = $$8$$.
   - Number of terms ($$ n$$):
     $$
     n = \frac{	ext{Last Term} - 	ext{First Term}}{	ext{Common Difference}} + 1
     $$
     Substituting values:
     $$
     n = \frac{494 - 102}{8} + 1 = 50.
     $$

6. **Sum of the AP**:
   - Sum of an AP is given by:
     $$
     S_n = \frac{n}{2} (T_1 + T_n)
     $$
   - Substituting values ($$ n=50, T_1=102, T_n=494$$):
     $$
     S_{50} = \frac{50}{2} (102 + 494) = (25)(596) = 14900.
     $$

Thus, the sum of all terms common to both sequences is **14900**, corresponding to option **B**.

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