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

The first series goes as follows:

46, 54, 62, 70, 78, 86, 94, 102,....

The second series goes as follows:

98, 102, 106, 110,...

The first common term is 102 (first term of the common terms) and the common difference between them will be hcf(4,8) = 8

=> The required sequence is 102, 110, 118,..... (last term should be less than 468 (100th term of second series))

=> 102 + (n-1)(8) $$\le$$ 498

=> n is less than or equal to 50.5  =>  n = 50

Using the summation of A.P. formula:

Required sum = $$\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)=\dfrac{50}{2}\left(2\times\ 102+49\times\ 8\right)=14900$$