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

The correct option is 14900.

Let us write down the two sequences given in the problem statement:

The first sequence is given by: $a_n=46+8n$ for $n\le 100$ where $n\in \mathbb{N}$ (i.e., $n=1,2,3,\dots ,100$).

The second sequence is given by: $b_m=98+4m$ for $m\le 100$ where $m\in \mathbb{N}$ (i.e., $m=1,2,3,\dots ,100$).

To find the terms that are common to both sequences, we equate their general terms:

$46+8n=98+4m$

Simplifying the equation:

$8n-4m=98-46$

$8n-4m=52$

Dividing the entire equation by 4:

$2n-m=13$

$m=2n-13$

We are given that $m$ must be a natural number and $1\le m\le 100$. We can find the range of valid values for $n$ using this condition:

1. For the lower bound of $m$:

$m\ge 1\Rightarrow 2n-13\ge 1$

$2n\ge 14\Rightarrow n\ge 7$

2. For the upper bound of $m$:

$m\le 100\Rightarrow 2n-13\le 100$

$2n\le 113\Rightarrow n\le 56.5$

Since $n$ must be an integer, we have $n\le 56$.

Thus, the valid range of $n$ for the common terms is $7\le n\le 56$.

Let us determine the total number of common terms ($k$):

$k=56-7+1=50$

Now, we can find the first and last common terms:

First common term (for $n=7$):

$a=46+8\left(7\right)=46+56=102$

Last common term (for $n=56$):

$l=46+8\left(56\right)=46+448=494$

These common terms form an Arithmetic Progression (AP) with a first term of 102, a last term of 494, and a total of 50 terms.

The sum of an arithmetic progression is given by the formula:

$S=\frac{k}{2}(a+l)$

Substituting the values of $k$, $a$, and $l$:

$S=\frac{50}{2}(102+494)$

$S=25\times 596$

$S=14900$

Therefore, the sum of all common terms is 14900.