Answer:
H
Step-by-step explanation:
Let represent A and B
A and B is the sum of the first 50 consecutive multiples of 3 and 6, this is the same as a arithmetic sequence because arithmetic sequences have a common sum.
We can represent this as
[tex]a _{n} = a {}^{1} + d(n - 1)[/tex]
where a^1 is the first term, d is the common sum, n is the number of multiples we adding.
The first term for both A and B respectively is 3 and 6.
Multiples implies that the Difference between A and B are 3 and 6 respectively. There are 50 consecutively. multiples.
Plug in the known parts for both.
[tex]3 + 3(49) = 150[/tex]
[tex]6 + 6(49) = 300[/tex]
We need to what percent of find 150% of 300.
B is twice as A, so this means that we need to multiply 2 by it original or 100% of it self.
[tex]2 \times 100 = 200[/tex]
So 200% is the answer.
