Final answer:
The common ratio of sequence B is 4.
Step-by-step explanation:
To find the common ratio of sequence B, we need to understand that the sum of two convergent geometric sequences is such that when subtracted, the result is 0. We are given that the first term of sequence A is 3 and the first term of sequence B is 2. We are also given that the common ratio of sequence B is four times the common ratio of sequence A. Let's represent the common ratio of sequence A as 'r'.
The sum of sequence A can be expressed as:
A = 3 + 3r + 3r^2 + 3r^3 + ...
The sum of sequence B can be expressed as:
B = 2 + 2(4r) + 2(4r)^2 + 2(4r)^3 + ...
Since A + B = B + A = 0, we can write:
3 + 3r + 3r^2 + 3r^3 + ... + 2 + 2(4r) + 2(4r)^2 + 2(4r)^3 + ... = 0
Combining the terms with the same powers of r, we get:
(3 + 2) + (3r + 8r) + (3r^2 + 16r^2) + (3r^3 + 64r^3) + ... = 0
5 + 11r + 19r^2 + 67r^3 + ... = 0
This is a geometric series with a common ratio of 'r' and we know that a geometric series with a common ratio between -1 and 1 converges to a finite value. Thus, for the sum to be 0, the terms must individually be 0.
This means that 5 = 11r = 19r^2 = 67r^3 = ... = 0
Since the common ratio of sequence A cannot be zero (otherwise all terms will be zero), we can divide the terms of sequence B by 4 to find the common ratio of sequence B:
5/4 = 11r/4 = 19r^2/4 = 67r^3/4 = ... = 0
Simplifying, we get:
5/4 = 11r/4 = 19r^2/4 = 67r^3/4 = ... = 0
Thus, the common ratio of sequence B is 4 (Option A).