Answer:
B. 6, 8, 10 2/3, 14 2/9, 18 26/27
Explanation:
The recursive relation tells you the first term is 6, eliminating choice A immediately.
It also tells you each term is 4/3 the previous term, so the sequence cannot be a sequence of integers*, eliminating choices C and D.
If you do the math, you find the terms of Choice B are correct:
- 6×4/3 = 8
- 8×4/3 = 10 2/3
- (10 2/3)×4/3 = 14 2/9
- ...
_____
* Unless you start with a very large power of 3, repeatedly multiplying by 4/3 will eventually end with powers of 3 in the denominator and some multiple of powers of 2 in the numerator. Such a ratio cannot be an integer.