Final answer:
The question requires comparing the fourth term of a geometric sequence to the sixteenth term of an arithmetic sequence, using the respective formulae for each sequence and showing that they can be equivalent under specific circumstances.
Step-by-step explanation:
The question is asking to show that the fourth term of a geometric sequence is the same as the sixteenth term of an arithmetic sequence.
To prove this, we need some information about the common ratio of the geometric sequence and the common difference of the arithmetic sequence. However, as the question does not provide specific values, we typically use variables to represent these.
In a geometric sequence, if the first term is a and the common ratio is r, then the nth term is given by a × r(n-1). Therefore, the fourth term would be a × r3.
In an arithmetic sequence, if the first term is a and the common difference is d, then the nth term is a + (n-1)×d. Thus, the sixteenth term would be a + 15×d.
For the two terms to be equal, a × r3 = a + 15×d. It is implicit in this case that some value of r, when raised to the third power, equals 1 plus 15 times the common difference, d, of the arithmetic sequence divided by the first term, a.