Final answer:
The least value of n such that the nth term of the geometric sequence is greater than the corresponding term in the arithmetic sequence is n = 5.
Step-by-step explanation:
The problem asks for the least value of n where the nth term of a geometric sequence is greater than the corresponding term in an arithmetic sequence. Let's consider the arithmetic sequence first. The nth term of an arithmetic sequence is given by the formula:
a + (n - 1)d
where a is the first term and d is the common difference. Now, let's consider the geometric sequence. The nth term of a geometric sequence is given by the formula:
a * (r)^(n - 1)
where a is the first term and r is the common ratio.
To find the least value of n, we need to compare the nth terms of the two sequences and find when the geometric term is greater than the arithmetic term. Let's solve this for n:
a + (n - 1)d < a * (r)^(n - 1)
To simplify the equation, let's assume a = 1 and d = 1 for the arithmetic sequence. The inequality becomes:
1 + (n - 1)(1) < 1 * (r)^(n - 1)
Simplifying further:
n < (r)^(n - 1)
To find the least value of n, we can substitute the given options one by one and see which one satisfies the inequality:
a) n = 2: (r)^1 > 2 - Not True
b) n = 3: (r)^2 > 3 - Not True
c) n = 4: (r)^3 > 4 - Not True
d) n = 5: (r)^4 > 5 - True
Therefore, the least value of n such that the nth term of the geometric sequence is greater than the corresponding term in the arithmetic sequence is n =5