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What is the n^th terms of 6,11,18,27,38

To find the nth term of the sequence 6, 11, 18, 27, 38, we can use the method of finite differences.

First, we calculate the differences between consecutive terms:

11 - 6 = 5 18 - 11 = 7 27 - 18 = 9 38 - 27 = 11

We can see that the second differences are all equal to 2. This tells us that the nth term is a quadratic function of n.

Next, we calculate the first differences between consecutive second differences:

7 - 5 = 2 9 - 7 = 2 11 - 9 = 2

Since the first differences between the second differences are all equal to 2, this tells us that the quadratic function is of the form:

an^2 + bn + c

where a = 1/2, since the second differences are all equal to 2.

To find the values of b and c, we can use the first two terms of the sequence:

When n = 1, the term is 6, so: a + b + c = 6

When n = 2, the term is 11, so: 4a + 2b + c = 11

Solving these two equations simultaneously, we get: b = 5/2 c = 3

Therefore, the nth term of the sequence is given by: an^2 + (5/2)n + 3/2

To find the nth term, we simply substitute the value of n into this formula. For example, to find the 6th term:

a(6)^2 + (5/2)(6) + 3/2 = 6(1/2)(36) + 15 + 3/2 = 18 + 15 + 3/2 = 36.5

Therefore, the 6th term of the sequence is 36.5.

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