These are the first 5 terms of a quadratic sequence:
1 3 7 13 21
1st difference: 2 4 6 8
2nd difference: 2 2 2
1n^2: 1 4 9 16 25
adjustment: ? ? ? ? ?
Thank you very much.
well, check the differences:
1-1=0
4-3=1
9-7=2
16-13=3
25-21=4
So, how does that relate to n?
You're welcome! To find the adjustment for the given quadratic sequence, we need to determine the pattern between the terms.
First, let's look at the differences between consecutive terms. We calculate the 1st difference by subtracting each term from the next one:
3 - 1 = 2
7 - 3 = 4
13 - 7 = 6
21 - 13 = 8
Next, we calculate the 2nd difference by subtracting each term of the 1st difference from the next one:
4 - 2 = 2
6 - 4 = 2
8 - 6 = 2
Since the 2nd difference is constant (2), we know that the quadratic sequence is of the form an^2, where "a" is a constant.
To find the specific values of "a", we can substitute the first five terms of the quadratic sequence (1, 3, 7, 13, 21) into the formula an^2:
a(1)^2 = 1
a(2)^2 = 3
a(3)^2 = 7
a(4)^2 = 13
a(5)^2 = 21
Solving these equations will give us the value of "a" and help us find the adjustment for the sequence. Let's solve them one by one:
1 x a = 1
4 x a = 3
9 x a = 7
16 x a = 13
25 x a = 21
Now, divide each equation by the coefficient of "a" to solve for "a":
1 / 1 = a
3 / 4 = a
7 / 9 = a
13 / 16 = a
21 / 25 = a
Simplifying each division, we find:
a = 1
a = 0.75
a = 0.777...
a = 0.8125
a = 0.84
Based on these values of "a," we can conclude that the adjustment for the quadratic sequence is approximately 1, 0.75, 0.777..., 0.8125, and 0.84.
Please note that the values of "a" are approximations since the sequence does not follow a perfect quadratic pattern.