Answer:
y = 2x - 1
n = 6
Explanation:
Hello!
This is an arithmetic sequence.
An Arithmetic sequence, also known as an arithmetic progression, is a string of numbers that follow a pattern where the difference between two terms is always the same (constant).
An Arithmetic sequence is modeled by the explicit function:
- t(n) = output (y - value)
- CD = common difference (slope)
- n = input ( x-value)
- t(0) = starting value (y-intercept)
Let's go step by step to solve these equations:
Step 1: Common Difference
The common difference of a sequence is similar to the slope of a line. The slope formula is given as
.
We can input values of x and y to find the common difference
We have the CD, 2!
Step 2: Starting Value
The starting value can be shown as the y-intercept of the line or the origin point of the sequence.
To find the starting value or t(0), we can input an x and y value for "n" and "t(n)"
- t(n) = (CD)n + t(0)
- 5 = 2(3) + t(0)
- 5 = 6 + t(0)
- -1 = t(0)
We know have the starting value, -1!
Step 3: The Equation
We now have all the values for our equation. let's bring our attention back to the input and output variables. Since we know that "t(n)" is the same as "y" and "n" is the same as "x", we can plug that instead of t(n) and n.
- t(n) = (CD)n + t(0)
- t(n) = 2n -1
- y = 2x - 1
Solve for n:
With our equation, we can plug in 11 as the output and solve for n.
- y = 2x - 1
- 11 = 2x - 1
- 12 = 2x
- 6 = x
n = 6
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Another way to solve this is to find the Recursive Equation
The Recursive Equation is only meant to find the next term, it doesn't do so well in finding the terms in the long run.
The basic form of a recursive equation is t(n + 1) = t(n) + (CD); where t(0) is ____
I'm not going to go in-depth with this, but you can see the values that we solved above can be implemented here.
Our recursive equation is t(n + 1) = t(n) + 2; where t(0) is -1
The port outside of the semi-colon represents where the sequence starts. If only t(n+1) = t(n) + 2 is given, we could start at 5, and go as 5, 7, 9...etc.
We would start by finding t(1) by doing:
- t(0 + 1) = t(0) + 2; where t(0) = -1
- t(1) = -1 + 2
- t(1) = 1
We can confirm this using our explicit rule: y = 2x - 1
- y = 2(1) - 1
- y = 2 - 1
- y = 1