For the arithmetic sequence (2 - x),
(-6 + 2x), (x + 2), solve for x and find t10.
Would I have to start off like this below:::
an = a1 + (n - 1) d
d = (-6 + 2x)-(2 - x) = (x + 2)-(-6 + 2x)
correct, just solve
(-6 + 2x)-(2 - x) = (x + 2)-(-6 + 2x)
....
I got
x = 4
so the sequence is
-2, 2, and 6
Yes, you're on the right track. To start solving for x in the given arithmetic sequence, you can use the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
In this case, the first term (a1) is 2 - x and the common difference (d) is the difference between consecutive terms. Let's determine the common difference first:
d = (-6 + 2x) - (2 - x)
= -6 + 2x - 2 + x
= -8 + 3x
Now, we can substitute the values of a1, d, and n into the formula to solve for x and find t10. But before we do that, we need to confirm what t10 refers to in this sequence. Are you referring to the 10th term?
To solve for x in the arithmetic sequence (2 - x), (-6 + 2x), (x + 2), you can use the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1) * d
In this case, a1 is the first term (2 - x), and d is the common difference.
To find d, you can subtract the second term from the first term:
d = (-6 + 2x) - (2 - x)
Simplifying this expression, you would get:
d = -6 + 2x - 2 + x
= -8 + 3x
Then, you can set up the equation for the nth term (in this case, t10):
t10 = a1 + (10 - 1) * d
Substituting a1 = (2 - x) and d = -8 + 3x, the equation becomes:
t10 = (2 - x) + (10 - 1) * (-8 + 3x)
Simplifying and solving for x, you can proceed with algebraic manipulations.