There is an arithmetic progression if
2t-6 -(t-2) = (4t-8)-(2t-6)
t -4 = 2t -2
t = -2
in which case the sequence is -4, -10, -16
Could someone at least start me off?
2t-6 -(t-2) = (4t-8)-(2t-6)
t -4 = 2t -2
t = -2
in which case the sequence is -4, -10, -16
The common difference (d) is the difference between any two consecutive terms in an arithmetic sequence. In this case, the first term is "t-2," the second term is "2t-6," and the third term is "4t-8."
To find the common difference, we subtract the second term from the first term and the third term from the second term.
Step 1: Find the difference between the second term and the first term:
(2t-6) - (t-2) = 2t-6-t+2 = t-4
Step 2: Find the difference between the third term and the second term:
(4t-8) - (2t-6) = 4t-8-2t+6 = 2t-2
Since we have the same common difference, t-4 and 2t-2, we can equate them:
t-4 = 2t-2
Now we can solve for t by simplifying and isolating t on one side:
t - 2t = -2 + 4
-t = 2
t = -2
Therefore, the value of t that makes the three terms form an arithmetic sequence is t = -2.