For what values of t will t-2, 2t-6, and 4t-8, in this order, form an arithmetic sequence?
Could someone at least start me off?
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
Well calculated
t-2,2t-6,4t-8 =(4t-8)=(2t-6)=4t-2t=-8+6=2t=-2=(t=-1)is the same:2t-6=t-2(t=-3)(-3= t=-1)(t=-3+1)t=-2 where is -4,-10,-16
To determine the values of t that would make the given numbers "t-2," "2t-6," and "4t-8" form an arithmetic sequence, we need to find the common difference between consecutive terms.
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.