To find tn for a geometric sequence, we need to know the first term t1 and the common ratio r. We can use the given terms t3 and t9 to set up two equations and solve for t1 and r.
From the formula for the nth term of a geometric sequence, we have:
t3 = t1 * r^2 (equation 1)
t9 = t1 * r^8 (equation 2)
We can divide equation 2 by equation 1 to eliminate t1:
t9/t3 = (t1 * r^8)/(t1 * r^2)
t9/t3 = r^6
Now we can solve for r:
r = (t9/t3)^(1/6)
r = (1536/24)^(1/6)
r = 2
Substituting r = 2 into equation 1, we have:
24 = t1 * 2^2
Solving for t1, we get:
t1 = 6
So the first term of the sequence is t1 = 6, and the common ratio is r = 2.
Now we can use the formula for the nth term of a geometric sequence to find tn:
tn = t1 * r^(n-1)
Substituting t1 = 6 and r = 2, we have:
tn = 6 * 2^(n-1)
To find tn, we need to know the value of n. We know that t3 = 24, so we can substitute n = 3 to find:
t3 = 6 * 2^(3-1) = 6 * 2^2 = 24
This confirms that our values for t1 and r are correct.
We can also use t9 = 1536 to find another value of n:
t9 = 6 * 2^(9-1) = 6 * 2^8 = 1536
So, tn can be 6 * 2^(n-1) for n = 3 or n = 9.
The reason there are two answers is because a geometric sequence has an infinite number of terms, and any power of the common ratio r will give us a valid value of tn. In this case, since r = 2, we can find tn for any value of n using the formula tn = 6 * 2^(n-1). However, since we only