the following sequence is geometric:

4/3, 4/9, 4/27....
determine the general term, tn and the term t10

so the general term is:
(4/3)x (1/3) ^n

so the tenth term is:
4/177147

is this correct? thanks!

You have one too many factors of three in the denominator.
a = 4/3, r=1/3 so t10=(4/3)*1/39 = 4/310

To determine the general term (tn) of a geometric sequence, you need to identify the common ratio (r) of the sequence first. In this case, the common ratio is found by dividing any term by its preceding term. Let's calculate:

r = (4/9) / (4/3) = (4/9) * (3/4) = 1/3

Now that we know the common ratio (r = 1/3), we can write the general term (tn) for the sequence. It is given by the formula:

tn = a * r^(n-1)

where a is the first term, r is the common ratio, and n is the term number.

In our case, the first term (a) is 4/3, and the term number (n) is 10. Let's substitute these values into the formula:

t10 = (4/3) * (1/3)^(10-1) = (4/3) * (1/3)^9

Simplifying this expression, we have:

t10 = (4/3) * (1/3)^9 = (4/3) * (1/3)^9 = 4/3^10 = 4/59049

So the tenth term (t10) of the given geometric sequence is 4/59049.

To summarize:
- The general term (tn) of the sequence is (4/3) * (1/3)^(n-1).
- The tenth term (t10) is 4/59049.