Question

I'm extremely bad at doing these

Answer 1

Term 2 is 21, and term 3 is 147.

Explanation:

The geometric sequence is 3, 3r, 3r^2, 1029.

`r = \sqrt[3]{ (1029)/(3) }`

`r = \sqrt[3]{343} = 7`

So we have 3, 21, 147, 1029.

Answer 2

The first few terms of a geometric sequence with first term
`a` and common ratio
`r` look like

`a, ar , ar^2, ar^3, \ldots`

and so on. Notice that the
`n`-th term (where
`n` is a natural number) is
`ar^(n-1)`.

For this particular sequence, the first term is

`a=3`

and the fourth term is

`ar^(4-1) = ar^3 = 1029`

Substitute
`a=3` into the second equation and solve for
`r`.

`3r^3 = 1029 \implies r^3 = 343 \implies r = \sqrt[3]{343} = \sqrt[3]{7^3} = 7`

Then the two terms between the 1st and 4th - i.e. the 2nd and 3rd terms - are

`ar = 3*7 = \boxed{21}`

and

`ar^2 = 3*7^2 = \boxed{147}`