Question

If the sum of the first three terms of a geometric sequence is `52`, and the common ratio is `3`, what are the first and sixth terms?

Answer 1

`t_1 = 4`
`t_6 = 972`

Explanation:

I'll answer the first question and leave the other ones up to other contributors.

The formula for the sum of the first `n` terms of a geometric sequence with first term `a` and common ratio `r` is

`s_n = (a(1- r^n))/(1- r)`

We know the sum of the first `3` terms is `52` and that the common ratio is `3`, therefore:

`52 = (a(1 - 3^3))/(1 - 3)`

Solving, we get:

`52(-2) = a(1- 27)`

`-104 = -26a`

`a = 4`

Recall that the nth term of a geometric sequence is given by `t_n = a r^(n - 1)`. This means that `t_6 = 4(3)^(5) = 972`

Hopefully this helps!