In a GP the 3rd term is equal to four times the first term and the sixth term is equal to three times the fourth term plus 32. determine the sequence.

ar^2 = 4a

ar^5 = 3(ar^3)+32

Now crank it out

To determine the sequence in a geometric progression (GP), we need to find the common ratio (r) and the first term (a).

Let's denote the first term as "a" and the common ratio as "r".

Given information:
- The 3rd term is equal to four times the first term.
- The 6th term is equal to three times the fourth term plus 32.

Using these conditions, we can set up equations:

1. The 3rd term is four times the first term:
a * r^2 = 4a

2. The 6th term is three times the fourth term plus 32:
a * r^5 = 3(a * r^3) + 32

We have two equations with two unknowns, so we can solve them simultaneously.

From the first equation, we can divide both sides by "a":
r^2 = 4

Taking the square root of both sides:
r = ±2

Since the common ratio cannot be negative, we take r = 2.

Substituting r = 2 into the second equation:
a * 2^5 = 3(a * 2^3) + 32
32a = 24a + 32

Simplifying:
8a = 32
a = 4

Now that we have found the first term (a = 4) and the common ratio (r = 2), we can determine the sequence.

The sequence in the geometric progression is:
4, 8, 16, 32, 64, 128, ...

Each term can be obtained by multiplying the previous term by 2.