We are given that the second term of a geometric sequence is 4 and the fourth term is 8. We are asked to find the common ratio, the first term, and the tenth term.

1. Problem Description

We are given that the second term of a geometric sequence is 4 and the fourth term is

8. We are asked to find the common ratio, the first term, and the tenth term.

2. Solution Steps

Let $a$ be the first term of the geometric sequence and $r$ be the common ratio.

The $n$th term of a geometric sequence is given by the formula:

$a_n = ar^{n-1}$

We are given that the second term is 4, so

$a_2 = ar^{2-1} = ar = 4$ (1)

We are also given that the fourth term is 8, so

$a_4 = ar^{4-1} = ar^3 = 8$ (2)

We want to find $a$, $r$, and $a_{10}$.

We can divide equation (2) by equation (1):

$\frac{ar^3}{ar} = \frac{8}{4}$

$r^2 = 2$

$r = \pm\sqrt{2}$

Case 1: $r = \sqrt{2}$

From equation (1), $a(\sqrt{2}) = 4$, so $a = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

The tenth term is $a_{10} = ar^9 = (2\sqrt{2})(\sqrt{2})^9 = 2(\sqrt{2})^{10} = 2(2^5) = 2(32) = 64$.

Case 2: $r = -\sqrt{2}$

From equation (1), $a(-\sqrt{2}) = 4$, so $a = \frac{4}{-\sqrt{2}} = \frac{-4\sqrt{2}}{2} = -2\sqrt{2}$.

The tenth term is $a_{10} = ar^9 = (-2\sqrt{2})(-\sqrt{2})^9 = (-2\sqrt{2})(-1)^9(\sqrt{2})^9 = (2\sqrt{2})(\sqrt{2})^9 = 2(\sqrt{2})^{10} = 2(2^5) = 2(32) = 64$.

Therefore, the common ratio is $\sqrt{2}$ or $-\sqrt{2}$, the first term is $2\sqrt{2}$ or $-2\sqrt{2}$ respectively, and the tenth term is

6

4.

3. Final Answer

The common ratio is $\pm\sqrt{2}$.

The first term is $\pm 2\sqrt{2}$.

The tenth term is

6

4.