The problem describes a geometric sequence where the second term is equal to the fourth term. We are given that the value of the fourth term is 8. We are asked to find the common ratio, the first term, and the tenth term of the geometric sequence.

1. Problem Description

The problem describes a geometric sequence where the second term is equal to the fourth term. We are given that the value of the fourth term is

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

2. Solution Steps

Let the first term of the geometric sequence be $a$ and the common ratio be $r$. The $n$th term of a geometric sequence is given by:

$a_n = a * r^(n-1)$

We are given that the second term is equal to the fourth term. So, $a_2 = a_4 = 8$. Using the general formula for the $n$th term, we can write:

$a_2 = a * r^(2-1) = a * r = 8$

$a_4 = a * r^(4-1) = a * r^3 = 8$

Since $ar = 8$ and $ar^3 = 8$, we can divide the second equation by the first:

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

$r^2 = 1$

$r = \pm 1$

Case 1: $r = 1$

If $r = 1$, then $a * 1 = 8$, so $a = 8$.

The tenth term is $a_{10} = a * r^(10-1) = 8 * 1^9 = 8$.

Case 2: $r = -1$

If $r = -1$, then $a * (-1) = 8$, so $a = -8$.

The tenth term is $a_{10} = a * r^(10-1) = -8 * (-1)^9 = -8 * (-1) = 8$.

Therefore, we have two possible geometric sequences.

If $r=1$, then $a=8$, and $a_{10} = 8$.

If $r=-1$, then $a=-8$, and $a_{10} = 8$.

3. Final Answer

Case 1:

Common ratio: $r = 1$

First term: $a = 8$

Tenth term: $a_{10} = 8$

Case 2:

Common ratio: $r = -1$

First term: $a = -8$

Tenth term: $a_{10} = 8$