Certainly! Let's solve the problem step-by-step.
Given:
- The first term of the geometric sequence (a) is 3.
- 16 times the 4th term is equal to the 8th term.
Let's denote:
- The common ratio by [tex]\( r \)[/tex].
### Step 1: Establish the relationship between the terms
In a geometric sequence, the [tex]\( n \)[/tex]-th term is given by:
[tex]\[ a_n = a \times r^{(n-1)} \][/tex]
For the 4th term:
[tex]\[ a_4 = a \times r^{(4-1)} = a \times r^3 \][/tex]
For the 8th term:
[tex]\[ a_8 = a \times r^{(8-1)} = a \times r^7 \][/tex]
### Step 2: Set up the equation given by the problem
According to the problem, 16 times the 4th term is equal to the 8th term:
[tex]\[ 16 \times a_4 = a_8 \][/tex]
Substituting the expressions:
[tex]\[ 16 \times (a \times r^3) = a \times r^7 \][/tex]
### Step 3: Simplify the equation
Since [tex]\( a \)[/tex] is positive and non-zero, we can divide both sides by [tex]\( a \)[/tex]:
[tex]\[ 16 \times r^3 = r^7 \][/tex]
### Step 4: Solve for [tex]\( r \)[/tex]
Divide both sides by [tex]\( r^3 \)[/tex] (assuming [tex]\( r \neq 0 \)[/tex]):
[tex]\[ 16 = r^4 \][/tex]
Taking the fourth root of both sides:
[tex]\[ r = 2 \][/tex]
### Step 5: Find the 15th term
The 15th term of the geometric sequence is given by:
[tex]\[ a_{15} = a \times r^{(15-1)} = a \times r^{14} \][/tex]
Substituting [tex]\( a = 3 \)[/tex] and [tex]\( r = 2 \)[/tex]:
[tex]\[ a_{15} = 3 \times 2^{14} \][/tex]
[tex]\[ a_{15} = 3 \times 16384 \][/tex]
[tex]\[ a_{15} = 49152 \][/tex]
### Final Answer
The 15th term of the geometric sequence is:
[tex]\[ 49152 \][/tex]
