To find the position [tex]\( k \)[/tex] of the term in the geometric sequence [tex]\( 2, 4, 8, \ldots \)[/tex] where the term is 128, we follow these steps:
Given:
- The first term [tex]\( a \)[/tex] of the geometric sequence is 2.
- The common ratio [tex]\( r \)[/tex] is 2.
We need to determine the position [tex]\( k \)[/tex] such that the term is 128.
The formula for the [tex]\( n \)[/tex]-th term of a geometric sequence is:
[tex]\[ a_n = a \cdot r^{n-1} \][/tex]
where [tex]\( a \)[/tex] is the first term, [tex]\( r \)[/tex] is the common ratio, and [tex]\( n \)[/tex] is the term number.
We are given:
[tex]\[ a_n = 128 \][/tex]
[tex]\[ a = 2 \][/tex]
[tex]\[ r = 2 \][/tex]
Substituting these values into the formula, we get:
[tex]\[ 128 = 2 \cdot 2^{k-1} \][/tex]
Now, we solve for [tex]\( k \)[/tex]:
First, divide both sides of the equation by 2:
[tex]\[ 64 = 2^{k-1} \][/tex]
Next, we recognize that 64 can be expressed as a power of 2:
[tex]\[ 64 = 2^6 \][/tex]
Therefore, we have:
[tex]\[ 2^{k-1} = 2^6 \][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[ k - 1 = 6 \][/tex]
Adding 1 to both sides, we find:
[tex]\[ k = 7 \][/tex]
Thus, [tex]\( k = 7 \)[/tex]. So, the term 128 is the 7th term in the geometric sequence [tex]\( 2, 4, 8, \ldots \)[/tex].
