To solve for the general term of a geometric progression (G.P.), we start with the essential formula for the [tex]\(n\)[/tex]-th term [tex]\(G(n)\)[/tex] of a geometric progression:
[tex]\[ G(n) = G_1 \cdot r^{n-1} \][/tex]
where:
- [tex]\(G_1\)[/tex] is the first term of the G.P.
- [tex]\(r\)[/tex] is the common ratio.
- [tex]\(n\)[/tex] is the term number.
Given:
- The first term [tex]\(G_1\)[/tex] is [tex]\(\frac{1}{2}\)[/tex].
- The common ratio [tex]\(r\)[/tex] is [tex]\(\frac{1}{2}\)[/tex].
Let's substitute these values into our formula:
[tex]\[ G(n) = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^{n-1} \][/tex]
To simplify this expression, we need to combine the fractions:
[tex]\[ G(n) = \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^{n-1} \][/tex]
[tex]\[ G(n) = \frac{1}{2} \cdot \frac{1^{n-1}}{2^{n-1}} \][/tex]
[tex]\[ G(n) = \frac{1}{2} \cdot \frac{1}{2^{n-1}} \][/tex]
[tex]\[ G(n) = \frac{1}{2 \cdot 2^{n-1}} \][/tex]
[tex]\[ G(n) = \frac{1}{2^n} \][/tex]
This simplifies further to:
[tex]\[ G(n) = 2^{-n} \][/tex]
Thus, the general term [tex]\(G(n)\)[/tex] for this geometric progression is [tex]\(2^{-n}\)[/tex].
Therefore, the correct answer is:
A. [tex]\(2^{-n}\)[/tex]
