Let's denote the first term of the geometric progression (GP) as \( a \) and the common ratio as \( r \).
In a geometric progression, the terms can be represented as follows:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- Fourth term: \( ar^3 \)
- Fifth term: \( ar^4 \)
According to the problem, we know the following:
- The second term \( ar = 16 \)
- The fifth term \( ar^4 = 2 \)
We can set up a system of equations:
\[ ar = 16 \quad (1) \] \[ ar^4 = 2 \quad (2) \]
From equation (1), we can express \( a \) in terms of \( r \):
\[ a = \frac{16}{r} \]
Now, substitute \( a \) into equation (2):
\[ \left(\frac{16}{r}\right) r^4 = 2 \]
This simplifies to:
\[ 16r^3 = 2 \]
Next, divide both sides by 16:
\[ r^3 = \frac{2}{16} = \frac{1}{8} \]
Now, take the cube root of both sides to solve for \( r \):
\[ r = \sqrt[3]{\frac{1}{8}} = \frac{1}{2} \]
Now that we have the common ratio \( r \), we can substitute \( r \) back into equation (1) to find \( a \):
\[ ar = 16 \] \[ a \left(\frac{1}{2}\right) = 16 \]
Multiply both sides by 2:
\[ a = 16 \times 2 = 32 \]
Thus, the first term \( a \) is \( 32 \) and the common ratio \( r \) is \( \frac{1}{2} \).
Summary:
- First term \( a = 32 \)
- Common ratio \( r = \frac{1}{2} \)