To solve this problem, let's use the properties of a geometric sequence.
1. Recall the general formula for any term in a geometric sequence:
[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.
Given:
- The 4th term, [tex]\( a_4 = \frac{5 \sqrt{2}}{4} \)[/tex]
- The 6th term, [tex]\( a_6 = \frac{5 \sqrt{2}}{8} \)[/tex]
2. Express these terms using the general formula:
[tex]\[ a_4 = a \cdot r^3 = \frac{5 \sqrt{2}}{4} \][/tex]
[tex]\[ a_6 = a \cdot r^5 = \frac{5 \sqrt{2}}{8} \][/tex]
3. To find the common ratio [tex]\( r \)[/tex], divide the 6th term by the 4th term:
[tex]\[ \frac{a \cdot r^5}{a \cdot r^3} = \frac{\frac{5 \sqrt{2}}{8}}{\frac{5 \sqrt{2}}{4}} \][/tex]
[tex]\[ r^2 = \frac{\frac{5 \sqrt{2}}{8}}{\frac{5 \sqrt{2}}{4}} \][/tex]
Simplify the right-hand side:
[tex]\[ r^2 = \frac{5 \sqrt{2} / 8}{5 \sqrt{2} / 4} = \frac{5 \sqrt{2}}{8} \cdot \frac{4}{5 \sqrt{2}} = \frac{4}{8} = \frac{1}{2} \][/tex]
Thus:
[tex]\[ r^2 = \frac{1}{2} \quad \Rightarrow \quad r = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \][/tex]
4. Now that we have [tex]\( r = \frac{\sqrt{2}}{2} \)[/tex], we find the first term [tex]\( a \)[/tex] using the 4th term:
[tex]\[ a \cdot r^3 = \frac{5 \sqrt{2}}{4} \][/tex]
[tex]\[ a \cdot \left( \frac{\sqrt{2}}{2} \right)^3 = \frac{5 \sqrt{2}}{4} \][/tex]
[tex]\[ a \cdot \left( \frac{(\sqrt{2})^3}{2^3} \right) = \frac{5 \sqrt{2}}{4} \][/tex]
[tex]\[ a \cdot \left( \frac{2\sqrt{2}}{8} \right) = \frac{5 \sqrt{2}}{4} \][/tex]
[tex]\[ a \cdot \left( \frac{\sqrt{2}}{4} \right) = \frac{5 \sqrt{2}}{4} \][/tex]
Solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{5 \sqrt{2}}{4} \cdot \frac{4}{\sqrt{2}} \][/tex]
[tex]\[ a = 5 \][/tex]
Hence, the first term [tex]\( a \)[/tex] of the sequence is [tex]\( 5 \)[/tex].
