Answer:
The first term of the sequence is 512 or -512
Step-by-step explanation:
Geometric Progression
The general term n of a geometric progression of first term a1 and common ratio r is:
[tex]a_n =a_1\cdot r^{n-1}[/tex]
We are given:
[tex]a_4=8[/tex]
[tex]\displaystyle a_6=\frac{1}{2}[/tex]
Applying the equation for n=4:
[tex]a_4 =a_1\cdot r^{4-1}[/tex]
[tex]a_4 =a_1\cdot r^{3}[/tex]
We have:
[tex]a_1\cdot r^{3}=8\qquad\qquad[1][/tex]
Applying the equation for n=6:
[tex]a_6 =a_1\cdot r^{6-1}[/tex]
[tex]a_6 =a_1\cdot r^{5}[/tex]
We have:
[tex]\displaystyle a_1\cdot r^{5}=\frac{1}{2} \qquad\qquad[2][/tex]
Dividing [2] by [1]:
[tex]\displaystyle \frac{r^{5}}{r^{3}}=\frac{\frac{1}{2}}{8}[/tex]
Operating:
[tex]\displaystyle r^{2}=\frac{1}{16}[/tex]
Taking the square root:
[tex]\displaystyle r=\sqrt{\frac{1}{16}}=\pm \frac{1}{4}[/tex]
There are two possible solutions:
[tex]\displaystyle r=\frac{1}{4}[/tex]
[tex]\displaystyle r=-\frac{1}{4}[/tex]
From [1]:
[tex]\displaystyle a_1 =\frac{8}{r^{3}}[/tex]
This gives also two possibles solutions for a1:
[tex]\displaystyle a_1 =\frac{8}{\left(\frac{1}{4}\right)^{3}} =512[/tex]
[tex]\displaystyle a_1 =\frac{8}{\left(-\frac{1}{4}\right)^{3}} =-512[/tex]
Thus, the first term of the sequence is 512 or -512
