Answer:
[tex]u_1=-128[/tex]
[tex]r=\dfrac{1}{4}[/tex]
Step-by-step explanation:
The general formula for the nth term of a geometric sequence is:
[tex]\boxed{\begin{array}{l}\underline{\sf Geometric \;sequence}\\\\u_n=u_1 \cdot r^{n-1}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$u_n$ is the $n$th term.}\\\phantom{ww}\bullet \;\textsf{$u_1$ is the first term.}\\\phantom{ww}\bullet\;\textsf{$r$ is the common ratio.}\\\phantom{ww}\bullet\;\textsf{$n$ is the position of the term.}\end{array}}[/tex]
Given that the second term of a geometric sequence is -32 and the fourth term is -2, then:
[tex]\begin{aligned}u_2&=-32\\u_1 \cdot r^{2-1}&=-32\\u_1 \cdot r&=-32\end{aligned}[/tex]
[tex]\begin{aligned}u_4&=-2\\u_1 \cdot r^{4-1}&=-2\\u_1 \cdot r^3&=-2\end{aligned}[/tex]
So, we have created two equations:
[tex]\begin{cases}u_1 \cdot r=-32\\u_1 \cdot r^3=-2\end{cases}[/tex]
Divide the second equation by the first equation to eliminate u₁, then solve for r:
[tex]\begin{aligned}\dfrac{u_1 \cdot r^3}{u_1 \cdot r}&=\dfrac{-2}{-32}\\\\\dfrac{r^3}{r}&=\dfrac{1}{16}\\\\r^{3-1}&=\dfrac{1}{16}\\\\r^2&=\dfrac{1}{16}\\\\r&=\pm\sqrt{\dfrac{1}{16}}\\\\r&=\pm\dfrac{1}{4}\end{aligned}[/tex]
In order for all terms to be negative, the first term (u₁) must be negative and the common ratio (r) must be positive. Therefore:
[tex]r=\dfrac{1}{4}[/tex]
Note that if r = -1/4, then r would be positive for the odd terms and negative for the even terms, implying that not all terms could be negative.
To find the value of the first term (u₁), we can substitute r = 1/4 into one of the equations:
[tex]\begin{aligned}u_1 \cdot \dfrac{1}{4}&=-32\\\\u_1 &=-128\end{aligned}[/tex]
Therefore, the values of u₁ and r are:
[tex]\Large\boxed{\boxed{u_1=-128\;\;\textsf{and}\;\;r=\dfrac{1}{4}}}[/tex]
