Answer:
There are two sets of solutions:
Step-by-step explanation:
The question states that [tex]a[/tex] and [tex]r[/tex] should represent the first term and the common ratio of this geometric sequence, respectively. Therefore:
Let [tex]n[/tex] denote a positive whole number. In general, the [tex]n\![/tex]-th term of this geometric sequence would be [tex]r^{n-1}\, a[/tex].
The sixth term of this sequence would thus be [tex]r^{6-1}\, a = r^5\, a[/tex].
The fact that the second term is [tex]2[/tex] and the sixth term is [tex]32[/tex] gives two equations about [tex]a[/tex] and [tex]r[/tex]:
[tex]\displaystyle \left\lbrace\begin{aligned}& r\, a = 2 \\ & r^5\, a= 32\end{aligned}\right.[/tex].
Take the quotient of these two equations to eliminate [tex]a[/tex]:
[tex]\displaystyle \frac{r^5\, a}{r\, a} = \frac{32}{2}[/tex].
[tex]r^4 = 16[/tex].
There are two possible roots: either [tex]r = 2[/tex] or [tex]r = -2[/tex].
When [tex]r = 2[/tex], substitute back to the first equation and solve for [tex]a[/tex]: [tex]a = 1[/tex].
On the other hand, when [tex]r = -2[/tex], substituting back and solving for [tex]a[/tex] would give [tex]a = -1[/tex].
Substitute these values into the second equation. Either set of values will work.
