Answer:
Either [tex]\displaystyle \frac{-1522}{\sqrt{41}}[/tex] (approximately [tex]-238[/tex]) or [tex]\displaystyle \frac{1522}{\sqrt{41}}[/tex] (approximately [tex]238[/tex].)
Step-by-step explanation:
Let [tex]a[/tex] denote the first term of this geometric series, and let [tex]r[/tex] denote the common ratio of this geometric series.
The first five terms of this series would be:
- [tex]a[/tex],
- [tex]a\cdot r[/tex],
- [tex]a \cdot r^2[/tex],
- [tex]a \cdot r^3[/tex],
- [tex]a \cdot r^4[/tex].
First equation:
[tex]a\, r^4 - a = 150[/tex].
Second equation:
[tex]a\, r^3 - a\, r = 48[/tex].
Rewrite and simplify the first equation.
[tex]\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}[/tex].
Therefore, the first equation becomes:
[tex]a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150[/tex]..
Similarly, rewrite and simplify the second equation:
[tex]\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}[/tex].
Therefore, the second equation becomes:
[tex]a\, r\, \left(r^2 - 1\right) = 48[/tex].
Take the quotient between these two equations:
[tex]\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}[/tex].
Simplify and solve for [tex]r[/tex]:
[tex]\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}[/tex].
[tex]8\, r^2 - 25\, r + 8 = 0[/tex].
Either [tex]\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}[/tex] or [tex]\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}[/tex].
Assume that [tex]\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}[/tex]. Substitute back to either of the two original equations to show that [tex]\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75[/tex].
Calculate the sum of the first five terms:
[tex]\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}[/tex].
Similarly, assume that [tex]\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}[/tex]. Substitute back to either of the two original equations to show that [tex]\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75[/tex].
Calculate the sum of the first five terms:
[tex]\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}[/tex].
