To find the fifteenth term of the given geometric sequence [tex]\( -\sqrt{5}, \sqrt{10}, -2\sqrt{5}, \ldots \)[/tex]:
1. Identify the first term ([tex]\(a_1\)[/tex]) of the sequence:
[tex]\[
a_1 = -\sqrt{5}
\][/tex]
2. Identify the second term ([tex]\(a_2\)[/tex]) of the sequence:
[tex]\[
a_2 = \sqrt{10}
\][/tex]
3. Identify the third term ([tex]\(a_3\)[/tex]) of the sequence:
[tex]\[
a_3 = -2\sqrt{5}
\][/tex]
4. Calculate the common ratio ([tex]\(r\)[/tex]) of the sequence using the first two terms:
[tex]\[
r = \frac{a_2}{a_1} = \frac{\sqrt{10}}{-\sqrt{5}}
\][/tex]
Simplify the fraction:
[tex]\[
\frac{\sqrt{10}}{-\sqrt{5}} = \frac{\sqrt{2} \cdot \sqrt{5}}{-\sqrt{5}} = \sqrt{2} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \sqrt{2} \cdot (-1) = -\sqrt{2}
\][/tex]
So, the common ratio [tex]\(r\)[/tex] is:
[tex]\[
r = -\sqrt{2}
\][/tex]
5. Use the formula for the [tex]\(n\)[/tex]-th term of a geometric sequence to find the fifteenth term ([tex]\(a_{15}\)[/tex]):
[tex]\[
a_n = a_1 \cdot r^{n-1}
\][/tex]
For the fifteenth term ([tex]\(n = 15\)[/tex]):
[tex]\[
a_{15} = a_1 \cdot r^{15-1} = a_1 \cdot r^{14}
\][/tex]
Substituting the known values:
[tex]\[
a_{15} = -\sqrt{5} \cdot (-\sqrt{2})^{14}
\][/tex]
6. Simplify [tex]\((- \sqrt{2})^{14}\)[/tex]:
Since the exponent is even, the negative sign will be squared out, and we can write:
[tex]\[
(-\sqrt{2})^{14} = (\sqrt{2})^{14}
\][/tex]
Calculate [tex]\((\sqrt{2})^{14}\)[/tex]:
[tex]\[
(\sqrt{2})^{14} = (2^{1/2})^{14} = 2^{7} = 128
\][/tex]
7. Combine the terms to find [tex]\(a_{15}\)[/tex]:
[tex]\[
a_{15} = -\sqrt{5} \cdot 128 = -128\sqrt{5}
\][/tex]
Therefore, the fifteenth term of the sequence is:
[tex]\[
\boxed{-128 \sqrt{5}}
\][/tex]
