Find a general term a n for the geometric sequence.
-$$a _ { 2 } = - 1 , a _ { 9 } = \frac { 1 } { 128 }$$
A) $a _ { n } = 6 \left( - \frac { 1 } { 4 } \right) ^ { n - 1 }$
B) $a _ { n } = 2 \left( \frac { 1 } { 2 } \right) ^ { n - 1 }$
C) $a _ { n } = 2 \left( - \frac { 1 } { 2 } \right) ^ { n - 1 }$
D) $a _ { n } = \frac { 1 } { 2 } ( - 2 ) ^{ n - 1}$

Question

Quizplus · Accepted Answer

Given $a_2 = -1$ and $a_9 = \frac{1}{128}$, we can use the formula for the nth term of a geometric sequence, $a_n = a_1 \cdot r^{n-1}$, where $a_1$ is the first term and $r$ is the common ratio. From $a_2 = a_1 \cdot r$, we find $r$ by using the given terms. The ratio $r$ can be found by dividing $a_9$ by $a_8$, or equivalently, using the given values and solving for $r$. The correct ratio that fits the given terms is $-\frac{1}{2}$, and the first term $a_1$ can be calculated as $2$ by back-calculating from $a_2 = -1$. Therefore, the general term is $a_n = 2 \left( -\frac{1}{2} \right)^{n-1}$.

Find a General Term An for the Geometric Sequence a2=−1,a9=1128a _ { 2 } = - 1 , a _ { 9 } = \frac { 1 } { 128 }a2​=−1,a9​=1281​

Find a General Term A_n for the Geometric Sequence $a _ { 2 } = - 1 , a _ { 9 } = \frac { 1 } { 128 }$