Find a general term a n for the geometric sequence.
-$$a _ { 1 } = \frac { 1 } { 6 } , r = - 3$$
A) $a _ { n } = \frac { 1 } { 6 } - \frac { 1 } { 3 } ( n - 1 )$
B) $a _ { n } = \frac { 1 } { 6 } - \frac { 2 } { 9 } ( n - 1 )$
C) $a _ { n } = \left( \frac { 1 } { 6 } \right) ^ { n - 1 }$
D) $a _ { n } = \frac { 1 } { 6 } \cdot ( - 3 ) ^ { n - 1 }$

Question

Quizplus · Accepted Answer

The general term for a geometric sequence is given by $a_n = a_1 \cdot r^{n-1}$, where $a_1$ is the first term and $r$ is the common ratio. Given $a_1 = \frac{1}{6}$ and $r = -3$, the formula becomes $a_n = \frac{1}{6} \cdot (-3)^{n-1}$.

Find a General Term An for the Geometric Sequence a1=16,r=−3a _ { 1 } = \frac { 1 } { 6 } , r = - 3a1​=61​,r=−3

Find a General Term A_n for the Geometric Sequence $a _ { 1 } = \frac { 1 } { 6 } , r = - 3$