Find the nth term of the geometric sequence.
-$$\frac { 1 } { 2 } , \frac { 1 } { 6 } , \frac { 1 } { 18 } , \ldots ; n = 6$$
A) $\frac { 1 } { 486 }$

B) $\frac { 1 } { 36 }$

C) $\frac { 1 } { 162 }$

D) $\frac { 1 } { 243 }$

Question

Find the nth term of the geometric sequence.
-$$\frac { 1 } { 2 } , \frac { 1 } { 6 } , \frac { 1 } { 18 } , \ldots ; n = 6$$
A) $\frac { 1 } { 486 }$

B) $\frac { 1 } { 36 }$

C) $\frac { 1 } { 162 }$

D) $\frac { 1 } { 243 }$

Quizplus · Accepted Answer

The common ratio (r) of the sequence is $\frac{1}{6} \div \frac{1}{2} = \frac{1}{3}$. The first term (a) is $\frac{1}{2}$. The nth term of a geometric sequence is given by $a \cdot r^{(n-1)}$. Substituting $a = \frac{1}{2}$, $r = \frac{1}{3}$, and $n = 6$, we get the 6th term as $\frac{1}{2} \cdot (\frac{1}{3})^{(6-1)} = \frac{1}{2} \cdot \frac{1}{243} = \frac{1}{486}$.

Find the Nth Term of the Geometric Sequence 12,16,118,…;n=6\frac { 1 } { 2 } , \frac { 1 } { 6 } , \frac { 1 } { 18 } , \ldots ; n = 621​,61​,181​,…;n=6

Find the Nth Term of the Geometric Sequence $\frac { 1 } { 2 } , \frac { 1 } { 6 } , \frac { 1 } { 18 } , \ldots ; n = 6$