Find the nth term of the geometric sequence.
-$2,1 , \frac { 1 } { 2 } , \frac { 1 } { 4 } , \frac { 1 } { 8 }$
A) $a _ { n } = 2 \cdot \left( \frac { 1 } { 2 } \right) ^ { n + 1 }$

B) $a _ { n } = 2 \cdot \left( \frac { 1 } { 4 } \right) ^ { n - 1 }$

C) $\mathrm { a } _ { \mathrm { n } } = 2 \cdot \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } - 1 }$

D) $a _ { n } = 2 \cdot \left( \frac { 1 } { 2 } \right) ^ { n }$

Question

Find the nth term of the geometric sequence.
-$2,1 , \frac { 1 } { 2 } , \frac { 1 } { 4 } , \frac { 1 } { 8 }$
A) $a _ { n } = 2 \cdot \left( \frac { 1 } { 2 } \right) ^ { n + 1 }$

B) $a _ { n } = 2 \cdot \left( \frac { 1 } { 4 } \right) ^ { n - 1 }$

C) $\mathrm { a } _ { \mathrm { n } } = 2 \cdot \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } - 1 }$

D) $a _ { n } = 2 \cdot \left( \frac { 1 } { 2 } \right) ^ { n }$

Quizplus · Accepted Answer

The sequence is a geometric sequence with the first term $a_1 = 2$ and common ratio $r = \frac{1}{2}$. The nth term of a geometric sequence is given by $a_n = a_1 \cdot r^{n-1}$. Substituting $a_1 = 2$ and $r = \frac{1}{2}$ gives $a_n = 2 \cdot \left( \frac{1}{2} \right)^{n-1}$.

Find the Nth Term of the Geometric Sequence 2,1,12,14,182,1 , \frac { 1 } { 2 } , \frac { 1 } { 4 } , \frac { 1 } { 8 }2,1,21​,41​,81​

Find the Nth Term of the Geometric Sequence $2,1 , \frac { 1 } { 2 } , \frac { 1 } { 4 } , \frac { 1 } { 8 }$