A geometric sequence is given. Find the common ratio and write out the first four terms.
-$$\left\{ 2 \left( \frac { 1 } { 2 } \right) ^ { n - 1 } \right\}$$
A) $\mathrm { a } _ { \mathrm { n } } = 2 \cdot ( 2 ) ^ { \mathrm { n } - 1 }$
$r = 2 ; 2,4,8,16 , \ldots$

B) $a _ { n } = 2 \left( \frac { 1 } { 4 } \right) ^ { n - 1 }$
$r = \frac { 1 } { 4 } ; 2 , \frac { 1 } { 2 } , \frac { 1 } { 8 } , \frac { 1 } { 32 } , \ldots$

C) $a _ { n } = 2 \left( \frac { 1 } { 2 } \right) ^ { n - 1 }$
$\mathrm { r } = \frac { 1 } { 2 } ; 2,1 , \frac { 1 } { 2 } , \frac { 1 } { 4 } , \ldots$

D) $a _ { n } = \frac { 1 } { 2 } ( 2 ) ^ { n - 1 }$
$r = \frac { 1 } { 2 } ; 2,4,8,16 , \ldots$

Question

A geometric sequence is given. Find the common ratio and write out the first four terms.
-$$\left\{ 2 \left( \frac { 1 } { 2 } \right) ^ { n - 1 } \right\}$$
A) $\mathrm { a } _ { \mathrm { n } } = 2 \cdot ( 2 ) ^ { \mathrm { n } - 1 }$
$r = 2 ; 2,4,8,16 , \ldots$

B) $a _ { n } = 2 \left( \frac { 1 } { 4 } \right) ^ { n - 1 }$
$r = \frac { 1 } { 4 } ; 2 , \frac { 1 } { 2 } , \frac { 1 } { 8 } , \frac { 1 } { 32 } , \ldots$

C) $a _ { n } = 2 \left( \frac { 1 } { 2 } \right) ^ { n - 1 }$
$\mathrm { r } = \frac { 1 } { 2 } ; 2,1 , \frac { 1 } { 2 } , \frac { 1 } { 4 } , \ldots$

D) $a _ { n } = \frac { 1 } { 2 } ( 2 ) ^ { n - 1 }$
$r = \frac { 1 } { 2 } ; 2,4,8,16 , \ldots$

Quizplus · Accepted Answer

The given sequence is explicitly defined as $a_n = 2 \left( \frac{1}{2} \right)^{n-1}$, which matches choice C. The common ratio $r$ is obtained by the base of the exponent, $\frac{1}{2}$, and the first four terms are calculated by substituting $n = 1, 2, 3, 4$ into the formula, resulting in $2, 1, \frac{1}{2}, \frac{1}{4}$, respectively.

A Geometric Sequence Is Given {2(12)n−1}\left\{ 2 \left( \frac { 1 } { 2 } \right) ^ { n - 1 } \right\}{2(21​)n−1}

A Geometric Sequence Is Given $\left\{ 2 \left( \frac { 1 } { 2 } \right) ^ { n - 1 } \right\}$