Find $a _ { 1 }$ and and r for a geometric sequence { $\left\{ a _ { n } \right\}$ } from the given information.
-$a _ { 2 } = - 6$ and $a _ { 6 } = - \frac { 243 } { 128 }$
A) $a _ { 1 } = 8$ and $r = \frac { 3 } { 4 }$
B) $a _ { 1 } = - \frac { 32 } { 3 }$ and $r = - \frac { 3 } { 4 }$
C) $a _ { 1 } = 8$ and $r = - \frac { 3 } { 4 }$
D) $a _ { 1 } = 8$ and $r = - \frac { 4 } { 3 }$

Question

Quizplus · Accepted Answer

Given $a_2 = -6$ and $a_6 = -\frac{243}{128}$, we can use the formula for the $n$th term of a geometric sequence, $a_n = a_1 \cdot r^{n-1}$, to find $a_1$ and $r$. For $a_2$, we have $a_1 \cdot r = -6$, and for $a_6$, we have $a_1 \cdot r^5 = -\frac{243}{128}$. Dividing the second equation by the first gives $r^4 = \frac{-\frac{243}{128}}{-6} = \frac{243}{128} \cdot \frac{1}{6} = \frac{243}{768} = \left(\frac{3}{4}\right)^4$, indicating $r = \frac{3}{4}$ or $r = -\frac{3}{4}$. Since $a_2$ is negative and $a_6$ is also negative, $r$ must be negative to maintain the sign through multiplication, so $r = -\frac{3}{4}$. Substituting $r$ back into $a_1 \cdot r = -6$ gives $a_1 \cdot (-\frac{3}{4}) = -6$, solving for $a_1$ gives $a_1 = 8$. Therefore, $a_1 = 8$ and $r = -\frac{3}{4}$.

Find a1a _ { 1 }a1​ And and R for a Geometric Sequence

Find $a _ { 1 }$ And and R for a Geometric Sequence