Question

find t(3)
t(n)=6(\frac{1}{2})^n;\frac{3}{4}
find the first term.
t(n)= - 8(3)^n; - 24

Explanation:

Step1: Recall geometric - sequence formula

The general formula for a geometric sequence is $t(n)=t(0)\cdot r^{n}$, where $t(0)$ is the zeroth - term, $r$ is the common ratio, and $n$ is the term number.

Step2: Find $t(3)$ for $t(n)=6(\frac{1}{2})^{n}$

Substitute $n = 3$ into the formula $t(n)=6(\frac{1}{2})^{n}$. Then $t(3)=6\times(\frac{1}{2})^{3}=6\times\frac{1}{8}=\frac{6}{8}=\frac{3}{4}$.

Step3: Find the first - term for $t(n)=-8(3)^{n}$

The first - term corresponds to $n = 1$. Substitute $n = 1$ into the formula $t(n)=-8(3)^{n}$. Then $t(1)=-8\times3=-24$.

Answer:

For $t(n)=6(\frac{1}{2})^{n}$, $t(3)=\frac{3}{4}$; for $t(n)=-8(3)^{n}$, the first - term is $-24$.