A geometric sequence goes from one term to the next by always multiplying or dividing by the constant value except 0. The constant number multiplied (or divided) at each stage of a geometric sequence is called the common ratio (r).

We have:

$a_1=\dfrac{1}{2}\\\\a_2=2\\\\a_3=8\\\\a_4=32\\\vdots$

Find the common ratio:

$\boxed{r=\dfrac{a_n}{a_{n-1}}}\Rightarrow r=\dfrac{2}{\frac{1}{2}}=\dfrac{8}{2}=\dfrac{32}{8}=...\\\\\boxed{r=4}$

<h3>The formula of the n-th term of a geometric sequence:</h3>

$\boxed{a_n=a_1\cdot r^{n-1}}$

Substitute:

$a_1=\dfrac{1}{2},\ r=4,\ n=8\\\\a_8=\dfrac{1}{2}\cdot4^{8-1}=\dfrac{1}{2}\cdot4^7=\dfrac{1}{2}\cdot16384\\\\\huge\boxed{a_8=8192}$

<h3>Method step by step:</h3>

$a_4=32,\ r=4\\\\a_5=a_4\cdot r\Rightarrow a_5=32\cdot4=128\\\\a_6=a_5\cdot r\Rightarrow a_6=128\cdot4=512\\\\a_7=a_6\cdot r\Rightarrow a_7=512\cdot4=2048\\\\a_8=a_7\cdot r\Rightarrow a_8=2048\cdot4=8192$

3 0

2 years ago

What is the 9th term in the following sequence?

Dmitry_Shevchenko [17]

The answer is 59 just add 6 to each number

6 0

2 years ago