Question

Which term of the sequence 2 , 1 , 1/2 , 1/4 , ......is 1/128.

Answer 1

This is a Geometric Sequence
Each term is the form [tex]ar^{n-1[tex]a = 2 [/tex]}[/tex]

for n = 1
$a = 2$
for n = 2
[tex] 2r = 1 [/tex]
$r= \frac{1}{2}$

therefore
$ar^{n-1}=2(\frac{1}{2} )^{n-1}= \frac{1}{128}= \frac{1}{2^7}$
$n=9$
So its the 9th term

Answer 2

Answer:

9th term

Step-by-step explanation:

Sequence = $2 ,1, \frac{1}{2}, \frac{1}{4},....,$

So, Common Ratio = r = $\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{1}{2}$

First term = a = 2

Since it is an G.P.

Formula of nth term :$a_n=ar^{n-1}$

n is no. of term

Now to calculate which term of the sequence is $\frac{1}{128}$

So, $\frac{1}{128}=(2)(\frac{1}{2}^{n-1})$

$\frac{1}{256}=(\frac{1}{2}^{n-1})$

$\frac{1}{2})^8=(\frac{1}{2}^{n-1})$

On comparing 8 = n-1

So, n = 9

So, $\frac{1}{128}$ is the 9th term of sequence .