Question

The following Geometric Sequence is Given:

2, 2^(x+1), 2^(2x+1), 2^(3x+1),...

i need to find 3 things.

1) the nth term An.
2)the fifth term A5.
3) the eight term A8.

in order to find A5 and A8 i need to find the common ratio.

The formula to find the common ratio is :

r = a(k+1)/ak

so far i have r = a2/a1 = 2^(x+10) / 2

my problem is that i don't know how to solve

r = 2^(x+1) / 2

Answer 1

Alright, so we have the sequence given as:\[2,~~2^{x+1},~~2^{2x+1},~~2^{3x+1}~~...\]
Which can also be written as: \[2,~~2^x.2,~~2^{2x}.2,~~2^{3x}.2~~...\]
Which again can be written as: \[2,~~~(2^x).2,~~~{(2^x)}^2.2,~~~{(2^x)}^3.2~~...\]
NOTE: \(2^{ax} = {(2^x)}^a = {(2^a)}^x\)

Now, this should give you a sort of intuitive feel about the common ratio. What is being multiplied to the previous term to get the next term?

It is: \(2^x\)

Which can also be calculated using the formula: \[r = \frac{a_{n}}{a_{n-1}} = \frac{a_{n+1}}{a_{n}}\]
Now you have to calculate:\[a_n = a.r^{n-1}\]


Can you do that now? :)

Answer 2

How did you get \[2^x.2\] from \[2^{x+1}\]?

Answer 3

That's what it means.
Using the law of indices.

Let's say we have: \[a^2.a^3\]
This equals: \[a^{2+3} = a^5\]