HallsofIvy said:Okay, so solve it! First, divide both sides by 3. If that sequence is actually geometric, you should be able to identify 1536/3 as a power of 2. I suggest you just calculate powers of 2: 1, 2, 4, 8, 16, ... until you get to that number.
There are errors in your work. Also, you need parentheses around your exponent expressions.nae99 said:1536= (3) (2)^n-1
1536/3 = 6^n-1[itex]/[/itex]3
512 = 2^n-1
512 = 2^10-1
512 = 2^9
n = 10
The above is incorrect. 3*2^(n - 1) [itex]\neq[/itex] 6^(n - 1)nae99 said:1536= (3) (2)^(n-1)
1536/3 = 6^n-1[itex]/[/itex]3
The above is also incorrect. [6^(n - 1)]/3 [itex]\neq[/itex] 2^(n - 1)nae99 said:512 = 2^n-1
nae99 said:512 = 2^10-1
512 = 2^9
n = 10
A geometric progression is a sequence of numbers where each term is found by multiplying the previous term by a constant value called the common ratio. For example, in the sequence 2, 6, 18, 54, the common ratio is 3.
In an arithmetic progression, each term is found by adding a constant value to the previous term. In a geometric progression, each term is found by multiplying a constant value to the previous term. Also, in a geometric progression, the common ratio does not have to be a whole number, unlike in an arithmetic progression.
The formula for finding the nth term of a geometric progression is: an = a1 * rn-1, where an represents the nth term, a1 is the first term, and r is the common ratio.
Geometric progressions can be used to model many real-life situations, such as population growth, compound interest, and the depreciation of assets. They can also be used in various branches of science, such as physics, chemistry, and biology, to describe processes that exhibit exponential growth or decay.
The sum of a finite geometric progression can be found using the formula: Sn = a1 * (1 - rn) / (1 - r), where Sn represents the sum of the first n terms, a1 is the first term, and r is the common ratio. If the common ratio is less than 1, the sum will approach a finite number as n approaches infinity.