Geometric Progression nth Term: Formula and Working Example

  • Thread starter ghostbuster25
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In summary, the expression for the nth term of the given geometric sequence with first three terms a1=100, a2=106, a3=112.36 and common ratio r=1.06 is un=100(1.06)(n-1). This can be confirmed by checking that 100(1.06)(n-1) gives the correct values for the first three terms and it is not equivalent to 106(n-1).
  • #1
ghostbuster25
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just a check of my work please.

I have to write an expression for the nth term of this geometric sequence.

a1=100
a2=106
a3=112.36

I've worked out the ratio to be r=1.06

I am using the formula un=ar(n-1)

so the expression i have come up with is un=100(1.06)(n-1)

Is this correct? I have checked and it does work. I tried multiplying 100(1.06) on its own first to give me the formula
106(n-1)
but that doesn't work...hence me doubting myself.

Any comments much appreciated
 
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  • #2
hi ghostbuster25! :wink:
ghostbuster25 said:
… so the expression i have come up with is un=100(1.06)(n-1)

Is this correct? I have checked and it does work. I tried multiplying 100(1.06) on its own first to give me the formula
106(n-1)
but that doesn't work...hence me doubting myself.

yup, un=100(1.06)(n-1) is fine :smile:

(i don't understand how you got 106(n-1) :confused:)
 
  • #3
ghostbuster, if what you were trying to do is convert [tex]100(1.06)^{n-1}[/tex] into [tex]106^{n-2}[/tex], this is wrong because what you should be doing is taking out the factor of 1.06 to give [tex]100(1.06)(1.06)^{(n-1)-1}=106(1.06)^{n-2}[/tex].

106n grows much faster than 1.06n.
 

What is a geometric progression?

A geometric progression is a sequence of numbers in which each term after the first is obtained by multiplying the previous term by a fixed number called the common ratio. For example, in the sequence 1, 2, 4, 8, 16, the common ratio is 2.

What is the formula for finding the nth term of a geometric progression?

The formula for finding the nth term of a geometric progression is: an = a1 * r(n-1), where an is the nth term, a1 is the first term, and r is the common ratio.

How do you use the formula for finding the nth term of a geometric progression?

To use the formula, plug in the values for a1, r, and n. Then, solve for an. For example, if a geometric progression has a first term of 2 and a common ratio of 3, the formula for the nth term would be an = 2 * 3(n-1).

Can you provide a working example of finding the nth term of a geometric progression?

Yes, let's take the geometric progression 2, 4, 8, 16, 32. To find the 6th term, we can use the formula an = a1 * r(n-1). In this case, a1 = 2, r = 2, and n = 6. Therefore, the 6th term would be a6 = 2 * 2(6-1) = 2 * 32 = 64. So, the 6th term in this geometric progression is 64.

What are some real-life applications of geometric progressions?

Geometric progressions are commonly used in finance, population growth, and other fields involving exponential growth. For example, interest rates in compound interest can be represented by a geometric progression, and population growth can follow a geometric progression if the population growth rate is constant.

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