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Question

Question: How do I find the equation of a geometric sequence?...

How do I find the equation of a geometric sequence?

Explanation

Solution

The general formula for an nth term of a geometric sequence is arn1a{{r}^{n-1}}. Here, a is the first term of the geometric series, and r is the common ratio of the series. We can find the common ratio by taking the ratio of a term with its previous term. By substituting the values for a, r, and n we can find the desired term that we want. We will take an example to make things more understandable.

Complete step by step solution:
We are given the infinite geometric series 25,75,225,675,...25,75,225,675,.... Here, the first term is 25, so a=25a=25. To find the common ratio, we need to take a ratio of a term with its previous term. Hence, we get the ratio as
r=7525r=\dfrac{75}{25},
Here, numerator and denominator have 25 as their highest common factor, cancelling out the common factors, we get r=3r=3.
Now, we have the first term and the common ratio. Substituting their values in the formula for the nth term of an geometric series. We get
arn1a{{r}^{n-1}}
25×3n125\times {{3}^{n-1}}
By substituting the different values of n, we can get the terms of the geometric progression

Note: For a general geometric series the formula for the sum of n terms is, a(1rn)1r\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r} for r<1\left| r \right|<1, and a(rn1)r1\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1} for r>1\left| r \right|>1. We can find the sum of infinite series only if the absolute value of the common ratio is less than one, that is r<1\left| r \right|<1.
We can derive the formula for infinite geometric series as,
limna(1rn)1r\displaystyle \lim_{n \to \infty }\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}
As r<1\left| r \right|<1, we can say that rn0r^{n} \to 0 . Using this in the above limit, we get the summation formula as
limna(1rn)1r=a(10)1r=a1r\displaystyle \lim_{n \to \infty }\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}=\dfrac{a\left( 1-0 \right)}{1-r}=\dfrac{a}{1-r}