How do you find the 6th term in the geometric sequence \[25,75,225,675,...\]?

Hint: The general formula for a nth term of a geometric sequence is \[a{{r}^{n-1}}\]. 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.

Complete step by step solution:
We are given the infinite geometric series \[25,75,225,675,...\]. Here, the first term is 25, so \[a=25\]. To find the common ratio, we need to take a ratio of a term with its previous term. Hence, we get the ratio as
\[r=\dfrac{75}{25}\], cancelling out the common factors, we get \[r=3\].
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
\[a{{r}^{n-1}}\]
As we want the 6th term to find, substituting \[n=6,r=3\And a=25\] in the above equation, we get \[\left( 25 \right){{\left( 3 \right)}^{6-1}}\]. Simplifying this expression, we get 6075.
Hence, the 6th term is 6075.

Note: For a general geometric series the formula for the sum of n terms is, \[\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\] for \[\left| r \right|<1\], and \[\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}\] for \[\left| r \right|>1\]. We can find the sum of infinite series only if the absolute value of the common ratio is less than one, that is \[\left| r \right|<1\].
We can derive the formula for infinite geometric series as,
\[\displaystyle \lim_{n \to \infty }\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\]
As \[\left| r \right|<1\], we can say that $r^{n}\to0$. Using this in the above limit, we get the summation formula as
\[\displaystyle \lim_{n \to \infty }\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}=\dfrac{a\left( 1-0 \right)}{1-r}=\dfrac{a}{1-r}\]