First we need to find the common ratio, which we can do by dividing the second term by the first:

$\displaystyle -1,449,459 \div 1,771,561 = 0.\overline{81}$

The first term is $\displaystyle 1,771, 561 (-0.\overline{81} ) ^ 0$, the second term is $\displaystyle 1,771, 561 (-0.\overline{81})^ 1$, so the 26th term is $\displaystyle 1,771, 561 (-0.\overline{81})^{ 25} = -11,738.206$

Find the common ratio for this geometric series:

$\displaystyle 1,7,49,343,2401$

The common ratio of a geometric series can be found by dividing any term by the term before it. More generally:

$\displaystyle r=\frac{N_t}{N_{t-1}}$

So, do try the following:

$\displaystyle r=\frac{343}{49}=7$

So our common ratio is 7

Find the next term in this geometric series:

$\displaystyle 1,7,49,343,2401$

to find the next term, we must first find the common ratio.

The common ratio of a geometric series can be found by dividing any term by the term before it. More generally:

$\displaystyle r=\frac{N_t}{N_{t-1}}$

So, do try the following:

$\displaystyle r=\frac{343}{49}=7$

So our common ratio is 7

Next, find the next term in the series by multiplying our last term by 7

$\displaystyle N_{t+1}=2401*7=16807$

Making our next term 16807

Use the geometric series summation formula.

$\displaystyle \sum_{i=1}^{n}a_i=a(\frac{1-r^n}{1-r})$

Substitute $\displaystyle S_2 = \frac{2}{3}$ into $\displaystyle \sum_{i=1}^{n}a_i$, and replace the $\displaystyle r$ and $\displaystyle n$ terms.  The value of $\displaystyle n$ is two.

$\displaystyle \frac{2}{3}=a(\frac{1-(\frac{1}{6})^2}{1-\frac{1}{6}})$

Simplify the terms on the right and solve for $\displaystyle a$.

$\displaystyle \frac{2}{3}=a(\frac{1-\frac{1}{36}}{1-\frac{1}{6}})$

$\displaystyle \frac{2}{3}=a(\frac{\frac{35}{36}}{ \frac{5}{6}} )$

Rewrite the complex fraction using a division sign.

$\displaystyle \frac{2}{3}=a(\frac{35}{36} \div \frac{5}{6})$

Change the sign from division to a multiplication sign and switch the second term.

$\displaystyle \frac{2}{3}=a(\frac{35}{36} \times \frac{6}{5})$

Simplify the terms inside the parentheses.

$\displaystyle \frac{2}{3}=a(\frac{7}{6})$

Isolate the variable by multiplying six-seventh on both sides.

$\displaystyle \frac{2}{3}\cdot \frac{6}{7}=a(\frac{7}{6}) \cdot \frac{6}{7}$

Simplify both sides.

$\displaystyle a=\frac{4}{7}$

The answer is:  $\displaystyle \frac{4}{7}$

Divide the second term with the first term, and the third term with the second term.

$\displaystyle \frac{4}{3}\div\frac{1}{3} = \frac{4}{3}\times \frac{3}{1} = 4$

$\displaystyle \frac{16}{3}\div\frac{4}{3}=\frac{16}{3}\times\frac{3}{4}=4$

The common ratio of this geometric sequence is four.

Multiply the third term by four.

$\displaystyle \frac{16}{3}\times 4 = \frac{64}{3}$

The answer is:  $\displaystyle \frac{64}{3}$

Write the formula to find the n-th term for a geometric sequence.

$\displaystyle a_n = a_1\cdot r^{n-1}$

Substitute the known values into the equation.

$\displaystyle a_{10} = (\frac{1}{3})\cdot (\frac{2}{5})^{10-1}= \frac{512}{5859375}$

This fraction is equivalent to:  $\displaystyle 8.738\times 10^{-5}$

The 19th term is:  $\displaystyle 8.738\times 10^{-5}$

Write the sum formula for a geometric series.

$\displaystyle \textup{Sum} = \frac{a_1(1-r^n)}{1-r}$

$\displaystyle a_1 =\textup{ first term} = 10$

$\displaystyle r=\textup{common ratio} = \frac{1}{10}$

Identify the number of terms that exist.

$\displaystyle [10,1,\frac{1}{10},\frac{1}{100},\frac{1}{1,000},\frac{1}{10,000}]$

There are six existing terms:  $\displaystyle n=6$

Substitute the terms into the formula.

$\displaystyle \textup{Sum} = \frac{a_1(1-r^n)}{1-r} = \frac{10(1-(\frac{1}{10})^6)}{1-\frac{1}{10}} = \frac{10-\frac{1}{100,000}}{\frac{10}{10}-\frac{1}{10}}$

$\displaystyle = \frac{\frac{1,000,000}{100,000}-\frac{1}{100,000}}{\frac{10}{10}-\frac{1}{10}} = \frac{\frac{999,999}{100,000}}{\frac{9}{10}}$

Simplify the complex fraction.

$\displaystyle \frac{999,999}{100,000} \div \frac{9}{10} = \frac{999,999}{100,000} \times \frac{10}{9} = \frac{111,111}{10,000}$

The sum is:  $\displaystyle \frac{111,111}{10,000}$

Write the formula for the n-th term of the geometric sequence.

$\displaystyle a_n = a_1r^{n-1}$

Substitute the first term and common ratio.

The equation is:  $\displaystyle a_n = 4(\frac{2}{3})^{n-1}$

To find the fifth term, substitute $\displaystyle n=5$.

$\displaystyle a_5 = 4(\frac{2}{3})^{5-1} = 4(\frac{2}{3})^4 =4(\frac{2}{3})(\frac{2}{3})(\frac{2}{3})(\frac{2}{3})= 4(\frac{16}{81})$

The answer is:  $\displaystyle \frac{64}{81}$

The formula for geometric sequences is defined by:  

$\displaystyle ar^{n-1}$

The term $\displaystyle a$ represents the first term, while $\displaystyle r$ is the common ratio.  The term $\displaystyle n$ represents the terms.

Substitute the known values.

$\displaystyle 3(3)^{n-1}$

To determine the seventh term, simply substitute $\displaystyle n=7$ into the expression.

$\displaystyle 3(3)^{7-1}$

The answer is:  $\displaystyle 2187$

The $\displaystyle n$th term of a geometric sequence is 

$\displaystyle a_{n} = a_{1} r ^{n-1}$

where $\displaystyle a_{1}$ is its initial term and $\displaystyle r$ is the common ratio between the terms. 

$\displaystyle a_{1} = 6$

$\displaystyle r= \frac{a_{2}}{a_{1}} = \frac{18}{6} = 3$

Substituting, the expression becomes 

$\displaystyle a_{n} = 6 \cdot 3 ^{n-1}$

By factoring and remultiplying, this becomes

$\displaystyle a_{n} = 2 \cdot 3 \cdot 3 ^{n-1}$

$\displaystyle a_{n} = 2 \cdot 3^{1} \cdot 3 ^{n-1}$

$\displaystyle a_{n} = 2 \cdot 3^{1+n-1}$

$\displaystyle a_{n} = 2 \cdot 3^{ n }$