This is a sequence that features every other positive, odd integers.  The missing number in this case is 9.  

Find the next term in the following arithmetic series:

$\displaystyle 4,11,18,25,\square$

To find the next term in an arithmetic series, we need to find the common difference. To do so, find the difference between any two consecutive terms in the sequence:

$\displaystyle 11-4=18-11=25-18=7$

Our common difference is 7. Now we need to add that to the last term to get what we want

$\displaystyle 25+7=32$

So our next term is 32

What is the common difference of the following arithmetic series?

$\displaystyle 23,15,7,-1,-9,-17,-25$

To find the common difference, we need to find the difference between any  two consecutive terms.

Try with the first two:

$\displaystyle 15-23=-8$

To be sure, try it with the 2nd and 3rd

$\displaystyle 7-15=-8$

We keep getting the same thing, -8. It must be negative, because our sequence is decreasing. Therefore, we have our answer: -8

The sequence is decreasing by 3 each term. To get from the first term to the 16th term, you must subtract 3 fifteen times:

$\displaystyle 7-3(15)=-38$

Write the n-th term formula.

$\displaystyle a_n = a_1 + (n - 1)d$

The $\displaystyle a_1$ represents the first term, and $\displaystyle a_n$ is the last term.  

$\displaystyle a_1=2$

$\displaystyle a_n=44$

The $\displaystyle d$ is the common difference among the numbers.  

$\displaystyle d=2$ since each term increases by two.

$\displaystyle 44=2+(n-1)(2)$

Solve for $\displaystyle n$.

$\displaystyle 44=2+2n-2$

$\displaystyle 2n=44$

Divide by two on both sides.

$\displaystyle n=22$

The formula for n-terms in a arithmetic sequence is:

$\displaystyle \sum_{1}^{n}a_i=(\frac{n}{2})[a_1+a_n]$

Substitute the known terms.

$\displaystyle (\frac{n}{2})[a_1+a_n] = (\frac{22}{2})[2+44] = (11)(46)=506$

The answer is:  $\displaystyle 506$

Write the formula for the sum of an arithmetic series.

$\displaystyle Sum = n(\frac{a_1+a_n }{2})$

$\displaystyle \textup{n= number of terms}$

$\displaystyle a_ 1 =\textup{first term} = 4$

$\displaystyle a_ n =\textup{last term}=94$

To determine the value of $\displaystyle n$, use the formula:

$\displaystyle a_1 + (n - 1)d = a_n$

$\displaystyle \textup{d= common difference} = 5$

$\displaystyle 4 + (n - 1)5 = 94$

$\displaystyle 5(n-1)=90$

Divide by five on both sides.

$\displaystyle n-1=18$

$\displaystyle n=19$

Substitute all the terms into the sum formula.

$\displaystyle Sum = 19(\frac{4+94 }{2}) = 19(\frac{98}{2}) = 19(49)=931$

The answer is:  $\displaystyle 931$

Write the formula to determine the sum of an arithmetic series.

$\displaystyle S= n(\frac{a_1+a_n}{2})$

where $\displaystyle n$ is the number of terms, $\displaystyle a_1$ is the first term, and $\displaystyle a_n$ is the last term.

$\displaystyle a_1=2 , a_n =222$

Use the following formula to determine how many terms are in this series.

$\displaystyle a_1+(n-1)d =a_n$

The term $\displaystyle d$ is the common difference.  Since the numbers are spaced five units, $\displaystyle d=5$.

Substitute the known values and solve for n.

$\displaystyle 2+(n-1)5 =222$

Subtract two from both sides, and distribute the five through the binomial.

$\displaystyle 5n-5 =220$

Add five on both sides.

$\displaystyle 5n-5 +5=220+5$

$\displaystyle 5n=225$

Divide by five.

$\displaystyle n=45$

Plug this value and the other givens to the sum formula to determine the sum.

$\displaystyle S= n(\frac{a_1+a_n}{2}) = 45(\frac{2+222}{2}) = 45(112) = 5040$

The answer is:  $\displaystyle 5040$

This represents an arithmetic sequence.  Write the formula.  

$\displaystyle a_n=a_1+(n-1)d$

Substitute the first term and the common difference in the formula.

$\displaystyle a_n=4+(n-1)3$

Simplify the terms.

$\displaystyle a_n=4+3x-3$

The answer is:  $\displaystyle a_n=3n+1$

Write the formula for the arithmetic sequence.

$\displaystyle a_n = a_1+ d(n-1)$

The first term is:  $\displaystyle a_1=5$

The common difference is the same for each term, which is increasing by six every term:  $\displaystyle d=6$

Substitute and simplify the formula.

$\displaystyle a_n = 5+ 6(n-1)$

$\displaystyle a_n = 5+6n-6$

$\displaystyle a_n =6n-1$

To find the hundredth term, plug in $\displaystyle n=100$.

$\displaystyle a_{100} =6(100)-1 = 599$

The answer is:  $\displaystyle 599$

An arithmetic sequence is one in which each term is generated by adding the same number - the common difference - to the previous term. As can be seen here, the differences between each term and the previous term is not constant from term to term:

$\displaystyle 72 - 24 = 48$

$\displaystyle 24 - 16= 8$

The sequence cannot be arithmetic.

A geometric sequence is one in which each term is generated by multiplying the previous term by the same number - the common ratio. As can be seen here, the ratio of each term to the previous one is the same:

$\displaystyle \frac{24}{8} = 3$

$\displaystyle \frac{72}{24} = 3$

$\displaystyle \frac{216}{72} = 3$

The sequence could be geometric.