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# Arithmetic sequences review

Review arithmetic sequences and solve various problems involving them.

## Parts and formulas of arithmetic sequences

In arithmetic sequences, the difference between consecutive terms is always the same. We call that difference the common difference.
For example, the common difference of the following sequence is $+2$:
$+2\phantom{\rule{0.167em}{0ex}}↷$$+2\phantom{\rule{0.167em}{0ex}}↷$$+2\phantom{\rule{0.167em}{0ex}}↷$
$3,$$5,$$7,$$9,\text{…}$
Arithmetic sequence formulas give $a\left(n\right)$, the ${n}^{\text{th}}$ term of the sequence.
This is the explicit formula for the arithmetic sequence whose first term is $k$ and common difference is $d$:
$a\left(n\right)=k+\left(n-1\right)d$
This is the recursive formula of that sequence:
$\left\{\begin{array}{l}a\left(1\right)=k\\ \\ a\left(n\right)=a\left(n-1\right)+d\end{array}$

## Extending arithmetic sequences

Suppose we want to extend the sequence $3,8,13,\text{…}$ We can see each term is $+5$ from the previous term:
$+5\phantom{\rule{0.167em}{0ex}}↷$$+5\phantom{\rule{0.167em}{0ex}}↷$$+5\phantom{\rule{0.167em}{0ex}}↷$
$3,$$8,$$13,\text{…}$
So we simply add that difference to find that the next term is $18$:
$+5\phantom{\rule{0.167em}{0ex}}↷$$+5\phantom{\rule{0.167em}{0ex}}↷$$+5\phantom{\rule{0.167em}{0ex}}↷$
$3,$$8,$$13,$$18,\text{…}$
Problem 1
What is the next term in the sequence $-5,-1,3,7,\dots$?

Want to try more problems like this? Check out this exercise.

## Writing recursive formulas

Suppose we want to write a recursive formula for $3,8,13,\text{…}$ We already know the common difference is $+5$. We can also see that the first term is $3$. Therefore, this is a recursive formula for the sequence:
$\left\{\begin{array}{l}a\left(1\right)=3\\ \\ a\left(n\right)=a\left(n-1\right)+5\end{array}$
Problem 1
Find $k$ and $d$ in this recursive formula of the sequence $-5,-1,3,7,\dots$.
$\left\{\begin{array}{l}a\left(1\right)=k\\ \\ a\left(n\right)=a\left(n-1\right)+d\end{array}$
$k=$
$d=$

Want to try more problems like this? Check out this exercise.

## Writing explicit formulas

Suppose we want to write an explicit formula for $3,8,13,\text{…}$ We already know the common difference is $+5$ and the first term is $3$. Therefore, this is an explicit formula for the sequence:
$a\left(n\right)=3+5\left(n-1\right)$
Problem 1
Write an explicit formula for $-5,-1,3,7,\dots$
$a\left(n\right)=$

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• How do you write an explicit formula for the sequence (8,11,16,23,32)
• Notice that the sequence of differences is simply the sequence of odd numbers. So this is the sequence of squares shifted by a constant. Since 8-1=7, we have
a(n) = n^2 +7
• Why was the Arithmetic Sequence review so hard on my mind? It took me to look at the explanation to understand the answers to every darn question. And, I am yet so confused on this section.
• you can check ncert (cbse)textbook math grade 10 and find the chapter Arithmetic progression.That might help.
• how are the formulas for arithmetic sequences similar to functions?
• Arithmetic sequences are functions. The difference between a linear function and a arithmetic sequence is that the first is continuous and the second is discrete, but the continuous line would go through all the points of its equivalent sequence.
• How would you find the explicit formula for a problem such as:
"a" subscript 38 =-53.2, while the common difference is -1.1?
• You need to know a common difference and one 'value' (sequence) in order to find the general form (formula). Since the common difference is -1.1, it would look like this: -1.1(n-1) + b (if you are starting from n=1).
Then plug in "a" subscript 38 = -53.2 into the formula. Then it would be -1.1(38-1) + b = -53.2
All you have to do is solving for 'b' and it would be -12.5.
Therefore, the general form (formula) would be -1.1(n-1) - 12.5 if you are starting from n=1.
• When given an arithmetic sequence with a negative term, how do you calculate what term the number will be when it reaches 0?
• Some sequences do not even have 0 as one of the possibilities. So if you start with a negative and have a negative common difference, then there is no zero such as -2, -4, -6, -8, ... Others will have a common difference that skips over zero such as -4, 4, 12, 20, 28, ... If the sequence does include a zero, such as an initial value of - 9 and a common difference of 3, we get an equation f(n) = -9 + 3(n-1). set this equal to zero, 0 = -9 + 3(n-1) move the - 9 by adding and distribute to get 9 = 3n - 3, add 3 to both sides, 3n = 12., divide by 3 to get n = 4. Sequence would be -9, -6, -3, 0 ... which is what we were looking for. So if your equation skipped 0 as above f(n) = -4 +8(n-1) and did the same thing, then 0 = -4 + 8n - 8, add 12 to get 12 = 8n, n = 12/8 = 3/2, but a sequence does not have a 3/2 term, only 1, 2, 3 ... terms.
• how do you write a recursive formula for 1, 1, 0, -1, -1, 0, 1, 1
• Consider the arithmetic sequence 27, 13, -1, ...
The explicit rule for the sequence in terms of n is a(n)= 27-14(n-1). If the nth term is -841, find the value of n.
• If the nth term is -841, then you were just given the result. Basically a(n)=-841
So, change the equation into -841=27-14(n-1).
You can now solve for n.
Hope this helps.
(1 vote)
• How would you write a explicit formula when f(0)=5, f(x)=f(x-1)+(2x-1)?