# 9.2 Arithmetic sequences  (Page 2/8)

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${a}_{n}={a}_{1}+\left(n-1\right)d$

Given the first term and the common difference of an arithmetic sequence, find the first several terms.

1. Add the common difference to the first term to find the second term.
2. Add the common difference to the second term to find the third term.
3. Continue until all of the desired terms are identified.
4. Write the terms separated by commas within brackets.

## Writing terms of arithmetic sequences

Write the first five terms of the arithmetic sequence    with ${a}_{1}=17$ and $d=-3$ .

Adding $\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}$ is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term.

The first five terms are $\text{\hspace{0.17em}}\left\{17,\text{\hspace{0.17em}}14,\text{\hspace{0.17em}}11,\text{\hspace{0.17em}}8,\text{\hspace{0.17em}}5\right\}$

List the first five terms of the arithmetic sequence with ${a}_{1}=1$ and $d=5$ .

Given any the first term and any other term in an arithmetic sequence, find a given term.

1. Substitute the values given for ${a}_{1},{a}_{n},n$ into the formula $\text{\hspace{0.17em}}{a}_{n}={a}_{1}+\left(n-1\right)d\text{\hspace{0.17em}}$ to solve for $\text{\hspace{0.17em}}d.$
2. Find a given term by substituting the appropriate values for $\text{\hspace{0.17em}}{a}_{1},n,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$ into the formula ${a}_{n}={a}_{1}+\left(n-1\right)d.$

## Writing terms of arithmetic sequences

Given ${a}_{1}=8$ and ${a}_{4}=14$ , find ${a}_{5}$ .

The sequence can be written in terms of the initial term 8 and the common difference $d$ .

$\left\{8,8+d,8+2d,8+3d\right\}$

We know the fourth term equals 14; we know the fourth term has the form ${a}_{1}+3d=8+3d$ .

We can find the common difference $d$ .

Find the fifth term by adding the common difference to the fourth term.

${a}_{5}={a}_{4}+2=16$

Given ${a}_{3}=7$ and ${a}_{5}=17$ , find ${a}_{2}$ .

${a}_{2}=2$

## Using recursive formulas for arithmetic sequences

Some arithmetic sequences are defined in terms of the previous term using a recursive formula    . The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is 5, then each term is the previous term plus 5. As with any recursive formula, the first term must be given.

$\begin{array}{lllll}{a}_{n}={a}_{n-1}+d\hfill & \hfill & \hfill & \hfill & n\ge 2\hfill \end{array}$

## Recursive formula for an arithmetic sequence

The recursive formula for an arithmetic sequence with common difference $d$ is:

$\begin{array}{lllll}{a}_{n}={a}_{n-1}+d\hfill & \hfill & \hfill & \hfill & n\ge 2\hfill \end{array}$

Given an arithmetic sequence, write its recursive formula.

1. Subtract any term from the subsequent term to find the common difference.
2. State the initial term and substitute the common difference into the recursive formula for arithmetic sequences.

## Writing a recursive formula for an arithmetic sequence

Write a recursive formula    for the arithmetic sequence    .

The first term is given as $-18$ . The common difference can be found by subtracting the first term from the second term.

$d=-7-\left(-18\right)=11$

Substitute the initial term and the common difference into the recursive formula for arithmetic sequences.

Do we have to subtract the first term from the second term to find the common difference?

No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference.

show that the set of all natural number form semi group under the composition of addition
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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