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Progressions: (lesson 1 of 2)

## Arithmetic Progressions

Definition:

By an arithmetic progression of $m$ terms, we mean a finite sequence of the form

$$a, a + d, a + 2d, a + 3d, . . . , a + ( m - 1)d.$$

The real number $a$ is called the first term of the arithmetic progression, and the real number $d$ is called the difference of the arithmetic progression.

Example 1:

Consider the sequence of numbers

$$1,\; 3, \; 5, \; 7, \; 9, \; 11, \; 13, \; 15, \; 17, \; 19, \; 21, \; 23$$

The property of this sequence is that the difference between successive terms is constant and equal to 2.

Here we have: $a = 1$; $d = 2$.

Example 2:

Consider the sequence of numbers:

$$2,\; 5, \; 8, \; 11, \; 14, \; 17, \; 20, \; 23, \; 26, \; 29, \; 32$$

The property of this sequence is that the difference between successive terms is constant and equal to 3.

Here we have: $a = 2$; $d = 3$.

### General term of arithmetic progression:

The general term of an arithmetic progression with first term $a_1$ and common difference $d$ is:

$$\color{blue}{a_k = a_1 + (k - 1)d}$$

Example 3: Find the general term for the arithmetic sequence $-1, \; 3, \; 7, \; 11, \; . . .$ and then find $a_{12}$.

Solution:

Here $a_1 = 1$. To find $d$ subtract any two adjacent terms: $d = 7 - 3 = 4$. The general term is:

\begin{aligned} \color{blue}{a_k} &\color{blue}{=} \color{blue}{a_1 + ( k - 1 ) d} \\ a_k &= -1 + ( k - 1 ) \cdot 4 \\ a_k &= -1 + 4k - 4 \\ \color{blue}{a_k} &\color{blue}{=} \color{blue}{4k - 5} \end{aligned}

To find $a_{12}$, let $k = 12$.

$$a_{12} = 4 \cdot 12 - 5 = 43$$

Example 4: If $a_3 = 8$ and $a_6 = 17$, find $a_{14}$.

Solution:

Use the formula for $a_k$ with the given terms

\begin{aligned} a_3 &= a_1 + (3 - 1) \cdot d \\ 8 &= a_1 + 2d \\ \\ a_6 &= a_1 + (6 - 1) \cdot d \\ 17 &= a_1 + 5d \end{aligned}

This gives us a system of two equations with two variables. By solving them, we can find that $a_1 = 2$ and $d=3$.

Use the formula for $a_k$ to find $a_{14}$

\begin{aligned} a_k &= a_1 + (k - 1) \cdot d \\ a_{14} &= 2 + (14 - 1) \cdot 3 \\ \color{blue}{a_{14}} &\color{blue}{=} \color{blue}{41} \end{aligned}

Exercise:

Level 1

 $$\color{blue}{ a_1 = 3, \ d = 4 \\ a_6 = ?? }$$ $a_6 = 20$ $a_6 = 21$ $a_6 = 22$ $a_6 = 23$

Level 2

 $$\color{blue}{a_4 = 7, \ a_8 = 15 \\ a_{10} = ?? }$$ $a_{10} = 18$ $a_{10} = 19$ $a_{10} = 20$ $a_{10} = 21$

### Sum of an arithmetic progression:

The sum of the $n$ terms of an arithmetic progression with first term $a_1$ and common difference $d$ is:

$$\color{blue}{a_n = \frac{n}{2} [ 2{a_1} + (n - 1)d ]}$$

Also, the sum of an arithmetic progression is equal to

$$\color{blue}{S_n = \frac{n}{2} (a_1 + a_n)}$$

Example 5: Find the sum of the $10$ terms of the arithmetic progression if $a_1 = 5$ and $d = 4$.

Solution:

\begin{aligned} S_n &= \frac{n}{2} [ 2 a_1 + (n - 1) d ] \\ S_{10} &= \frac {10}{2} [ 2 \cdot 5 + ( 10 - 1 ) \cdot 4 ] \\ S_{10} &= 5 \cdot 46 \\ \color{blue}{S_{10}} &\color{blue}{=} \color{blue}{230} \end{aligned}

Example 6: Find $1 + 2 + 3 + . . . + 100$

Solution:

In this example, we have: $a_1 = 1$, $d = 1$, $n = 100$, $a_{100} = 100$. The sum is:

\begin{aligned} S_n &= \frac{n}{2} ( a_1 + a_n ) \\ S_{100} &= \frac {100}{2} ( 1 + 100 ) \\ \color{blue}{S_{100}} &\color{blue}{=} \color{blue}{5050} \end{aligned}

Exercise:

Level 1

 $$\color{blue}{ a_1 = -6, \ d = 2 \\ S_8 = ?? }$$ $S_8 = 0$ $S_8 = 10$ $S_8 = 20$ $S_8 = 30$

Level 2

 $$\color{blue}{a_4 = 11, \ a_7 = 20 \\ S_5 = ?? }$$ $S_5 = 20$ $S_5 = 30$ $S_5 = 40$ $S_5 = 50$