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Progressions: (lesson 1 of 2)

Arithmetic Progressions

Definition:

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

a,a+d,a+2d,a+3d,...,a+(m1)d.a, a + d, a + 2d, a + 3d, . . . , a + ( m - 1)d.

The real number aa is called the first term of the arithmetic progression, and the real number dd 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,  231,\; 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=1a = 1; d=2d = 2.

Example 2:

Consider the sequence of numbers:

2,  5,  8,  11,  14,  17,  20,  23,  26,  29,  322,\; 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=2a = 2; d=3d = 3.

General term of arithmetic progression:

The general term of an arithmetic progression with first term a1a_1 and common difference dd is:

ak=a1+(k1)d\color{blue}{a_k = a_1 + (k - 1)d}

Example 3: Find the general term for the arithmetic sequence 1,  3,  7,  11,  ... -1, \; 3, \; 7, \; 11, \; . . . and then find a12a_{12}.

Solution:

Here a1=1a_1 = 1. To find dd subtract any two adjacent terms: d=73=4d = 7 - 3 = 4. The general term is:

ak=a1+(k1)dak=1+(k1)4ak=1+4k4ak=4k5 \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 a12a_{12}, let k=12k = 12.

a12=4125=43a_{12} = 4 \cdot 12 - 5 = 43

Example 4: If a3=8a_3 = 8 and a6=17a_6 = 17, find a14a_{14}.

Solution:

Use the formula for aka_k with the given terms

a3=a1+(31)d8=a1+2da6=a1+(61)d17=a1+5d \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 a1=2a_1 = 2 and d=3d=3.

Use the formula for aka_k to find a14a_{14}

ak=a1+(k1)da14=2+(141)3a14=41 \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

a1=3, d=4a6=?? \color{blue}{ a_1 = 3, \ d = 4 \\ a_6 = ?? } a6=20 a_6 = 20
a6=21 a_6 = 21
a6=22 a_6 = 22
a6=23 a_6 = 23

Level 2

a4=7, a8=15a10=?? \color{blue}{a_4 = 7, \ a_8 = 15 \\ a_{10} = ?? } a10=18 a_{10} = 18
a10=19 a_{10} = 19
a10=20 a_{10} = 20
a10=21 a_{10} = 21

Sum of an arithmetic progression:

The sum of the nn terms of an arithmetic progression with first term a1a_1 and common difference dd is:

an=n2[2a1+(n1)d]\color{blue}{a_n = \frac{n}{2} [ 2{a_1} + (n - 1)d ]}

Also, the sum of an arithmetic progression is equal to

Sn=n2(a1+an)\color{blue}{S_n = \frac{n}{2} (a_1 + a_n)}

Example 5: Find the sum of the 1010 terms of the arithmetic progression if a1=5a_1 = 5 and d=4d = 4.

Solution:

Sn=n2[2a1+(n1)d]S10=102[25+(101)4]S10=546S10=230 \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+...+1001 + 2 + 3 + . . . + 100

Solution:

In this example, we have: a1=1a_1 = 1, d=1d = 1, n=100n = 100, a100=100a_{100} = 100. The sum is:

Sn=n2(a1+an)S100=1002(1+100)S100=5050 \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

a1=6, d=2S8=?? \color{blue}{ a_1 = -6, \ d = 2 \\ S_8 = ?? } S8=0 S_8 = 0
S8=10 S_8 = 10
S8=20 S_8 = 20
S8=30 S_8 = 30

Level 2

a4=11, a7=20S5=?? \color{blue}{a_4 = 11, \ a_7 = 20 \\ S_5 = ?? } S5=20 S_5 = 20
S5=30 S_5 = 30
S5=40 S_5 = 40
S5=50 S_5 = 50