Sequences and Series

Hong Kong

Stage 4 - Stage 5

Lesson

Recall that an arithmetic progression starts with a first term, commonly called $a$`a`, and then either increases or decreases by a constant amount called the common difference $d$`d`. The progression $-3,5,13,21$−3,5,13,21 for example is an arithmetic progression with $a=-3$`a`=−3 and $d=8$`d`=8.

The generating rule for arithmetic progressions is quite easily found. Using our example in the first paragraph,

we could say that the first term is given by $t_1=-3$`t`1=−3,

the second term is given by $t_2=5=-3+1\times8$`t`2=5=−3+1×8 ,

the third term is given by $t_3=13=-3+2\times8$`t`3=13=−3+2×8 ,

the fourth term $t_4=21=-3+3\times8$`t`4=21=−3+3×8 and so on.

The pattern starts to become clear and we could guess that the tenth term becomes $t_{10}=69=-3+9\times8$`t`10=69=−3+9×8 and the one-hundredth term $t_{100}=789=-3+99\times8$`t`100=789=−3+99×8

We could conclude that the generating rule for any arithmetic progression is given by $t_n=a+\left(n-1\right)d$`t``n`=`a`+(`n`−1)`d` where $t_n$`t``n` is the $n$`n`^{th} term of the sequence.

In our example we have that $t_n=-3+\left(n-1\right)\times8$`t``n`=−3+(`n`−1)×8 and this can be simplified by expanding the brackets and collecting like terms as follows:

$t_n$tn |
$=$= | $-3+\left(n-1\right)\times8$−3+(n−1)×8 |

$t_n$tn |
$=$= | $-3+8n-8$−3+8n−8 |

$t_n$tn |
$=$= | $8n-11$8n−11 |

Checking, we see that $t_1=8\times1-11=-3$`t`1=8×1−11=−3 and that $t_2=8\times2-11=5$`t`2=8×2−11=5 and that $t_{100}=8\times100-11=789$`t`100=8×100−11=789.

Trying another example, the *decreasing *arithmetic sequence $87,80,73,66...$87,80,73,66..., ... has $a=87$`a`=87 and $d=-7$`d`=−7. Using our generating formula we have $t_n=87+\left(n-1\right)\times-7$`t``n`=87+(`n`−1)×−7

This simplifies to $t_n=80-7n$`t``n`=80−7`n` and we can use this formula to find any term we like. For example, the first negative term in the sequence is given by $t_{12}=80-7\times12=-4$`t`12=80−7×12=−4.

Sometimes we are provided with two terms of an arithmetic sequence and then asked to find the generating rule. For example, suppose a certain arithmetic progression has $t_5=38$`t`5=38 and $t_9=66$`t`9=66. This mean we can write down two equations:

$a+4d=38$`a`+4`d`=38 (1)

$a+8d=66$`a`+8`d`=66 (2)

If we now subtract equation (1) from equation (2) the first term in each equation will cancel out to leave us with $\left(8d-4d\right)=66-38$(8`d`−4`d`)=66−38. This means $4d=28$4`d`=28 and so $d=7$`d`=7.

With the common difference found to be $7$7, then we know that, using equation (1) $a+4\times7=38$`a`+4×7=38 and so $a$`a` is clearly $10$10. The general term is given by $t_n=a+\left(n-1\right)d=10+\left(n-1\right)\times7$`t``n`=`a`+(`n`−1)`d`=10+(`n`−1)×7 and this simplifies to $t_n=3+7n$`t``n`=3+7`n`.

Checking, we see $t_5=3+7\times5=38$`t`5=3+7×5=38 and $t_9=3+7\times9=66$`t`9=3+7×9=66.

In an arithmetic progression where $a$`a` is the first term, and $d$`d` is the common difference, $T_7=43$`T`7=43 and $T_{14}=85$`T`14=85.

Determine $d$

`d`, the common difference.Determine $a$

`a`, the first term in the sequence.State the equation for $T_n$

`T``n`, the $n$`n`th term in the sequence.Hence find $T_{25}$

`T`25, the $25$25th term in the sequence.

The $n$`n`th term in an arithmetic progression is given by the formula $T_n=15+5\left(n-1\right)$`T``n`=15+5(`n`−1).

Determine $a$

`a`, the first term in the arithmetic progression.Determine $d$

`d`, the common difference.Determine $T_9$

`T`9, the $9$9th term in the sequence.