New Zealand
Level 7 - NCEA Level 2

# Terms in Arithmetic Progressions

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 $t_1=-3$t1=3,

the second term is given by $t_2=5=-3+1\times8$t2=5=3+1×8 ,

the third term is given by $t_3=13=-3+2\times8$t3=13=3+2×8 ,

the fourth term $t_4=21=-3+3\times8$t4=21=3+3×8  and so on.

The pattern starts to become clear and we could guess that the tenth term becomes $t_{10}=69=-3+9\times8$t10=69=3+9×8  and the one-hundredth term $t_{100}=789=-3+99\times8$t100=789=3+99×8

We could conclude that the generating rule for any arithmetic progression is given by $t_n=a+\left(n-1\right)d$tn=a+(n1)d where $t_n$tn is the $n$nth term of the sequence.

In our example we have that $t_n=-3+\left(n-1\right)\times8$tn=3+(n1)×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 $t_1=8\times1-11=-3$t1=8×111=3  and that $t_2=8\times2-11=5$t2=8×211=5 and that $t_{100}=8\times100-11=789$t100=8×10011=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 $t_n=87+\left(n-1\right)\times-7$tn=87+(n1)×7

This simplifies to $t_n=80-7n$tn=807n and we can use this formula to find any term we like. For example, the first negative term in the sequence is given by $t_{12}=80-7\times12=-4$t12=807×12=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 $t_5=38$t5=38 and $t_9=66$t9=66. This mean we can write down two equations:

$a+4d=38$a+4d=38          (1)

$a+8d=66$a+8d=66          (2)

If we now subtract equation (1) from equation (2) the first term in each equation will cancel out to leave us with $\left(8d-4d\right)=66-38$(8d4d)=6638. This means $4d=28$4d=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 $t_n=a+\left(n-1\right)d=10+\left(n-1\right)\times7$tn=a+(n1)d=10+(n1)×7 and this simplifies to $t_n=3+7n$tn=3+7n

Checking, we see $t_5=3+7\times5=38$t5=3+7×5=38 and $t_9=3+7\times9=66$t9=3+7×9=66.

#### Worked Examples

##### QUESTION 2

In an arithmetic progression where $a$a is the first term, and $d$d is the common difference, $T_7=43$T7=43 and $T_{14}=85$T14=85.

1. Determine $d$d, the common difference.

2. Determine $a$a, the first term in the sequence.

3. State the equation for $T_n$Tn, the $n$nth term in the sequence.

4. Hence find $T_{25}$T25, the $25$25th term in the sequence.

##### QUESTION 3

The $n$nth term in an arithmetic progression is given by the formula $T_n=15+5\left(n-1\right)$Tn=15+5(n1).

1. Determine $a$a, the first term in the arithmetic progression.

2. Determine $d$d, the common difference.

3. Determine $T_9$T9, the $9$9th term in the sequence.

### Outcomes

#### M7-3

Use arithmetic and geometric sequences and series

#### 91258

Apply sequences and series in solving problems