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+(n−1)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+(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 $t_1=8\times1-11=-3$t1=8×1−11=−3 and that $t_2=8\times2-11=5$t2=8×2−11=5 and that $t_{100}=8\times100-11=789$t100=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 $t_n=87+\left(n-1\right)\times-7$tn=87+(n−1)×−7
This simplifies to $t_n=80-7n$tn=80−7n 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=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 $t_5=38$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 $\left(8d-4d\right)=66-38$(8d−4d)=66−38. 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+(n−1)d=10+(n−1)×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 1
QUESTION 2
An arithmetic progression has a first term of $T_1=a$T1=a and a common difference of $d$d.
Two of the terms in the sequence are $T_7=43$T7=43 and $T_{14}=85$T14=85.
Determine $d$d, the common difference.
Determine $a$a, the first term in the sequence.
State the equation for $T_n$Tn, the $n$nth term in the sequence.
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(n−1).
Determine $a$a, the first term in the arithmetic progression.
Determine $d$d, the common difference.
Determine $T_9$T9, the $9$9th term in the sequence.