Terms in Arithmetic Progressions
Recall that an arithmetic progression starts with a first term, commonly called $$a, and then either increases or decreases by a constant amount called the common difference $$d. The progression $$−3,5,13,21 for example is an arithmetic progression with $$a=−3 and $$d=8.
The generating rule for arithmetic progressions is quite easily found. Using our example in the first paragraph,
we could say that the first term is given by $$t1=−3,
the second term is given by $$t2=5=−3+1×8 ,
the third term is given by $$t3=13=−3+2×8 ,
the fourth term $$t4=21=−3+3×8 and so on.
The pattern starts to become clear and we could guess that the tenth term becomes $$t10=69=−3+9×8 and the one-hundredth term $$t100=789=−3+99×8
We could conclude that the generating rule for any arithmetic progression is given by $$tn=a+(n−1)d where $$tn is the $$nth term of the sequence.
In our example we have that $$tn=−3+(n−1)×8 and this can be simplified by expanding the brackets and collecting like terms as follows:
| $$tn | $$= | $$−3+(n−1)×8 |
| $$tn | $$= | $$−3+8n−8 |
| $$tn | $$= | $$8n−11 |
Checking, we see that $$t1=8×1−11=−3 and that $$t2=8×2−11=5 and that $$t100=8×100−11=789.
Trying another example, the decreasing arithmetic sequence $$87,80,73,66..., ... has $$a=87 and $$d=−7. Using our generating formula we have $$tn=87+(n−1)×−7
This simplifies to $$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 $$t12=80−7×12=−4.
Sometimes we are provided with two terms of an arithmetic sequence and then asked to find the generating rule. For example, suppose a certain arithmetic progression has $$t5=38 and $$t9=66. This mean we can write down two equations:
$$a+4d=38 (1)
$$a+8d=66 (2)
If we now subtract equation (1) from equation (2) the first term in each equation will cancel out to leave us with $$(8d−4d)=66−38. This means $$4d=28 and so $$d=7.
With the common difference found to be $$7, then we know that, using equation (1) $$a+4×7=38 and so $$a is clearly $$10. The general term is given by $$tn=a+(n−1)d=10+(n−1)×7 and this simplifies to $$tn=3+7n.
Checking, we see $$t5=3+7×5=38 and $$t9=3+7×9=66.
Worked Examples
QUESTION 1
QUESTION 2
An arithmetic progression has a first term of $$T1=a and a common difference of $$d.
Two of the terms in the sequence are $$T7=43 and $$T14=85.
Determine $$d, the common difference.
Determine $$a, the first term in the sequence.
State the equation for $$Tn, the $$nth term in the sequence.
Hence find $$T25, the $$25th term in the sequence.
QUESTION 3
The $$nth term in an arithmetic progression is given by the formula $$Tn=15+5(n−1).
Determine $$a, the first term in the arithmetic progression.
Determine $$d, the common difference.
Determine $$T9, the $$9th term in the sequence.