Terms in Arithmetic Progressions

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 $n$n^th 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:

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

QUESTION 3