Question

In an arithmetic sequence, a_17 = -40 and a_28 = -73. Please explain how to use this information to write a recursive formula for this sequence!

Answer 1

An arithmetic sequence

$a_1,a_2,a_3,\ldots,a_n,\ldots$

is one in which consecutive terms of the sequence differ by a fixed number, call it d. This means that, given the first term $a_1$, we can build the sequence by simply adding d :

$a_2=a_1+d$

$a_3=a_2+d$

$a_4=a_3+d$

and so on, the general pattern governed by the recursive rule,

$a_n=a_{n-1}+d$

We can exploit this rule in order to write any term of the sequence in terms of the first one. For example,

$a_3=a_2+d=(a_1+d)+d=a_1+2d$

$a_4=a_3+d=(a_1+2d)+d=a_1+3d$

and so on up to

$a_n=a_1+(n-1)d$

In this case, we're not given the first term right away, but the 17th. But this isn't a problem; we can use the same exploit to get

$a_{18}=a_{17}+d$

$a_{19}=a_{17}+2d$

$a_{20}=a_{17}+3d$

and so on, up to the next term we know,

$a_{28}=a_{17}+11d=-40+11d$

(Notice how the subscript of a on the right and the coefficient of d add up to the subscript of a on the left.)

The 28th term is -73, so we can solve for d :

$-73=-40+11d\implies -33=11d\implies d=-3$

To get the first term of the sequence, we use the rule found above and either of the known values of the sequence. For instance,

$a_{17}=a_1+16d\implies-40=a_1-16\cdot3\implies a_1=8$

Then the recursive rule for this particular sequence is

$\begin{cases}a_1=8\\a_n=a_{n-1}-3&\text{for }n>1\end{cases}$