# How do you write the first five terms of the sequence defined recursively a_1=25, a_(k+1)=a_k-5, then how do you write the nth term of the sequence as a function of n?

Jul 17, 2017

We obtain general form for the equation of Arithmetic Progression from the reference:

${a}_{n} = {a}_{1} + \left(n - 1\right) d$

We are given ${a}_{1} = 25$ and the equation ${a}_{k + 1} = {a}_{k} - 5$ implies that $d = - 5$, therefore, the equation for the nth term is:

a_n = 25 + (n-1)(-5); {n in ZZ|n>0}

Jul 17, 2017

${a}_{k} = 30 - 5 k$, $k = 1 , 2 , . . . , n$

$\implies \underline{{a}_{n} = 30 - 5 n}$

Well, the ${a}_{1}$ term is given as $25$, and the successive terms ${a}_{k + 1}$ are five less than the current term, ${a}_{k}$. So, the sequence given is:

$25 , 20 , 15 , 10 , 5 , 0 , - 5 , . . .$

This is very similar to the graph $y = - 5 x$; the term index is represented by the $x$ value corresponding to the term represented by the $y$ value:

graph{-5x [-30.45, 42.63, -7.54, 28.97]}

In the graph, we analogously have:

$y = 25 , 20 , 15 , 10 , 5 , 0 , - 5 , . . .$
$x = - 5 , - 4 , - 3 , - 2 , - 1 , 0 , 1 , . . .$

This means the $k$th term, iterating $k$ from $1$ to $n$, can be found using the series:

$\textcolor{b l u e}{{a}_{k} = 30 - 5 k , k = 1 , 2 , 3 , . . . , n}$