Question

What is the 41st term of the sequence below? -13, -17, -21, -25....

Answer 1

Recursively, this sequence is given by

$\begin{cases}a_1=-13\\a_n=a_{n-1}-4&\text{for }n>1\end{cases}$

You can solve for the $n$th term explicitly:

$a_n=a_{n-1}-4=a_{n-2}-2\times4=a_{n-3}-3\times4=\cdots=a_1-(n-1)\times4$

So the explicit formula for this sequence would be

$a_n=-13-4(n-1)=-4n-9$

This means the 41st term is

$a_{41}=-4(41-1)-9=-173$