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*4=a_(n-3)-3*4=\cdots=a_1-(n-1)*4`

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`