Question

What is the zero turn for the arithmetic sequence 128,96,64,32

Answer 1

You are falling by
`32`,

`128, 128 - 32, 128 - 32\cdot2,\dots`

In general your sequence is defined as,

`a_n=128-32\cdot n` where
`0\leq n \lt\infty`.

The question is at which
`n` does the value
`a_n=0`.

If you divide
`128` with
`32` you get the number of steps needed to stuff
`128` with
`32`,
`4`.

If you plug in
`n=4`, you get
`a_4=128-32\cdot4`, since
`32\cdot4=128` you get
`a_4=0`.

The zero turn of the arithmetic sequence is thus at
`n=4`.

Hope this helps :)