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 :)