How many terms of the AP 12, 19, 26, 33, ... lie between 50 and 200?

$a_n=12+(n-1)7=7n+5$. For $a_n$ > 50 we need $n>6.43$, so the first such term is $a_7=54$; for $a_n$ < 200 we need $n<27.86$, so the last is $a_{27}=194$. That is 27 - 7 + 1 = 21 terms. Subtracting the term numbers without adding 1 gives 20, and 27 is the position of the last term, not the count.