If the first, second and last terms of the Arithmetic Progression are 5, 9, 101, respectively, how many total number of terms are there in the Progression?

The given progression is an Arithmetic Progression (AP). The first term is $a_1 = 5$. The second term is $a_2 = 9$. The last term is $a_n = 101$. The common difference $d$ of an AP is the difference between consecutive terms: $d = a_2 - a_1 = 9 - 5 = 4$. The formula for the $n$-th term of an AP is $a_n = a_1 + (n-1)d$. We are given $a_n = 101$, $a_1 = 5$, and $d=4$. Substitute these values into the formula to find $n$: $101 = 5 + (n-1)4$. Subtract 5 from both sides: $101 - 5 = (n-1)4$. $96 = (n-1)4$. Divide both sides by 4: $n-1 = \frac{96} {4}$. $n-1 = 24$. Add 1 to both sides: $n = 24 + 1 = 25$. There are 25 terms in the Arithmetic Progression.