If the numbers $a, b, c, d, e$ form an $A.P.$,then the value of $a - 4b + 6c - 4d + e$ is

Let $S = \{(a, b, c) \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} : a+b+c=21, a \leq b \leq c\}$ and $T = \{(a, b, c) \in \mathbb{N} \times \mathbb{N} \times \mathbb{N} : a, b, c \text{ are in } AP\}$, where $\mathbb{N}$ is the set of all natural numbers. Then, the number of elements in the set $S \cap T$ is:

If the $m^{th}$ terms of the series $63 + 65 + 67 + 69 + \dots$ and $3 + 10 + 17 + 24 + \dots$ are equal,then $m = $

Consider an $A$.$P$.: $a_1, a_2, \dots, a_n$, with $a_1 > 0$. If $a_2 - a_1 = -\frac{3}{4}$, $a_n = \frac{1}{4} a_1$, and $\sum_{i=1}^n a_i = \frac{525}{2}$, then $\sum_{i=1}^{17} a_i$ is equal to:

If $a_1, a_2, a_3, \dots, a_n$ are in $A.P.$ and $a_1 + a_4 + a_7 + \dots + a_{16} = 114$,then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

The $15^{th}$ term of the arithmetic progression $4 + 9 + 14 + 19 + \dots$ is......