If the first term of an $A.P.$ is $3$ and the sum of its first $25$ terms is equal to the sum of its next $15$ terms,then the common difference of this $A.P.$ is:

Which term of the sequence $(-8 + 18i), (-6 + 15i), (-4 + 12i), \dots$ is purely imaginary (in $^{th}$)?

If the $7^{th}$ term of an $A.P.$ is $40$,then the sum of the first $13$ terms 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:

In an arithmetic progression,the sum of the first and third terms is $12$,and the product of the first and second terms is $24$. Find the first term.

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