Let $a_1, a_2, a_3, \dots, a_n$ be in $A.P$. If $a_3 + a_7 + a_{11} + a_{15} = 72$,then the sum of its first $17$ terms is equal to:

If the sum of a certain number of terms of the $A.P.$ $25, 22, 19, \ldots$ is $116$,find the last term.

Let $a_1, a_2, \ldots, a_{2024}$ be an Arithmetic Progression such that $a_1 + (a_5 + a_{10} + a_{15} + \ldots + a_{2020}) + a_{2024} = 2233$. Then $a_1 + a_2 + a_3 + \ldots + a_{2024}$ is equal to . . . . . .

For $x \geq 0$,the least value of $K$,for which $4^{1+x}+4^{1-x}$,$\frac{K}{2}$,and $16^{x}+16^{-x}$ are three consecutive terms of an $A.P.$ is equal to :

Insert $6$ numbers between $3$ and $24$ such that the resulting sequence is an $A.P.$

Let $a_{1}, a_{2}, \ldots, a_{21}$ be an $A.P.$ such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}} = \frac{4}{9}$. If the sum of this $A.P.$ is $189$,then $a_{6} a_{16}$ is equal to: