The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,the number of terms is

If $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$,then

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

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:

Let $a_1, a_2, \ldots, a_n$ be in an $A$.$P$. If $a_5 = 2a_3$ and $a_{11} = 18$,then $12\left(\frac{1}{\sqrt{a_{10}}+\sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}}+\sqrt{a_{12}}} + \ldots + \frac{1}{\sqrt{a_{17}}+\sqrt{a_{18}}}\right)$ is equal to $..........$.

If $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in an arithmetic progression,then $x = \dots$