If $m$ arithmetic means $(A.Ms)$ and three geometric means $(G.Ms)$ are inserted between $3$ and $243$ such that the $4^{\text{th}}$ $A.M.$ is equal to the $2^{\text{nd}}$ $G.M.$,then $m$ is equal to:

Define a sequence $\langle a_n \rangle$ by $a_1 = 5, a_n = a_1 a_2 \dots a_{n-1} + 4$ for $n > 1$. Then,$\lim_{n \to \infty} \frac{\sqrt{a_n}}{a_{n-1}}$

If the value of $\left(1+\frac{2}{3}+\frac{6}{3^{2}}+\frac{10}{3^{3}}+\ldots \text{ to } \infty\right)^{\log_{(0.25)}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \text{ to } \infty\right)}$ is $l$,then $l^{2}$ is equal to $......$

Let $a_1=b_1=1$ and $a_n=a_{n-1}+(n-1)$,$b_n=b_{n-1}+a_{n-1}$,$\forall n \geq 2$. If $S =\sum \limits_{n=1}^{10} \frac{b_n}{2^n}$ and $T =\sum \limits_{n=1}^8 \frac{n}{2^{n-1}}$,then $2^7(2S - T)$ is equal to $........$.

Let $\{a_{n}\}_{n=0}^{\infty}$ be a sequence such that $a_{0}=0, a_{1}=0$ and $a_{n+2}=3a_{n+1}-2a_{n}+1$ for all $n \geq 0$. Then the value of $a_{25}a_{23}-2a_{25}a_{22}-2a_{23}a_{24}+4a_{22}a_{24}$ is equal to:

Let $S_{n} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \ldots$ up to $n$ terms. If the sum of the first six terms of an $A.P.$ with first term $-p$ and common difference $p$ is $\sqrt{2026 S_{2025}}$,then the absolute difference between the $20^{\text{th}}$ and $15^{\text{th}}$ terms of the $A.P.$ is