Let the first term of a series be $T_1=6$ and its $r^{\text{th}}$ term $T_r=3T_{r-1}+6^r$ for $r=2, 3, \ldots, n$. If the sum of the first $n$ terms of this series is $\frac{1}{5}(n^2-12n+39)(4 \cdot 6^n - 5 \cdot 3^n + 1)$,then $n$ is equal to:

Let $a = \min \{x^{2} + 2x + 3 : x \in R\}$ and $b = \lim_{\theta \rightarrow 0} \frac{1 - \cos \theta}{\theta^{2}}$. Then $\sum_{r=0}^{n} a^{r} b^{n-r}$ is

Let $a_{1}, a_{2}, \ldots, a_{10}$ be an $AP$ with common difference $-3$ and $b_{1}, b_{2}, \ldots, b_{10}$ be a $GP$ with common ratio $2$. Let $c_{k}=a_{k}+b_{k}, k=1, 2, \ldots, 10$. If $c_{2}=12$ and $c_{3}=13$,then $\sum_{k=1}^{10} c_{k}$ is equal to:

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = (-1)^{n-1} 5^{n+1}$.

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

Let $2^{\text{nd}}$,$8^{\text{th}}$,and $44^{\text{th}}$ terms of a non-constant $A.P.$ be respectively the $1^{\text{st}}$,$2^{\text{nd}}$,and $3^{\text{rd}}$ terms of a $G.P.$ If the first term of the $A.P.$ is $1$,then the sum of the first $20$ terms is equal to: