If $a_1 = a_2 = 2$ and $a_n = a_{n-1} - 1$ for $n > 2$,then $a_5$ is

If $t_n$ is the $n^{th}$ term of an arithmetic progression and $t_7 = 9$,what is the value of the common difference $d$ that minimizes the product $t_1 t_2 t_7$?

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

If $A_1, A_2$ are two arithmetic means,$G_1, G_2$ are two geometric means,and $H_1, H_2$ are two harmonic means between two numbers,then $\frac{A_1 + A_2}{H_1 + H_2} \cdot \frac{H_1 H_2}{G_1 G_2} = \dots$

If for $x, y \in \mathbb{R}, x > 0,$ $y = \log_{10} x + \log_{10} x^{1/3} + \log_{10} x^{1/9} + \dots$ up to $\infty$ terms and $\frac{2+4+6+\dots+2y}{3+6+9+\dots+3y} = \frac{4}{\log_{10} x}$,then the ordered pair $(x, y)$ is equal to:

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: