Write the first five terms of the following sequence and obtain the corresponding series:
$a_{1} = a_{2} = 2, a_{n} = a_{n-1} - 1, n > 2$

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:

If the first three terms of the sequence $\frac{1}{16}, a, b, \frac{1}{6}$ are in a geometric progression and the last three terms are in a harmonic progression,then the values of $a$ and $b$ are:

For every natural number $n$,$n(n + 1)$ is always

Let the sum of the first $n$ terms of an $A$.$P$. be $3n^2 + 5n$. Then the sum of squares of the first $10$ terms of the $A$.$P$. is:

The sum of the first ten terms of an $A$.$P$. is $160$ and the sum of the first two terms of a $G$.$P$. is $8$. If the first term of the $A$.$P$. is equal to the common ratio of the $G$.$P$. and the first term of the $G$.$P$. is equal to the common difference of the $A$.$P$.,then the sum of all possible values of the first term of the $G$.$P$. is: