Find the $n^{th}$ term of the following series: $5, 14, 29, 50, 77, \ldots$

If $x, y, z \in \mathbb{R}^+$ such that $x + y + z = 4$,then the maximum possible value of $xyz^2$ is

Let $a_1, a_2, \dots, a_{49}$ be in $A.P.$ such that $\sum_{k=0}^{12} a_{4k+1} = 416$ and $a_9 + a_{43} = 66$. If $a_1^2 + a_2^2 + \dots + a_{17}^2 = 140m$,then $m = \dots$

If $\log _{5} 2, \log _{5}(2^{x}-3)$ and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$,then the value of $x$ is:

The number of terms common to the two arithmetic progressions $3, 7, 11, \ldots, 407$ and $2, 9, 16, \ldots, 709$ is

The sum of three consecutive terms in a geometric progression is $14$. If $1$ is added to the first and the second terms and $1$ is subtracted from the third,the resulting new terms are in arithmetic progression. Then the lowest of the original terms is