Write the seventh term of the following $GP$: $39366, 13122, 4374, 1458, \ldots$

The sum of a few terms of a geometric progression is $728$. If the common ratio is $3$ and the last term is $486$,then the first term of the series will be:

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$. If $S_4 = 16$ and $S_6 = -48$,then $S_{10}$ is equal to

If $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ such that $a < b < c$ and $a+b+c = \frac{3}{4}$,then the value of $a$ is

$1 + \frac{1^3 + 2^3}{1 + 2} + \frac{1^3 + 2^3 + 3^3}{1 + 2 + 3} + \dots + \frac{1^3 + 2^3 + 3^3 + \dots + 15^3}{1 + 2 + 3 + \dots + 15} - \frac{1}{2}(1 + 2 + 3 + \dots + 15)$ is equal to

The $A.M., G.M.$ and $H.M.$ between two numbers are $\frac{144}{15}$,$15$ and $12$,but not necessarily in this order. Then $H.M., G.M.$ and $A.M.$ respectively are