The product of three consecutive terms of a $G.P.$ is $512$. If $4$ is added to each of the first and the second of these terms,the three terms now form an $A.P.$ Then the sum of the original three terms of the given $G.P.$ 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 $t_1 t_2 t_7$?

Write the first five terms of the following sequence and obtain the corresponding series:
$a_{1}=3, a_{n}=3a_{n-1}+2$ for all $n > 1$.

Three numbers are in an increasing geometric progression with common ratio $r$. If the middle number is doubled,then the new numbers are in an arithmetic progression with common difference $d$. If the fourth term of the $G.P.$ is $3r^{2}$,then $r^{2}-d$ is equal to:

If $[x]$ represents the greatest integer not greater than $x$,then $\left[\left(1+\frac{1}{100000}\right)^{100000}\right]=$

If $a, b, c$ are in $GP$ and $4a, 5b, 4c$ are in $AP$ such that $a + b + c = 70$,then the value of $a^3 + b^3 + c^3$ is