The ${4^{th}}$ term of a $G.P.$ is square of its second term, and the first term is $-3$ Determine its $7^{\text {th }}$ term.
Let $a$ be the first term and $r$ be the common ratio of the $G.P. $
$\therefore a=-3$
It is known that, $a_{n}=a r^{n-1}$
$\therefore a_{4}=a r^{3}=(-3) r^{3}$
$a_{2}=a r^{2}=(-3) r$
According to the given condition,
$(-3) r^{3}=[(-3) r]^{2}$
$\Rightarrow-3 r^{3}=9 r^{2} \Rightarrow r=-3 a_{7}=a r^{7-1}=a r^{6}=(-3)(-3)^{6}=-(3)^{7}=-2187$
Thus, the seventh term of the $G.P.$ is $-2187 .$
Evaluate $\sum\limits_{k = 1}^{11} {\left( {2 + {3^k}} \right)} $
The sum of the series $5.05 + 1.212 + 0.29088 + ...\,\infty $ is
The roots of the equation
$x^5 - 40x^4 + px^3 + qx^2 + rx + s = 0$ are in $G.P.$ The sum of their reciprocals is $10$. Then the value of $\left| s \right|$ is
Find the sum of the following series up to n terms:
$5+55+555+\ldots$
Three numbers are in $G.P.$ such that their sum is $38$ and their product is $1728$. The greatest number among them is