Find the $20^{\text {th }}$ and $n^{\text {th }}$ terms of the $G.P.$ $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots$
The given $G.P.$ is $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots .$
Here, $a=$ First term $=\frac{5}{2}$
$r=$ Common ratio $=\frac{5 / 4}{5 / 2}=\frac{1}{2}$
$a_{20}=a r^{20-1}=\frac{5}{2}\left(\frac{1}{2}\right)^{19}=\frac{5}{(2)(2)^{19}}=\frac{5}{(2)^{20}}$
$a_{n}=a r^{n-1}=\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}=\frac{5}{(2)(2)^{n-1}}=\frac{5}{(2)^{n}}$
If $n$ geometric means between $a$ and $b$ be ${G_1},\;{G_2},\;.....$${G_n}$ and a geometric mean be $G$, then the true relation is
If $2(y - a)$ is the $H.M.$ between $y - x$ and $y - z$, then $x - a,\;y - a,\;z - a$ are in
Find the sum to $n$ terms of the sequence, $8,88,888,8888 \ldots$
Suppose that the sides $a,b, c$ of a triangle $A B C$ satisfy $b^2=a c$. Then the set of all possible values of $\frac{\sin A \cot C+\cos A}{\sin B \cot C+\cos B}$ is
If $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0),$ then show that $a, b, c$ and $d$ are in $G.P.$