$0.\mathop {423}\limits^{\,\,\,\,\, \bullet \, \bullet \,} = $
$\frac{{419}}{{990}}$
$\frac{{419}}{{999}}$
$\frac{{417}}{{990}}$
$\frac{{417}}{{999}}$
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 $G.P.$
The sum of an infinite geometric series is $3$. A series, which is formed by squares of its terms, have the sum also $3$. First series will be
Find the $10^{\text {th }}$ and $n^{\text {th }}$ terms of the $G.P.$ $5,25,125, \ldots$
Find the sum of the products of the corresponding terms of the sequences $2,4,8,16,32$ and $128,32,8,2, \frac{1}{2}$
In a increasing geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is $25 .$ Then, the sum of $4^{\text {th }}, 6^{\text {th }}$ and $8^{\text {th }}$ terms is equal to