If $2^{10}+2^{9} \cdot 3^{1}+28 \cdot 3^{2}+\ldots+2 \cdot 3^{9}+3^{10}=S -211$ then $S$ is equal to
$\frac{3^{11}}{2}+2^{10}$
$3^{11}-2^{12}$
$3^{11}$
$2 \cdot 3^{11}$
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 $G.P.$
Find four numbers forming a geometric progression in which the third term is greater than the first term by $9,$ and the second term is greater than the $4^{\text {th }}$ by $18 .$
If $(y - x),\,\,2(y - a)$ and $(y - z)$ are in $H.P.$, then $x - a,$ $y - a,$ $z - a$ are in
Find a $G.P.$ for which sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.
The number $111..............1$ ($91$ times) is a