The remainder when the polynomial $1+x^2+x^4+x^6+\ldots+x^{22}$ is divided by $1+x+x^2+x^3+\ldots+x^{11}$ is
$0$
$2$
$1+x^2+x^4+\ldots+x^{10}$
$2\left(1+x^2+x^4+\ldots+x^{10}\right)$
Let $a_1, a_2, a_3, \ldots .$. be a sequence of positive integers in arithmetic progression with common difference $2$. Also, let $b_1, b_2, b_3, \ldots .$. be a sequence of positive integers in geometric progression with common ratio $2$ . If $a_1=b_1=c$, then the number of all possible values of $c$, for which the equality
$2\left(a_1+a_2+\ldots .+a_n\right)=b_1+b_2+\ldots . .+b_n$
holds for some positive integer $n$, is. . . . . . .
Find the sum to indicated number of terms in each of the geometric progressions in $\left.1,-a, a^{2},-a^{3}, \ldots n \text { terms (if } a \neq-1\right)$
The number of natural number $n$ in the interval $[1005, 2010]$ for which the polynomial. $1+x+x^2+x^3+\ldots+x^{n-1}$ divides the polynomial $1+x^2+x^4+x^6+\ldots+x^{2010}$ is
The sum of first three terms of a $G.P.$ is $16$ and the sum of the next three terms is
$128.$ Determine the first term, the common ratio and the sum to $n$ terms of the $G.P.$
If five $G.M.’s$ are inserted between $486$ and $2/3$ then fourth $G.M.$ will be