If $\frac{1}{2 \cdot 3^{10}}+\frac{1}{2^{2} \cdot 3^{9}}+\ldots+\frac{1}{2^{10} \cdot 3}=\frac{K}{2^{10} \cdot 3^{10}}$,then the remainder when $K$ is divided by $6$ is

Find the sum to the indicated number of terms in the geometric progression: $x^{3}, x^{5}, x^{7}, \ldots$ for $n$ terms (where $x \neq \pm 1$).

The sum of the first five terms of the series $3 + 4\frac{1}{2} + 6\frac{3}{4} + \dots$ will be

$A$ particle starts at the origin and moves $1$ unit horizontally to the right and reaches $P_{1}$, then it moves $\frac{1}{2}$ unit vertically up and reaches $P_{2}$, then it moves $\frac{1}{4}$ unit horizontally to the right and reaches $P_{3}$, then it moves $\frac{1}{8}$ unit vertically down and reaches $P_{4}$, then it moves $\frac{1}{16}$ unit horizontally to the right and reaches $P_{5}$ and so on. Let $P_{n} = (x_{n}, y_{n})$ and $\lim_{n \rightarrow \infty} x_{n} = \alpha$ and $\lim_{n \rightarrow \infty} y_{n} = \beta$. Then, $(\alpha, \beta)$ is

If $25, x - 6$,and $x - 12$ are consecutive terms of a geometric progression,then $x = \dots$

The product of three consecutive terms of a geometric progression is $216$ and the sum of the products of these terms taken two at a time is $156$. Find the terms.