If the mean deviation about the median of the numbers $\mathrm{k}, 2 \mathrm{k}, 3 \mathrm{k}, \ldots ., 1000 \mathrm{k}$ is 500 , then $\mathrm{k}^2$ is equal to :

Let $\mathrm{S}=\left\{z \in \mathbb{C}: 4 z^2+\bar{z}=0\right\}$. Then $\sum\limits_{z \in \mathrm{~S}}|z|^2$ is equal to:

Let $\mathrm{C}_{\mathrm{r}}$ denote the coefficient of $x^{\mathrm{r}}$ in the binomial expansion of $(1+x)^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}, 0 \leq \mathrm{r} \leq \mathrm{n}$. If

$P_n=C_0-C_1+\frac{2^2}{3} C_2-\frac{2^3}{4} C_3+\ldots . .+\frac{(-2)^n}{n+1} C_n$, then the value of $\sum\limits_{n=1}^{25} \frac{1}{P_{2 n}}$ equals.

Suppose $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P. and $\mathrm{a}^2, 2 \mathrm{~b}^2, \mathrm{c}^2$ are in G.P. If $\mathrm{a}<\mathrm{b}<\mathrm{c}$ and $\mathrm{a}+\mathrm{b}+\mathrm{c}=1$, then $9\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)$ is equal to $\_\_\_\_$ .