Class 11th

  ${\displaystyle \sum _{i=1}^n{a}_i}=192\Rightarrow a_1+a_2+a_3+......+a_n=192$

$\therefore n\left(a_1+a_n\right)=384..............(i)$

and ${\displaystyle \sum _{i=1}^{n/2}{a}_{2i}}=120\Rightarrow a_2+a_4+a_6+......+a_n=120$  

 n (a₂ + a_n) = 480 ……………… (ii)

n (a₂ – a₁) = 96

$\therefore n=96$

Number of moles of gas = $\frac{22.4}{22400}=\frac{1}{1000}moles$

Mass of nitrogen = $\left(\frac{1}{1000}*28\right)gm$

% of N in organic compound = $\frac{\left(\frac{28}{1000}\right)}{0.2}*100\mathrm{\%}=14\mathrm{\%}$

${\left(ta{n}^{-1}x\right)}^3+{\left(co{t}^{-1}x\right)}^3=k{\pi }^3,x\in R$

$\Rightarrow \left(ta{n}^{-1}x+co{t}^{-1}x\right)\left({\left(ta{n}^{-1}x+co{t}^{-1}x\right)}^2-3ta{n}^{-1}xco{t}^{-1}x\right)=k{\pi }^3$

$\Rightarrow \frac{1}{3}\le k<\frac{7}{8}$

$\begin{array}{l}PC{l}_5\left(g\right)?PC{l}_3\left(g\right)+C{l}_2\left(g\right)\\ t=05moles00\\ \underset{\_}{-xmole}\underset{\_}{+xmole}+\underset{\_}{xmole}\end{array}$

Eq.(5 – x) mole x molex mole

Total number of moles at equilibrium

$=\left(5-x\right)+x+x+n_{N_2}=\left(7+x\right)$

$K_p=\frac{\left(P_{PC{l}_3}\right)*{\left(PC{l}_2\right)}_{Cl}}{{\left(P_{PC{l}_5}\right)}_{eq}}=\frac{\left(\frac{3}{10}*2.46\right)*\left(\frac{3}{10}*2.46\right)}{\left(\frac{2}{10}*2.46\right)}=1.107atm$

= 1107 * 103 atm

$\alpha +\beta =-\frac{\lambda }{3},\alpha \beta =-\frac{1}{3}$

$\therefore \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=15\Rightarrow \lambda =\pm 3$

$\therefore 6{\left({\alpha }^3+{\beta }^3\right)}^2=6{\left({\left(\alpha +\beta \right)}^3-3\alpha \beta \left(\alpha +\beta \right)\right)}^2=6*{\left(-\left(1+1\right)\right)}^2=24$

${PV)}_{H_2}={\left(PV\right)}_{Gas}$

${\left(nRT\right)}_{H_2}={\left(nRT\right)}_{Gas}$

$n_{H_2}*T_{H_2}=n_{Gas}*T_{Gas}$

$\Rightarrow \frac{0.2}{2}*200=\frac{3}{{\left(MM\right)}_{Gas}}*300\Rightarrow {\left(MM\right)}_{Gas}=45gm/mole$

Equation of tangent to the circle at (2, 4) is

(4 + A) x + (8 + B) y + 2A + 2B + 2C = 0……. (i)

Also equation of tangent to parabola y = x² at (2, 4) is

4x – y = 4 ………. (ii)

Comparing (i) and (ii) we get, A + C = 16

(3 * 3 * 3 * ………2022 times) ¸ 5

Remainder = (-1) (-1) (-1) ….1011 times)

Remainder = -1 + 5 = 4

$A=\left\{z\in c:1\le \left|z-\left(1+i\right)\right|\le 2\right\}$

$\left|z-\left(1+i\right)\right|\ge 1$

and $$ $\left|z-\left(1+i\right)\right|\le 2$

and also $\left|z-\left(1-i\right)\right|=1$  $$

hence B has infinite set.