Class 11th

z₁ + z₂ = 5

$z_1^3+z_2^3=20+15i$            

  $z_1^3+z_2^3={\left(z_1+z_2\right)}^3-3z_1{z}_2\left(z_1+z_2\right)$           

$z_1^3+z_2^3=125-3z_1\cdot z_2\left(5\right)$            

 ⇒ 20 + 15i = 125 – 15z₁z₂

⇒ 3z₁z₂ = 25 – 4 – 3i

3z₁z₂ = 21– 3i

z₁⋅z₂ = 7 – i

(z₁ + z₂)² = 25

$z_1^2+z_2^2=25-27\left(7-i\right)$     

= 11 + 2i

  ${\left(z_{\text{?}1}^2+z_2^2\right)}^2$         = 121 − 4 + 44i

⇒   $z_1^4+z_2^4+2{\left(7-i\right)}^2=117+44i$

⇒   $z_1^4+z_2^4$ = 117 + 44i − 2(49 −1−14i )

= 21 + 72i

⇒   $\left|Z_1^4+Z_2^4\right|=75$

a + d, a + 7d and a + 43d are 1^st, 2^nd, 3^rd term of G.P.

$\frac{a+7d}{a+d}=\frac{a+43d}{a+7d}$              

⇒ (a + 7d)² = (a + d) (a + 43d)

⇒ a² + 49d² + 14d = a² + 44ad + 43d³

⇒ 6d² = 30ad

⇒ d² = 5d

⇒ d = 0, 5

a = 1, d = 5

  $S_{20}=\frac{20}{2}\left[2+\left(19\right)5\right]$          

= 10 [95 + 2]

= 970

  $\frac{a+b+68+44+40+60}{6}=55$

212 + a + b = 330

⇒ a + b = 118

$\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\overline{x}\right)}^2=194$          

$\frac{a^2+b^2+{\left(68\right)}^2+{\left(44\right)}^2+{\left(40\right)}^2+{\left(60\right)}^2}{6}={\left(55\right)}^2=194$

= 3219

11760 + a² + b² = 19314

⇒ a² + b² = 19314 – 11760

= 7554

(a + b)² –2ab = 7554

From here b = 41.795

a + b = 118

⇒ a + b + 2b = 118 + 83.59

= 201.59

a = sin⁻¹ (sin5) = 5 − 2π

and b = cos⁻¹ (cos5) = 2π − 5

∴    a² + b² = (5 − 2π)² + (2π − 5)²

= 8π² − 40π + 50

$w=-nRTln\frac{V_2}{V_1}$

$\left|w\right|=2.303nRTlog\frac{V_2}{V_1}$

$\left|w\right|=1*2.303*8.314*300log\frac{100}{10}$        

|w| = 5744 J

|w| = 5.744 kJ » 6 kJ

K₁₉ = 1s²2s²2p⁶3s²3p⁶4s¹

For 4s electron

n = 4

l = 0

m = 0

$s=\frac{1}{2}$

Polarisation power µ $\frac{Charge}{Size}$ for K⁺, polarising power is least and ionic character is maximum.