Class 11th

First element in 20^th group is equal to '20'.

(20, 21, ….upto 39 times)

$\therefore Sum=\frac{39}{2}\left(2*20+\left(39-1\right)*1\right)$        

(k-3)/ (h-2) * (k-0)/ (h-0) = -1
⇒ k (2k – 3) = -2 (h – 2)h
⇒ 2h² + 2k² – 4h – 3k = 0
2x² + 2y² – 4x – 3y = 0
(0,0)

$\stackrel{\stackrel{+4}{{\mathrm{N}\mathrm{O}}_2}}{2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{N}\mathrm{O}}_2}+\stackrel{+5}{\mathrm{H}\mathrm{N}\mathrm{O}}{}_3$

$\stackrel{0}{{\mathrm{C}\mathrm{l}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{-1}{\mathrm{H}\mathrm{C}\mathrm{l}}+\mathrm{H}\mathrm{O}\mathrm{C}\mathrm{l}$

$2\stackrel{+4}{{\mathrm{C}\mathrm{l}\mathrm{O}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_2}+\stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}$

$\stackrel{+6}{{\mathrm{C}\mathrm{l}}_2}{\mathrm{O}}_6+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}+\stackrel{+{+}^{+}}{{\mathrm{C}\mathrm{l}\mathrm{O}}_4}$

(5³)^740 * 5² = 5² (125)^740 = 5² (126 – 1)^740
So remainder = 4.

C.V. = (σ/x? ) * 100 ⇒ σ = (C.V. * x? ) / 100
∴ σ? = (50*30)/100 = 15 and σ? = (60*25)/100 = 15 ⇒ σ? - σ? = 0

Equation of tangent is (x/a) (1/2) + (y/b) (√3/2) = 1
Equation of auxiliary circle is x² + y² = a²
Homogenising (ii) with (i) and making coefficient of x² + coefficient of y² = 0
⇒ (3a²/4b²) - (7/4) = 0 ⇒ e = 2/√7

$N_1{V}_1=N_2{V}_2$

$0.01*V_1=0.01*20*3$

$V_1=60\mathrm{m}\mathrm{L}$

log? (abc) = 6 ⇒ abc = 6?
a = b/r & c = br
⇒ b = 36 and a = 36/r
⇒ r = 2,3,4,6,9,12,18
Also b - a = 36 (1 - 1/r) is a perfect cube.
∴ r = 4 ⇒ a + b + c = 9 + 36 + 144 = 189

β² – 6β + 12 = 0 β − 6 = -12/β
∴ Reqd. Expression is (α – 2)³)? + (-12)¹² / (αβ)¹²) - 1
= (α³ – 8 – 6α (α – 2)? + (12)¹² / (12)¹²) - 1 = (α (α² – 6α + 12) – 8)? = 8? = 2¹²

Solving, x² – 9 = kx² ⇒ x² (k − 1) + 9 = 0 ⇒ x? + x? = 0 and x? = 9 / (k-1)
|x? - x? | = 10 = √ (x? + x? )² - 4x? x? ) ⇒ k = 16/25