Sequences and Series

Kindly consider the following figure

S = tan? ¹ (1/3) + tan? ¹ (1/7) + tan? ¹ (1/13) + . upto 10 term
S = tan? ¹ (2-1)/ (1+1⋅2) + tan? ¹ (3-2)/ (1+2⋅3) + tan? ¹ (4-3)/ (1+3⋅4) + . + tan? ¹ (11-10)/ (1+11⋅10)
S = (tan? ¹2 - tan? ¹1) + (tan? ¹3 - tan? ¹2) + . + (tan? ¹11 - tan? ¹10)
S = tan? ¹11 - tan? ¹1
S = tan? ¹ (11) - π/4
tan (S) = 5/6

a, b, c are in Andhra Pradesh then
2b = a + c
28 = 3^ (2sin2θ-1) + 3^ (4-2sin2θ)
Put 3^ (2sin2θ) = x
28 = x/3 + 81/x ⇒ x² - 84x + 243 = 0
(x-3) (x-81) = 0
3^ (2sin2θ) = 3 or 3?
2sin2θ = 1 or 4
sin2θ = 1/2
terms are 1, 14, 27,
then T? = 1 + 5 (13)

S' = 2¹? + 2? ⋅ 3 + 2? ⋅ 3² + . + 2 ⋅ 3? + 3¹?
G.P. → a = 2¹? , r = 3/2, n = 11
S' = 2¹? ⋅ [ (3/2)¹¹ - 1)/ (3/2 - 1)] = 2¹¹ (3¹¹/2¹¹) - 1)
= 3¹¹ - 2¹¹

Let the first A.P. be a? , a? + d, a? + 2d.
a? = a? + 39d = -159
a? = a? + 99d = -399
Subtracting the equations, 60d = -240 ⇒ d = -4.
Substituting d back, a? + 39 (-4) = -159 ⇒ a? - 156 = -159 ⇒ a? = -3.
Now, for the second A.P. with first term b? and common difference D = d+2 = -2.
b? = a?
⇒ b? + 99D = a? + 69d
⇒ b? + 99 (-2) = -3 + 69 (-4)
⇒ b? - 198 = -3 - 276
⇒ b? = -279 + 198 = -81

Given,
300 = 1 + (N – 1)d
⇒ (N − 1)d = 299
∴ (N, d) = (24,13) is the only possible pair
∴ a? = 1 + 19 (13) = 248 and, S? = (1+248)/2 * 20
= 2490

1 + (1 - 2²⋅1) + (1 - 4²⋅3) + . + (1 - 20²⋅19)
= α - 220β
= 11 - (2²⋅1 + 4²⋅3 + . + 20²⋅19)
= 11 - 2² ⋅ Σ? (r=1) r² (2r-1) = 11 - 4 (110²/2) - 35 x 11)
= 11 - 220 (103)
⇒ α = 11, β = 103

3, A? , A? , A? , .A? , 243
d = (243-3)/ (m+1) = 240/ (m+1)
3, G? , G? , G? , 243
r = (243/3)¹/ (³? ¹) = (81)¹/? = 3
G? = A?
=> 3 (3)² = 3 + 4 (240/ (m+1)
=> 27 = 3 + 960/ (m+1)
=> m+1 = 40
=> m=39