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If x, 2y, 3z are in A.P. where the distinct numbers x, y, z are in G.P., then the common ratio of the G.P. is
Choice (c) is correct. Since x, 2y, 3z are in A.P. 2y = x + 3z/ 2⇒ 4y = x + 3z Again, x, y, z are in G.P. Let r be its common ratio. y = xr and z = yr = (xr)r = xr² Putting values of y and z from (2) in (1), we get 4xr = x +3xr² ⇒ 3r² - 4r + 1 = 0 ⇒ 3r² - 3r - r + 1= 0 ⇒ 3r(r - 1) - (r - 1) = 0Read more
Choice (c) is correct.
Since x, 2y, 3z are in A.P.
2y = x + 3z/ 2⇒ 4y = x + 3z
Again, x, y, z are in G.P.
Let r be its common ratio.
y = xr and z = yr = (xr)r = xr²
Putting values of y and z from (2) in (1), we get
4xr = x +3xr² ⇒ 3r² – 4r + 1 = 0 ⇒ 3r² – 3r – r + 1= 0
⇒ 3r(r – 1) – (r – 1) = 0 ⇒ (r – 1)(3r – 1) = 0 ⇒ r = 1 or r = 1/3
r = 1/3
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In a G.P. of positive terms, if any term is equal to the sum of the next two terms. Then the common ratio of the G.P. is
Choice (d) is correct. Let a and r be the first term and the common ratio of the G.P. Given, any term of the G.P. is equal to the sum of the next two terms. Tₙ = Tₙ ₊ ₁ + Tₙ ₊ ₂ ⇒ arⁿ ⁻ ¹ = arⁿ + arⁿ ⁺ ¹ ⇒ 1 = r + r² ⇒ r² + r - 1 = 0 ⇒ r = -1 ±√ 1² - 4 × 1 × (-1)/ 2 × 1 = -1 ± √5/2 ⇒ r =√5 - 1/2 ⇒ Read more
Choice (d) is correct.
Let a and r be the first term and the common ratio of the G.P.
Given, any term of the G.P. is equal to the sum of the next two terms.
Tₙ = Tₙ ₊ ₁ + Tₙ ₊ ₂ ⇒ arⁿ ⁻ ¹ = arⁿ + arⁿ ⁺ ¹ ⇒ 1 = r + r²
⇒ r² + r – 1 = 0
⇒ r = -1 ±√ 1² – 4 × 1 × (-1)/ 2 × 1 = -1 ± √5/2
⇒ r =√5 – 1/2
⇒ r = 2( √5 – 1/2) = 2 sin 18°
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The fourth term from the end of the G.P. 8, 4, 2, ….., 1/128 is
Choice (a) is correct. The series is a G.P. with a = 8, r = 1/2, l = 1/128 The nth term lₙ from end is given by lₙ = 1/rⁿ ⁻ ¹ The fourth term l₄ from end is given by l₄ = 1/28/(1/2)³ = 8/128 = 1/16 For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-8
Choice (a) is correct.
The series is a G.P. with a = 8, r = 1/2, l = 1/128
The nth term lₙ from end is given by lₙ = 1/rⁿ ⁻ ¹
The fourth term l₄ from end is given by l₄ = 1/28/(1/2)³ = 8/128 = 1/16
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Which term of the series 1 – 1/3 + 1/9 – 1/22 + …. is -(1/243)?
Choice (d) is correct. We have a = 1, r = -(1/3) Let nth term tₙ = -(1/243) ⇒ a.rⁿ ⁻ ¹ = -(1/243) ⇒ 1.(-1/3)ⁿ ⁻ ¹ = (-1/3)⁵ ⇒ n - 1 = 5 ⇒ n = 6 For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chapter-8
Choice (d) is correct.
We have a = 1, r = -(1/3)
Let nth term tₙ = -(1/243)
⇒ a.rⁿ ⁻ ¹ = -(1/243) ⇒ 1.(-1/3)ⁿ ⁻ ¹ = (-1/3)⁵
⇒ n – 1 = 5 ⇒ n = 6
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The sixth term from the beginning of the G.P. 8, 4, 2, … is
Choice (b) is correct. The given sequence is a G.P. with a = 8, r = 1/2 The nth term aₙ from the beginning is given by aₙ = arⁿ ⁻ ¹ The sixth term a₆ from beginning is given by a₆ = 8 × (1/2)⁵ = 8/32 = 1/4 For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/maths/#chaRead more
Choice (b) is correct.
The given sequence is a G.P. with a = 8, r = 1/2
The nth term aₙ from the beginning is given by aₙ = arⁿ ⁻ ¹
The sixth term a₆ from beginning is given by a₆ = 8 × (1/2)⁵ = 8/32 = 1/4
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