Morgan asked in Science & MathematicsMathematics · 11 months ago

The first term in a sequence is 3 and the seventh term is 139968, determine r and the nth term

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• Ray
Lv 7
11 months ago

a = 3

7th term = ar^6

7th term = 3r^6

3r^6 = 139968

r^6 = 46656

Using a calculator r = 6

nth term = ar^(n - 1)

nth term = 3.6^(n - 1)

• Mike G
Lv 7
11 months ago

139968 = 3r^6

r = 46656^(1/6) = 6

nth term = 3*6^(n-1)

• 11 months ago

a₁ = a = 3, a₇ = ar⁶ = 139968, so r⁶ = 46656

r⁶ - 6⁶ = 0 factors to (r - 6cis(0))(r - 6cis(π/3))(r - 6cis(2π/3))(r - 6cis(π))(r - 6cis(4π/3))(r - 6cis(5π/3)) = 0 in complex numbers.

r₁ = 6cis(0) = 6 yields aᵤ = 6ᵘ/2 {3, 18, 108, ...}

r₄ = 6cis(π) = -6 yields aᵤ = (-1)ᵘ⁻¹6ᵘ/2 {3, -18, 108, ...}

r₂ = 6cis(π/3) = 3+3√3i yields aᵤ = 3(3+3√3i)ᵘ⁻¹ = 6ᵘcis((u-1)π/3)/2

r₃ = 6cis(2π/3) = -3+3√3i yields aᵤ = 3(-3+3√3i)ᵘ⁻¹ = 6ᵘcis(2(u-1)π/3)/2

r₅ = 6cis(4π/3) = -3-3√3i yields aᵤ = 3(-3-3√3i)ᵘ⁻¹ = 6ᵘcis(4(u-1)π/3)/2

r₆ = 6cis(5π/3) = 3-3√3i yields aᵤ = 3(3-3√3i)ᵘ⁻¹ = 6ᵘcis(5(u-1)π/3)/2

• 11 months ago

a * r = 3

a * r^7 = 139968

a * r^7 / (a * r) = 139968 / 3

r^6 = 46656

r^6 = 6^6 * 1^6

r^6 = 6^6 * (cos(2pi * k) + i * sin(2pi * k))^6

r = 6 * (cos(pi * k / 3) + i * sin(pi * k / 3))

r = 6 * (cos(0) + i * sin(0)) , 6 * (cos(pi/3) + i * sin(pi/3)) , 6 * (cos(2pi/3) + i * sin(2pi/3)) , 6 * (cos(pi) + i * sin(pi)) , 6 * (cos(4pi/3) + i * sin(4pi/3)) , 6 * (cos(5pi/3) + i * sin(5pi/3))

r = 6 , -6 , 6 * (1/2 + i * sqrt(3)/2) , -6 * (1/2 + i * sqrt(3)/2) , 6 * (-1/2 + i * sqrt(3)/2) , -6 * (-1/2 + i * sqrt(3)/2)

r = 6 , -6 , 3 + 3 * sqrt(3) * i , -3 - 3 * sqrt(3) * i , -3 + 3 * sqrt(3) * i , 3 - 3 * sqrt(3) * i

a * r = 3

a * r^n = 3 * r^(n - 1)

3 * 6^(n - 1)

3 * (-6)^(n - 1)

3 * (3 + 3 * sqrt(3) * i)^(n - 1)

3 * (-3 - 3 * sqrt(3) * i)^(n - 1)

3 * (3 - 3 * sqrt(3) * i)^(n - 1)

3 * (-3 + 3 * sqrt(3) * i)^(n - 1)

Take your pick. With the given information, they're all correct responses.