Question

Find the nth term of the geometric sequence.

Answer 1

$a_2 = ar^{2-1} =ar = 6~~ ...(i)\\\\a_4 = ar^{4-1} = ar^3 = 2186 ~~....(ii)\\\\\\(ii)\div (i):\\\\\dfrac{ar^3}{ar} = \dfrac{2186}6\\\\\implies r^2 = \dfrac{2186}6 = \dfrac{1093}3\\\\\implies r = \pm \sqrt{\dfrac{1093}3}\\\\a= \pm \dfrac{6}{ \sqrt{\dfrac{1093}3}}\\\\\text{nth term} = ar^{n-1} = \pm \dfrac{6}{ \sqrt{\dfrac{1093}3}} \left( \pm{ \sqrt{\dfrac{1093}3} \right)^{n-1} = \dfrac{6}{ \sqrt{\dfrac{1093}3}} \left( { \sqrt{\dfrac{1093}3} \right)^{n-1}$

$= 6\left(\dfrac{1093}3 \right)^{\tfrac{n-2}2}$

Answer 2

Answer:

• aₙ = (6/19)*19ⁿ⁻¹

Step-by-step explanation:

Given:

• a₂ = 6
• a₄ = 2166

Equation of nth term of GP:

• aₙ = a₁rⁿ⁻¹

Substitute for 2nd and 4th terms:

• a₂ = a₁r = 6
• a₄ = a₁r³ = 2166

Substitute the value of a₂ into a₄ and find the value of r:

• a₁r³ = a₁r * r² = 6r²
• 6r² = 2166
• r² = 2166/6
• r² = 361
• r = √361
• r = 19

Find the first term:

• a₁r = 6
• a₁*19 = 6
• a₁ = 6/19

The equation of nth term is:

• aₙ = (6/19)*19ⁿ⁻¹