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The last 3 terms of the sequence 4;x;y;18 form a geometric progression whilst the first 3 terms form arithmetic progression.

1. Calculate the values of x and y if both are natural numbers.
2. Hence calculate T15 of the geometric.

Respuesta :

The value of x and y are 8 and 12 respectively.

The value of T₁₅ of the geometric progression is 1167.72

4, x, y, 18

4, x, y

The first three terms forms an arithmetic progression. Therefore,

AP formula

  • aₙ = a + (n - 1)d

where

a = first term

d = common difference

n= number of term

Therefore,

3rd term

y = 4 +(3 - 1)(x-4)

y = 4 + 2(x - 4)

y = 4 + 2x - 8

y - 2x = -4

y = 2x - 4

x, y, 18

GP formula

  • aₙ = arⁿ⁻¹

where

a = first term

r = common ratio

Therefore,

a = x

r = 18 / y

18 = x(18 / y)²

18 = 324x / y²

18y² = 324x

x = 18y² / 324

x = y² / 18

Combine the equation

y = 2(y² / 18) - 4

y = y² / 9 - 4

y - y² / 9 = -4

9y - y²/9 = -4

-y² + 9y = -36

y² - 9y - 36 = 0

y² + 3y - 12y - 36 = 0

(y - 12)(y + 3)

y = 12 or  - 3

x = 12² / 18 = 144 / 18 = 8

a₁₅ = ar¹⁵⁻¹

a₁₅ = 4 × (12 / 8)¹⁴

a₁₅ = 4 × (3 / 2)¹⁴

a₁₅ = 4 × 291.929260254

a₁₅ = 1167.71704102

a₁₅ = 1167.72

learn more on arithmetic and geometric progression here:https://brainly.com/question/21093837