Question

A large quantity of waste material contains metal. At the successive passes through a recovery
process the mass of the metal recovered is:
32kg at the first pass,
24kg at the second pass;
18kg at the third pass;
13,5kg at the fourth pass, and so on, to form a geometric sequence,
1. How much metal would be recovered at the eighth pass?
(Round off the answer to TWO decimal places)
​

Answer 1

Answer:

The amount of metal recovered at the eight pass is approximately 4.27 kg

Step-by-step explanation:

The mass of metal recovered formed the following geometric sequence;

32 kg, 24 kg, 18 kg, 13.5 kg

∴ The first term, a = 32 kg

The common ratio in the geometric sequence is found as follows;

r = a·r/a = 24 kg/(32 kg) = 0.75

r = a·r²/a·r = 18 kg/(24 kg) = 0.75

r = a·r³/a·r² = 13.5 kg/(18 kg) = 0.75

The mass of the metal recovered in the eight pass will be the 8th term of the geometric series given as follows;

nth term = a·rⁿ⁻¹

a = 32, r = 0.75, n = 8

∴ 8th term = 32·(0.75)⁸⁻¹ = 32·0.75⁷ = 2187/512 ≈ 4.27

The amount of metal recovered at the eight pass ≈ 4.27 kg.

Answer 2

The eight term of the geometric sequence is 4.27.

Geometric sequence

A geometric sequence is in the form:

aₙ = ar⁽ⁿ⁻¹⁾

where aₙ is the nth term, a is the first term and r is the common difference.

Given that a = 32, r = 24/32 = 0.75

The eight term is a₈ = 32(0.75)⁸⁻¹ = 4.27

The eight term of the geometric sequence is 4.27.

Find out more on Geometric sequence at: https://brainly.com/question/9300199