Answer:
34 cm
Step-by-step explanation:
The nth term of a geometric progression Tₙ = arⁿ⁻¹ where a = first term, and r = common ratio.
Given that our first term, T₁ = ar¹⁻¹ = a = 121 cm and our third term T₃ = ar³⁻¹ = ar² = 64 cm
Then, T₃/T₁ = ar²/a = r²
r² = T₃/T₁
r² = 64 cm/121 cm
r = √64/121
r = 8/11
So, the height of our fifth bounce T₅ = ar⁵⁻¹
= ar⁴
= 121 cm × (8/11)⁴
= 121 × 4096/14641
= 4096/121 cm
= 33.9 cm
≅ 34 cm to the nearest cm
Using a geometric sequence, it is found that the height of the fifth bounce is of 33.85 cm.
In a geometric sequence, the quotient between consecutive terms is always the same, and it is called common ratio r.
The nth term of a geometric sequence is given by:
[tex]a_n = a_1r^{n-1}[/tex]
In which [tex]a_1[/tex] is the first term.
In this problem:
[tex]a_n = a_1r^{n-1}[/tex]
[tex]64 = 121r^{2}[/tex]
[tex]r^2 = \sqrt{\frac{64}{121}}[/tex]
[tex]r = \frac{8}{11}[/tex]
Hence:
[tex]a_n = 121\left(\frac{8}{11}\right)^{n-1}[/tex]
The height of the fifth bounce is of [tex]a_5[/tex], hence:
[tex]a_5 = 121\left(\frac{8}{11}\right)^{4} = 33.85[/tex]
The height of the fifth bounce is of 33.85 cm.
You can learn more about geometric sequences at https://brainly.com/question/11847927
