PLEASE HELP! I'M ON A TIMER A rope is swinging in such a way that the length of the arc traced by a knot at its bottom end is decreasing geometrically. If the third arc is 18 ft. long and the seventh arc is 8 ft. long, what is the length of the arc on the sixth swing? Round your answer to the nearest tenth of a foot.

The correct answer is 9.8 ft.

Explanation:
This is a geometric sequence, which follows the explicit formula 
[tex]g_n=g_1\times r^{n-1}[/tex] 

where g₁ is the first term, r is the common ratio and n is the term number.

We know that the third term is 18; this gives us 18=g₁×r³⁻¹ or 18=g₁×r².

We also know the seventh term is 8, which gives us 8=g₁×r⁷⁻¹ or 8=g₁×r⁶.

Solving for g₁ in the third term gives us g₁=18/r², and solving for g₁ in the seventh term gives us g₁=8/r⁶. They both equal g₁ so we set them equal to each other:

18/r² = 8/r⁶.
Multiply both sides by r⁶, which gives us
18r⁶/r² = 8.

Using our properties of exponents, we have 18r⁴ = 8. Divide both sides by 18, which gives us
r⁴=8/18.

We can find the fourth root by taking the square root twice:
taking the square root gives us r² = √8/√18.

Simplifying √8 we get 2√2, and simplifying √18 gives us 3√2; we now have

r²=2√2)/3√2.

The √2 will cancel, leaving r²=2/3. Taking the square root again, we have

r=√2/√3; simplifying this gives us r=√6/3.

We can now work backward to find the sixth term using the seventh one; Divide 8 by √6/3. Dividing by a fraction means multiplying by the reciprocal, so we multiply 8 by 3/√6; this gives us 24/√6, and in a calculator that gives us 9.8 ft.