When the same constant is added to the numbers 60, 100, and 180, a three-term geometric sequence arises. What is the common ratio of the resulting sequence?

\(\phantom{60, 100, and 160}\)

Guest Jun 26, 2022

#2**0 **

Let x be the constant that is added to all 3 numbers, and let d be the common ratio

We have: \((60+x)d = (100 + x)\)

Solving for d gives us: \(d = {100 + x \over 60 + x }\)

We also know from the second and final term that \((100 + x) d = (180 + x)\)

Substituting the known value of d, we have: \((100 + x) \times {100 + x \over 60 + x} = (180 + x)\)

Cross multiplying gives: \((x + 100)^2 = (180 + x)(60 + x)\)

Simplify both sides to: \(x^2+200x+10000=x^2+240x+10800 \)

Solving for x gives us: \(x = -20\)

This means that the geometric series is \(40, 80, 160\).

Can you find the common ratio from here?

BuilderBoi Jun 26, 2022

#2**0 **

Best Answer

Let x be the constant that is added to all 3 numbers, and let d be the common ratio

We have: \((60+x)d = (100 + x)\)

Solving for d gives us: \(d = {100 + x \over 60 + x }\)

We also know from the second and final term that \((100 + x) d = (180 + x)\)

Substituting the known value of d, we have: \((100 + x) \times {100 + x \over 60 + x} = (180 + x)\)

Cross multiplying gives: \((x + 100)^2 = (180 + x)(60 + x)\)

Simplify both sides to: \(x^2+200x+10000=x^2+240x+10800 \)

Solving for x gives us: \(x = -20\)

This means that the geometric series is \(40, 80, 160\).

Can you find the common ratio from here?

BuilderBoi Jun 26, 2022