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# Algebra

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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?

$$\phantom{60, 100, and 160}$$

Jun 26, 2022

#2
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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?

Jun 26, 2022

#1
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The common ratio is 11/3.

Jun 26, 2022
#2
+2448
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