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When the same constant is added to the numbers $60,$ $110,$ and $165,$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?

 Mar 12, 2024
 #1
avatar+179 
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Let x be the constant that is added to each term. Then the three terms become 60+x, 110+x, and 165+x. Note that the common ratio is simply the middle term divided by the first term. Thus, the common ratio is $ \frac{110+x}{60+x}$.

 

We are given that these terms form a geometric sequence, so the ratio between the second and third terms must equal the ratio between the first and second terms.

 

Thus, we have the equation 110+x165+x​=60+x110+x​.

 

Multiplying both sides of the equation by (60+x)(110+x)(165+x) gives us (165+x)(110+x)=(110+x)(60+x).

 

Expanding both sides gives us 18150+775x+x2=6600+170x+x2. Simplifying the right side gives us 18150+775x+x2=6600+170x+x2.

 

Subtracting x2 from both sides gives us 18150+775x=6600+170x. Subtracting 170x from both sides gives us 17980+605x=6600.

 

Subtracting 17980 from both sides gives us 605x=−11380. Dividing both sides by 605 gives us x=−18.8.

 

Substituting −18.8 for x in our expression for the common ratio, we get a common ratio of $ \frac{110-18.8}{60-18.8} = \frac{91.2}{41.2} = \boxed{\tfrac{45.6}{20.6}}$.

 Mar 12, 2024
 #2
avatar+128796 
+1

Bosco gives a better answer here :

https://web2.0calc.com/questions/geometric-progression_12

 

cool cool cool

 Mar 12, 2024

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