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When the same constant is added to the numbers $60,$ $120,$ and $140,$ a three-term geometric sequence arises. What is the common ratio of the resulting sequence?

 Apr 2, 2024
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So, we have:

$60 + a$, $120 + a$, and $140 + a$

For a three-term geometric sequence, the ratio between consecutive terms is constant.

Thus, we can write:

$\frac{120 + a}{60 + a} = \frac{140 + a}{120 + a}$

Cross multiplying:

$(120 + a)^2 = (60 + a)(140 + a)$

Expanding both sides:

$14400 + 240a + a^2 = 8400 + 200a + 140a + a^2$

$14400 + 240a + a^2 = 8400 + 340a + a^2$

$6000 = 100a$

$a = 60$

Therefore, the common ratio of the resulting sequence is $60$.

 Apr 3, 2024

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