+1839 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?
+6 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 $60$.
