9.) If a-1, a+3, and a-12 are consecutive terms in a geometric sequence, find:
a) the value of a
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b) the value of r
(You get +10 points if your solution and answer is honest)
according to the definition,
(a+3)^2 = (a-1)(a-12)
a^2 + 6a + 9 = a^2 - 13a + 12
19a = 3
a = 3/19
your terms are 3/19-1 , 3/19+3 , and 3/19-12
or -16/9 , 60/19, -225/19
r = 60/19 ÷ -16/9 = -15/4
checking: -225/19 ÷ 60/19 = -15/4