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Given a2 = 6 and a5 = –384 of a geometric sequence, what is the recursive equation for the nth term?

an = –4(an – 1), a1 = –1.5
an = –4(an – 1), a1 = 1.5
an = 4(an – 1), a1 = –1.5
an = 4(an – 1), a1 = 1.5

User Francena
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2 Answers

5 votes

Please see the image that I've included. The answer is option A.

Given a2 = 6 and a5 = –384 of a geometric sequence, what is the recursive equation-example-1
User Dan Nissenbaum
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2 votes

Answer:


a_n = (-4)^(n - 1); a_1 = -1.5

Explanation:


a_1 = a_1


a_2 = a_1 * r


a_3 = a_1 * r^2


a_4 = a_1 * r^3


a_5 = a_1 * r^4


a_1 * r = 6


a_1 * r^4 = -384


r = (6)/(a_1)


a_1 * ((6)/(a_1))^4 = -384


r = (6)/(a_1)


(1296)/(a_1)^3} = -384


r = (6)/(a_1)


(1)/(a_1)^3} = -(384)/(1296)


r = (6)/(a_1)


a_1^3 = -(27)/(8)


r = (6)/(a_1)


a_1 = -1.5


r = (6)/(-1.5)


a_1 = -1.5


r = -4


a_1 = -1.5


a_n = (-4)^(n - 1); a_1 = -1.5

User Jimmy Zelinskie
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