How do you find the first 3 terms of the geometric sequence with a5 = -192 a9 = 1,536?

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Answer 1 · 2016-08-04T20:15:51

It seems there is a problem with this question. SHould it not be the 8th term which is 1536?

Divide the two terms - their general form and their values:

$a_9/a_5 = (ar^8)/(ar^4) = 1536/-192$

$(cancelar^8)/(cancelar^4) = 1536/-192$

$r^4 = -8 = (-2^3)$

$r = (-2)^(3/4)" sub r to find a"$

$a_5 = ar^4 = -192$

$a((-2)^(3/4))^4 = -192$

$a(-2)^3 = -192$

$a = -192/-8 = 24$

Now we have $a and r$