How do you find A1 in a geometric sequence if A3=-8 and A6=1?

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Answer 1 · 2015-11-08T10:07:10

$A_1 = -32$

$A_n = A_1r^(n -1)$

$A_3 = -8 = A_1r^(3-1)$

$=> -8 = A1r^2$

$A_6 = 1 = A_1r^(6-1)$

$=> 1 = A_1r^5$

$=> 1 = A_1r^2r^3$

But $-8 = A1r^2$

$=> 1 = -8r^3$

$=> -1/8 = r^3$

$=> r = -1/2$

$A_3 = A_1r^(3-1)$

$=> -8 = A_1*(-1/2)^2$

$=> -8 = A_1*1/4$

$=> -32 = A_1$