# Quotient

Determine the quotient and the second member of the geometric progression where a3=10, a1+a2=-1,6 a1-a2=2,4.

Result

q =  -5
a2 =  -2

#### Solution:

$\ \\ a_{ 3 } = 10 \ \\ a_{ 1 }+a_{ 2 } = -1.6 \ \\ a_{ 1 }-a_{ 2 } = 2.4 \ \\ a_{ 1 } = (-1.6+2.4)/2 = \dfrac{ 2 }{ 5 } = 0.4 \ \\ a_{ 2 } = -1.6-a_{ 1 } = -1.6-0.4 = -2 \ \\ a_{ 2 } = q a_{ 1 } \ \\ q = a_{ 2 }/a_{ 1 } = (-2)/0.4 = -5$
$a_{ 2 } = (-2) = -2$

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