# Quotient

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

Correct result:

q =  -5
a2 =  0

#### Solution:

$\ \\ a_{3}=10 \ \\ a_{1}+a_{2}=-1.6 \ \\ a_{1}-a_{2}=2.4 \ \\ a_{1}=(-1.6+2.4)/2=\dfrac{ 2 }{ 5 }=0.4 \ \\ a_{2}=-1.6-a_{1}=-1.6-0.4=-2 \ \\ a_{2}=q a_{1} \ \\ q=a_{2}/a_{1}=(-2)/0.4=-5$
${a}_{2}=0$

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