Geometric seq

Find the third member of geometric progression if a1 + a2 = 36 and a1 + a3 = 90. Calculate its quotient.

Result

c1 =  81
c2 =  373248
q1 =  3
q2 =  -0.5

Solution:

$a+b=36 \ \\ b=qa \ \\ a+c=90 \ \\ c=q \cdot \ b \ \\ a+qa=36 \ \\ a + q^2 \ a=90 \ \\ a=36/(1+q) \ \\ \ \\ 36 + q^2 \cdot \ 36=90 \cdot \ (1+q) \ \\ 36q^2 -90q -54=0 \ \\ \ \\ a=36; b=-90; c=-54 \ \\ D=b^2 - 4ac=90^2 - 4\cdot 36 \cdot (-54)=15876 \ \\ D>0 \ \\ \ \\ q_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ 90 \pm \sqrt{ 15876 } }{ 72 } \ \\ q_{1,2}=\dfrac{ 90 \pm 126 }{ 72 } \ \\ q_{1,2}=1.25 \pm 1.75 \ \\ q_{1}=3 \ \\ q_{2}=-0.5 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 36 (q -3) (q +0.5)=0 \ \\ a_{1}=36/(1+q_{1})=36/(1+3)=9 \ \\ a_{2}=36/(1+q_{2})=36/(1+(-0.5))=72 \ \\ c_{1}=a_{1} \cdot \ q_{1}^2=9 \cdot \ 3^2=81$

Checkout calculation with our calculator of quadratic equations.

$c_{2}=a_{2} \cdot \ a_{2}^2=72 \cdot \ 72^2=373248$
$q_{1}=3$
$q_{2}=(-0.5)=- \dfrac{ 1 }{ 2 }=-0.5$

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