Geometric seq

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

Correct answer:

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

Step-by-step explanation:

$$\begin{gathered} a+b=36 \\ b=qa \\ a+c=90 \\ c=q\cdot b \\ a+qa=36 \\ a+q^{2}a=90 \\ a=36\mathrm{/}(1+q) \\ 36+q^2\ast 36=90\ast (1+q) \\ 36+q^2\cdot 36=90\cdot (1+q) \\ 36q^2-90q-54=0 \\ a=36;b=-90;c=-54 \\ D=b^2-4ac=90^2-4\cdot 36\cdot (-54)=15876 \\ D>0 \\ {q}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{90\pm \sqrt{15876}}{72}} \\ {q}_{1,2}={\displaystyle \frac{90\pm 126}{72}} \\ {q}_{1,2}=1.25\pm 1.75 \\ {q}_1=3 \\ {q}_2=-0.5 \\ \text{ Factored form of the equation: } \\ 36(q-3)(q+0.5)=0 \\ {a}_1=36\mathrm{/}(1+q_1)=36\mathrm{/}(1+3)=9 \\ {a}_2=36\mathrm{/}(1+q_2)=36\mathrm{/}(1+(-0.5))=72 \\ {c}_1=a_1\cdot q_1^2=9\cdot 3^2=81 \end{gathered}$$

Our quadratic equation calculator calculates it.

$c_2=a_2\cdot a_2^2=72\cdot 72^2=373248$

$q_1=3$

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