Respuesta :

a₁ = (a₀/2) +3 = (a₀ +6)/2
a₂ = (a₁/2) +3 = (a₀/2 +3)/2 +3
  = (a₀ +18)/4

Then a₂/a₁ is
a₂/a₁ = ((a₀ +18)/4)/((a₀ +6)/2)
  = (a₀ +18)/(2a₀ +12) . . . . . . . . matches selection D
First Term = a₀

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Second term is 3 greater than 1/2 of First Term:
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[tex]\text {Second Term = } \dfrac{1}{2} a_0 + 3[/tex]

[tex]\text {Second Term = } \dfrac{a_0 + 6}{2} [/tex]

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Third term is 3 greater than 1/2 of Second Term:
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[tex]\text {Third Term = } \dfrac{1}{2} (\ \dfrac{a_0 + 6}{2} ) + 3[/tex]

[tex]\text {Third Term = } \ \dfrac{a_0 + 6}{4} + 3[/tex]

[tex]\text {Third Term = } \ \dfrac{a_0 + 6 + 12}{4} [/tex]

[tex]\text {Third Term = } \ \dfrac{a_0 + 18}{4} [/tex]

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Ratio of the third term to the second term:
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[tex] \dfrac{a_0 + 18}{4} : \dfrac{a_0 + 6}{2} [/tex]

[tex] \dfrac{a_0 + 18}{4} : \dfrac{2a_0 + 12}{4} [/tex]

[tex]a_0 + 18 : 2a_0 + 12[/tex]

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Put in in fraction form:
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[tex] \dfrac{a_0 + 18 }{2a_0 + 12} [/tex]

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Answer: D
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