Answer:
see explanation
Step-by-step explanation:
(a)
Using the recursive rule and a₁ = 4 , then
a₂ = 2([tex]\frac{1}{3}[/tex] + a₁ ) = 2([tex]\frac{1}{3}[/tex] + 4) = 2 × 4 [tex]\frac{1}{3}[/tex] = 2 × [tex]\frac{13}{3}[/tex] = [tex]\frac{26}{3}[/tex]
a₃ = 2([tex]\frac{1}{3}[/tex] + a₂) = 2([tex]\frac{1}{3}[/tex] + [tex]\frac{26}{3}[/tex] ) = 2 × [tex]\frac{27}{3}[/tex] = 2 × 9 = 18
a₄ = 2([tex]\frac{1}{3}[/tex] + a₃) = 2([tex]\frac{1}{3}[/tex] + 18) = 2 × 18 [tex]\frac{1}{3}[/tex] = 2 × [tex]\frac{55}{3}[/tex] = [tex]\frac{110}{3}[/tex]
a₅ = 2([tex]\frac{1}{3}[/tex] + a₄) = 2([tex]\frac{1}{3}[/tex] + [tex]\frac{110}{3}[/tex]) = 2 × [tex]\frac{111}{3}[/tex] = 2 × 37 = 74
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(b)
Using the recursive rule and a₁ = 6 , then
a₂ = [tex]\frac{2}{a_{1} }[/tex] = [tex]\frac{2}{6}[/tex] = [tex]\frac{1}{3}[/tex]
a₃ = [tex]\frac{3}{a_{2} }[/tex] = [tex]\frac{3}{\frac{1}{3} }[/tex] = 9
a₄ = [tex]\frac{4}{a_{3} }[/tex] = [tex]\frac{4}{9}[/tex]
a₅ = [tex]\frac{5}{a_{4} }[/tex] = [tex]\frac{5}{\frac{4}{9} }[/tex] = 5 × [tex]\frac{9}{4}[/tex] = [tex]\frac{45}{4}[/tex]
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(c)
The sequence has a common ratio
r = [tex]\frac{\frac{2}{3} }{\frac{1}{6} }[/tex] = [tex]\frac{\frac{8}{3} }{\frac{2}{3} }[/tex] = 4
Multiply the previous terms by 4 , then
a₄ = 4 × [tex]\frac{8}{3}[/tex] = [tex]\frac{32}{3}[/tex]
a₅ = 4 × [tex]\frac{32}{3}[/tex] = [tex]\frac{128}{3}[/tex]
