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Question Number 87648 by john santu last updated on 05/Apr/20

the sequence a_1 ,a_2 ,a_3 , ... satisfies the relation a_(n+1) = a_n +a_(n−1) , for n>1. given that a_(20) = 6765 and a_(18) = 2584 what is a_(16)

$$\mathrm{the}\:\mathrm{sequence}\:\mathrm{a}_{\mathrm{1}} ,\mathrm{a}_{\mathrm{2}} ,\mathrm{a}_{\mathrm{3}} ,\:...\:\mathrm{satisfies} \\ $$ $$\mathrm{the}\:\mathrm{relation}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} \:=\:\mathrm{a}_{\mathrm{n}} +\mathrm{a}_{\mathrm{n}−\mathrm{1}} \:,\:\mathrm{for} \\ $$ $$\mathrm{n}>\mathrm{1}.\:\mathrm{given}\:\mathrm{that}\:\mathrm{a}_{\mathrm{20}} \:=\:\mathrm{6765}\:\mathrm{and} \\ $$ $$\mathrm{a}_{\mathrm{18}} \:=\:\mathrm{2584}\:\mathrm{what}\:\mathrm{is}\:\mathrm{a}_{\mathrm{16}} \\ $$

Commented byjohn santu last updated on 05/Apr/20

a_(20) = a_(19) + a_(18) = a_(18) +a_(17) +a_(18) = 3×a_(18) −a_(16) ⇒a_(16) = 3×a_(18) −a_(20) ⇒a_(16) = 3×2584−6765 = 987

$$\mathrm{a}_{\mathrm{20}} \:=\:\mathrm{a}_{\mathrm{19}} \:+\:\mathrm{a}_{\mathrm{18}} \\ $$ $$=\:\mathrm{a}_{\mathrm{18}} +\mathrm{a}_{\mathrm{17}} +\mathrm{a}_{\mathrm{18}} \\ $$ $$=\:\mathrm{3}×\mathrm{a}_{\mathrm{18}} −\mathrm{a}_{\mathrm{16}} \\ $$ $$\Rightarrow\mathrm{a}_{\mathrm{16}} \:=\:\mathrm{3}×\mathrm{a}_{\mathrm{18}} −\mathrm{a}_{\mathrm{20}} \\ $$ $$\Rightarrow\mathrm{a}_{\mathrm{16}} \:=\:\mathrm{3}×\mathrm{2584}−\mathrm{6765}\:=\:\mathrm{987} \\ $$

Commented byjohn santu last updated on 05/Apr/20

so for this question x^2 −x−1=0 x = ((1 ± (√5))/2) a_n = A(((1+(√5))/2))^n +B(((1−(√5))/2))^n

$$\mathrm{so}\:\mathrm{for}\:\mathrm{this}\:\mathrm{question}\: \\ $$ $$\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$ $$\mathrm{x}\:=\:\frac{\mathrm{1}\:\pm\:\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$ $$\mathrm{a}_{\mathrm{n}} =\:\mathrm{A}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{n}} +\mathrm{B}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{n}} \\ $$

Commented bymr W last updated on 05/Apr/20

yes

$${yes} \\ $$

Commented byjohn santu last updated on 05/Apr/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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