Menu Close

the-sequence-a-1-a-2-a-3-satisfies-the-relation-a-n-1-a-n-a-n-1-for-n-gt-1-given-that-a-20-6765-and-a-18-2584-what-is-a-16-




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 by john 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 by john 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 by mr W last updated on 05/Apr/20
yes
$${yes} \\ $$
Commented by john santu last updated on 05/Apr/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *