Question Number 10489 by Saham last updated on 13/Feb/17

The fifth, nineth, sixteenth terms of a linear   sequence are consecutive terms of an exponential  sequence .  (1) Find the common difference of the linear   sequence in terms of the first term  (2) Show that 21^(th) , 37^(th) , 65^(th)  terms of the linear  sequence are consecutive terms of an exponential  sequence whose common ratio is (3/4)

Answered by mrW1 last updated on 14/Feb/17

(1)  a_5 =a+4d  a_9 =a+8d  a_(16) =a+15d  ((a+15d)/(a+8d))=((a+8d)/(a+4d))  let x=(d/a)  ((1+15x)/(1+8x))=((1+8x)/(1+4x))  1+19x+60x^2 =1+16x+64x^2   4x^2 −3x=0  4x(x−(3/4))=0  x=0 or  x=(3/4)  x=(d/a)=(3/4)  d=(3/4)a  (2)  a_(21) =a+20d=a+20×(3/4)a=16a  a_(37) =a+36d=a+36×(3/4)a=28a  a_(65) =a+64d=a+64×(3/4)a=49a  (a_(65) /a_(37) )=((49a)/(28a))=(7/4)  (a_(37) /a_(21) )=((28a)/(16a))=(7/4)=(a_(65) /a_(37) )

Commented bySaham last updated on 14/Feb/17

God bless you sir.