Question

the sequence of positive whole numbers is given such that A1<A2 and An+2=An+1+an for n>=1.if a7=120.find A8.​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given :  the sequence of positive whole numbers is given such that a₁ <a₂ and aₙ₊₂=aₙ₊₁ +aₙ for  n≥1.  if a₇=120

To find : a₈

Solution:

let say a₁ = a   and a₂ = a + k   ∵  a₁ < a₂

where a , k are positive whole numbers

a₃ = a₂ + a₁   =  a + k + a = 2a + k

a₄ = a₃ + a₂  = 2a+k + a + k  = 3a + 2k

a₅ = a₄ + a₃ = 3a + 2k + 2a + k  = 5a + 3k

a₆ = a₅ + a₄ = 5a + 3k + 3a + 2k  = 8a + 5k

a₇ = a₆ + a₅ = 8a + 5k + 5a + 3k  = 13a + 8k

a₈ = a₇ + a₆ = 13a + 8k + 8a + 5k  = 21a + 13k

a₇ = 120

a₇  = 13a + 8k

Equating both :

13a + 8k = 120

5a + 8a + 8k =  8 * 15

=> 5a = 8(15 - a - k )

a = 8   and  5 = 15 - a - k   => 5 = 15 - 8 - k  => k = 2

k = 2    

Hence

a₈ =  21a + 13k  = 21(8) + 13(2)  = 168 + 26  =   194

a₈  = 194

Series is  a₁ = 8   and a₂ = 8 + 2 = 10

8  , 10  , 18 , 28  ,  46  , 74  , 120  ,   194  and so on

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