SOLUTION: Let a_1, a_2, a_3, \dots be a sequence. If a_n = a_{n - 1} + a_{n - 2} for all n \ge 3, and a_{11} = 4 and a_{10} = 1, then find a_6.

Algebra ->  Sequences-and-series -> SOLUTION: Let a_1, a_2, a_3, \dots be a sequence. If a_n = a_{n - 1} + a_{n - 2} for all n \ge 3, and a_{11} = 4 and a_{10} = 1, then find a_6.      Log On


   

Question 1209400: Let a_1, a_2, a_3, \dots be a sequence. If
a_n = a_{n - 1} + a_{n - 2}
for all n \ge 3, and a_{11} = 4 and a_{10} = 1, then find a_6.
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

a3 = a2+a1
a4 = a3+a2
a5 = a4+a3
a6 = a5+a4
... etc ...
a11 = a10+a9
The indices decrease when reading left to right.

Let's rearrange each equation so that the last term on the right hand side is by itself.
a1 = a3-a2
a2 = a4-a3
a3 = a5-a4
a4 = a6-a5
... etc ...
a9 = a11-a10
The template is a_n = a_{n+2} - a_{n+1}

Since we are given a11 = 4 and a10 = 1, we can then say,
a9 = a11-a10
a9 = 4-1
a9 = 3
Then,
a8 = a10-a9
a8 = 1-3
a8 = -2
And,
a7 = a9-a8
a7 = 3-(-2)
a7 = 3+2
a7 = 5
And,
a6 = a8-a7
a6 = -2-5
a6 = -7

There might be a faster more efficient approach.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I doubt if there is a faster way to solve the problem than what is shown by the other tutor... but his response has a lot of unnecessary discussion at the beginning.

The given recursive formula tells us how to get the next numbers in the sequence from the previous numbers in the sequence:

a_n = a_(n-1) + a_(n-2)

Turn that recursive formula around so it tells you how to find PREVIOUS numbers in the sequence from known numbers in the sequence:

a_(n-2) = a_n - a_(n-1)

Then, starting with the given a_11 = 4 and a_10 = 1...

a_9 = a_11 - a_10 = 4-1 = 3

a_8 = a_10 - a_9 = 1-3 = -2

a_7 = a_9 - a_8 = 3-(-2) = 5

a_6 = a_8 - a_7 = -2-5 = -7

ANSWER: a_6 = -7