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.Com
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) (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) (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
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Dear math teacher,
I am having difficulties with the following problem:
4 times nC2 (answered by Theo)
1. A(n)=-3+(n-1)(5)
2. A(n)=-11+(n-1)(2)
3. A(n)=9+(n-1)(8)
4. A(n)=0.5+(n-1)(3.5)
(answered by tommyt3rd)
1. A(n)=-3+(n-1)(5)
2. A(n)=-11+(n-1)(2)
3. A(n)=9+(n-1)(8)
4. A(n)=0.5+(n-1)(3.5)
(answered by tommyt3rd)
1. A(n)=-3+(n-1)(5)
2. A(n)=-11+(n-1)(2)
3. A(n)=9+(n-1)(8)
4. A(n)=0.5+(n-1)(3.5)
(answered by tommyt3rd)