Maximizing elements in a series

if the first 2 positive integer is a & b

the the sequence will be a, b, a+b, a+2b, 2a+3b

now, 2a+3b = 2012

=> a + 1.5b = 1006

if b = 2 then a will be the maximum. so the first term is 1003

In a sequence of positive integers, every term after the first two terms is the sum of the two previous terms in the sequence. If the fifth term is 2012, what is the maximum possible value of the first term? (This problem is #13 from this year's state Mathcounts sprint round)

Let's say the sequence of five positive integers is:

A, B, C, D, 2012

Since every term after the first two terms is the sum of the previous terms then

C + D = 2012

B + C = D

Substituting gives:

C + B + C = 2012

B + 2C = 2012

A + B = C

Substituting gives:

B + 2(A + B) = 2012

2A + 3B = 2012

So for A to be the maximum then B must be a minimum. The smallest positive integer is 1.

2A + 3(1) = 2012

2A = 2009

A = 1004.5

So does A = 1004 or 1005?

Say A = 1004: 1004, 1, 1005, 1006, 2011 Nope.

Say A = 1005: 1005, 1, 1006, 1007, 2013 Nope.

What? Well let's try B = 2, the next smallest positive integer.

2A + 3(2) = 2012

2A = 2006

A = 1003

1003, 2, 1005, 1007, 2012 Yeah!

So the max. value for the first term is 1003.