Numbers

The question should have stated that the forty values are positive integers, because I can always use this sequence:

$0.5, 1.5, 2.5, 3.5, \ldots, 39.5$

The requirements are satisfied, and I will get an answer of $39.5$ .

Anyway, since the answer only takes integers, I'll assume that the sequence contains only integers

For two positive integers to add up to $2$ , there is only one possibility: $1+1=2$

The third number would then be: $2 \times 2 - 1 = 3$

The fourth number is: $3 \times 2 - 3 = 3$

5th number: $4 \times 2 - 3 = 5$

6th number: $5 \times 2 - 5 = 5$

From here, we can see that the numbers form a sequence of odd numbers, where each odd number is repeated before going to the next odd number:

$1, 1, 3, 3, 5, 5, \ldots$

We want to find $T_{40}$ , and if you noticed, the 2nd, 4th, 6th numbers and so on form an arithmetic progression with $20$ terms:

$1, 3, 5, \ldots \quad\quad\quad a=1, d = 2$

$T_{40}$ is found to be the last term, $T_{20}$ , of the second progression, therefore

$T_{40} = 1 + (20 -1)(2) = \boxed{39}$

I solved this problem in this way, Take the numbers as a, b, c, d, e.... Its given, (a+b)/2 =1 or a+b+2
(b+c)/2=2 or b+c=4
Likewise, c+d = 6
and d+e=8
Now subtracting, we get
c-a=2
And, d-b=2, e-c=2
The sequence can then be written as, a, b, a+2, b+2, a+4, b+4 .......
Hence, the 40th term is b+38
But a+b=2
0<b<2
so, 38< last term< 40