Note that the $n$ th term is $n^{2}+50$ for any positive integer $n$ . Suppose that $x$ is a possible GCD. Then:

$x|n^{2}+50, x|(n+1)^{2}+50$

Now we subtract to get $x|(n+1)^{2}+50-n^{2}-50=2n+1$ .

Thus $x|(2n+1)(2n-1)=4n^{2}-1$

However, $x|4 (n^{2}+50)=4n^{2}+200$ .

From the last two equations, subtracting gives $x|201$ . Since $x$ is a positive integer, $x=1, 3, 67, 201$ .

It suffices to show that all of them work.

1 and 3 have been shown to work in the question, the 33rd and 34th term have a GCD of 67 and the 100th and 101st term have a GCD of 201. Thus the answer is $1+3+67+201=272$ .

I wrote a little bit of Haskell code for this one, where binaryGCD comes from the arithmoi package and nub comes from Data.List (nub returns a list's unique elements):

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nums = 51 : 54 : zipWith (\a b -> b + (b - a) + 2) nums (tail nums)
gcds = zipWith binaryGCD nums (tail nums)
nub gcds

Of course, this assumes that no new GCD values appear when the numbers get extremely large. Output: [3, 1, 67, 201] .

Very simple, the for loop extends to much larger numbers as well:

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a = zeros(1,100);

for n = 1:100
    a(n) = gcd((50+n^2),(50+(n+1)^2));
end

A = unique(a);
sum(sum(A))