J-uby, 16 bytes

:-&(:+&:+)|~:%&2

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Japt, 7 bytes

gVpVx¹v

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JCram, 11 bytes

TⒻ⍅∆E⍁ζ9Vi∨

Translation of the Python answer.

Rust, 23 bytes

Port of xnor's answer, which you should upvote!

|a,b,n|[a,b,a+b][n%3]%2

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Rust is Language of the month for December, but it doesn't have a ton of answers, so I'm posting a trivial one to remind y'all that Rust can sometimes beat Python too.

Factor, 27 bytes

[ [ tuck + ] times . odd? ]

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Well, executing the Fibonacci recurrence directly is way shorter than any other "smart" methods. Takes three items a0 a1 n on the stack, and outputs t for odd and f for even. Prints an unused value as a side effect.

For golfing, . is shorter than drop, and odd? is shorter than 2 mod or even?.

[                   ! ( a0 a1 n )
  [ tuck + ] times  ! ( an an+1 ) Execute Fibonacci recurrence n times
  . odd?            ! drop an+1 by printing and test if an is odd
]

R + pryr, 30 bytes

pryr::f(c(x,y,x+y)[n%%3+1]%%2)

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Calculates parity of modular index of (x,y,x+y); returns 0 for even and 1 for odd. Function with arguments n,x and y, where x and y represent the initial two fibonacci numbers.

The f() function from the pryr library is a short way to define a function. In code-golf, it's sometimes used simply because pryr::f has one less character than function. However, it also tries to 'guess' the function arguments from the body of the function if they are not declared (using a mysterious and not well-documented algorithm). This is particularly useful here, where declaring x,y,n in a conventional function definition would cost an extra 5 bytes.
Luckily, in this case f somehow manages to guess the arguments perfectly.

Jelly, 3 bytes

+¡Ḃ

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Outputs 0 for even and 1 for odd

How it works

+¡Ḃ - Main link. Takes a on the left and b on the right and n on the command line
 ¡  - Run the following dyad f(x,y) n times, starting with l = a and r = b.
      After each step, update l and r to l = r, r = f(l, r):
+   - f(x,y): x+y
      This is a generalised n'th Fibonacci "builtin"
  Ḃ - Bit; Return 1 for odd, 0 for even

Perl 5 -MList::Util=sum -pla, 21 bytes

A port of @xnor's python answer as a Perl program.

$_=(@F,sum@F)[<>%3]%2

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Inputs the two starting sequence numbers on the first line and the element to find on the second. Outputs 1 for odd parity and 0 for even.

CJam, 9 bytes

q~_:++=2%

Even is zero, odd is one.

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Explanation

q~        e# Read input and evaluate
  _       e# Duplicate initial tuple
   :+     e# Sum initial values
     +    e# Append sum to initial tuple
      =   e# Take the n-th element (modular indexing)
       2% e# Modulo 2

Mathematica, 38 36 bytes

OddQ@({#,#2,+##}&@@#)[[#2~Mod~3+1]]&

Anonymous function, based off of @xnor's Python approach. Takes a list and an integer, and outputs either True for odd or False for even. Not much beyond that.

Actually, 12 bytes

╗│+k3╜%@E2@%

This uses the same strategy as xnor's Python answer. Input is a b n, with the spaces replaced by newlines. Output is 0 for even parity and 1 for odd parity.

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Explanation:

╗│+k3╜%@E2@%
              implicit input ([a, b, n])
╗             save n in reg0 ([a, b])
 │            duplicate stack ([a, b, a, b])
  +k          add, push stack as list ([[a, b, a+b]])
    3╜%       n mod 3 ([[a, b, a+b], n%3])
       @E     element in list ([[a, b, a+b][n%3])
         2@%  mod 2 ([[a, b, a+b][n%3]%2)

Julia, 23 bytes

x\n=[x;sum(x)][n%3+1]%2

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Jelly, 4 bytes

^¡^Ḃ

Prints 0 for even, 1 for odd. Try it online!

How it works

The quick ¡ takes a link it applies repeated to the previous output, and optionally a second link that specifies the number of iterations. If that second link is absent (which is the case here), the number of iterations is taken from the last command-line argument.

When the first link is dyadic (which is the case with ^), ¡ applies that link to the previous return value (which is initialized as the left argument) and the right argument, then updates the return value with the result and the right argument with the previous return value. After doing so as many times as specified in the number of iterations, it returns the last return value.

The quicklink ^¡ is called with the left argument (first command-line argument) and the result of ^ (bitwise XOR of the left and right argument, i.e., the first and second command-line argument). XORing is necessary since the right argument is not among the return values, meaning that the right argument has to be the x(-1) in order to return x(n) with x(0) ^ n ¡ x(1).

To actually calculate x(n), we'd use the code +¡_@ (+¡ instead of ^¡ and _@ – left argument subtracted from right argument – instead of ^), but since we're only interested in the parities, any combination of +, _ and ^ will do.

Finally, Ḃ (bit) computes and returns the parity of x(n).

Perl 6,  24  23 bytes

{(|@^a,sum @a)[$^n%3]%2}
{(|@^a,&[+]...*)[$^n]%2}
{(|@^a,*+*...*)[$^n]%2}

Test:

use v6.c;
use Test;

constant even = 0;
constant odd = 1;
my @test = (
  ([0, 1], 4)    => odd,
  ([2, 1], 9)    => even,
  ([14, 17], 24) => even,
  ([3, 9], 15)   => odd,
  ([2, 4], 10)   => even,
);

my &seq-parity = {(|@^a,*+*...*)[$^n]%2}

for @test -> ( :key($input), :value($expected) ) {
  is seq-parity(|$input), $expected, ($input => <even odd>[$expected]).gist
}

ok 1 - ([0 1] 4) => odd
ok 2 - ([2 1] 9) => even
ok 3 - ([14 17] 24) => even
ok 4 - ([3 9] 15) => odd
ok 5 - ([2 4] 10) => even

Explanation:

{
  (
    # generate the Fibonacci integer sequence
    # ( @^a is the first two values )
    |@^a, *+* ... *
  )[ $^n ] # get the one at the appropriate index
  % 2 # modulo 2
}

MATL, 10 7 bytes

tshwQ)o

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Explanation:

t            #Duplicate the array of numbers
 s           #Sum them
  h          #and add that to the end of the array
   w         #Swap inputs so that the *N* is on top
    Q        #Increment *N* (Since MATL uses 1-based indexing)
     )       #Get that element of the array. MATL uses modular indexing.
      o      #Return it's parity

Pyke, 9 bytes

3%QDs+@2%

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    s+    -   input+sum(input)
      @   -  ^[V]
3%        -   line_2 % 3
       2% - ^ % 2

J, 15 bytes

2|0{(]{:,+/)^:[

Simple method. Computes the n^th term, and takes it modulo 2 to find its parity. Even parity is represented by 0 and odd parity by 1.

Usage

   f =: 2|0{(]{:,+/)^:[
   4 f 0 1
1
   9 f 2 1
0
   24 f 14 17
0
   15 f 3 9
1

Explanation

2|0{(]{:,+/)^:[  Input: n on LHS, and initial values [a, b] on RHS
    ( .... )^:[  Repeat the function n times
         +/      Find the sum of a+b
      {:         Get the value b from the tail of [a, b]
        ,        Join the values to get [b, a+b]
     ]           Return those values as the new [a, b]
  0{             Select the first value at index 0 from the list (after n applications)
2|               Take that value modulo 2 and return

Python, 29 bytes

lambda a,b,n:[a,b,a+b][n%3]%2

The generalized Fibonacci sequence modulo 2 has cycle length of 3. So, we reduce n mod 3 and take that element of the first three.

31 bytes:

lambda a,b,n:0<a%2*2+b%2!=n%3+1

True for odd, False for even

Python, 38 bytes

lambda a,b,n:(b*(n%3>0)+a*(~-n%3>0))%2