, 2013

The parity of an integer is whether it is even or odd.

Since the Fibonacci sequence begins $1, 1, 2, 3, 5, 8, 13, 21, \ldots$, then the parities of the first eight terms are Odd, Odd, Even, Odd, Odd, Even, Odd, Odd.

In the sequence, if $x$ and $y$ are consecutive terms, then the next term is $x+y$.

In general, suppose that $x$ and $y$ are integers.

If $x$ is even and $y$ is even, then $x+y$ is even. If $x$ is even and $y$ is odd, then $x+y$ is odd.

If $x$ is odd and $y$ is even, then $x+y$ is odd. If $x$ is odd and $y$ is odd, then $x+y$ is even.

Therefore, the parities of two consecutive terms $x$ and $y$ in the Fibonacci sequence determine the parity of the following term $x+y$.

Also, once there are two consecutive terms whose parities match the parities of two earlier consecutive terms in the sequence, then the parities will repeat in a cycle.

In particular, the parities of the fourth and fifth terms (Odd, Odd) are the same as the parities of the first and second terms (Odd, Odd).

Therefore, the parities in the sequence repeat the cycle Odd, Odd, Even.

This cycle has length 3.

Therefore, the 99th term in the Fibonacci sequence ends one of these cycles, since 99 is a multiple of 3.

In particular, the 99th term ends the 33rd cycle.

Each cycle contains two odd terms.

Therefore, the first 99 terms in the sequence include $2\times 33 = 66$ odd terms.

Finally, the 100th term in the sequence begins a new cycle, so is odd.

Therefore, the first 100 terms include $66+1=67$ odd terms.
Suppose that the first term in the given sequence is $a$ and the common difference is $d$.

Then the first four terms are $a,a+d,a+2d,a+3d$.

From the given information, $a+(a+2d)=6$ and $(a+d)+(a+3d)=20$.

The first equation simplifies to $2a+2d=6$ or $a+d=3$.

The second equation simplifies to $2a+4d=20$ or $a+2d=10$.

Therefore, $(a+2d)-(a+d)=10-3$ or $d=7$.

Since $a+d=3$ and $d=7$, then $a=-4$.

Therefore, the tenth term in the sequence is $a+9d=-4+9(7)=59$.