So little information!!!

First, we need to find out if $n$ is odd or even.

Case : If $n$ is odd

We get

$\left(T_2 + T_4 + T_6 + T_8 + \ldots + T_{n-1}\right) - \left(T_1 + T_3 + T_5 + T_7 + \ldots + T_n\right) = 501\\ \left((a+d) + (a+3d) + \ldots + (a+(n-2)d)\right) - \left(a+(a+d)+ \ldots + (a+(n-1)d)\right) = 501$

The left side has $\dfrac{n-1}{2}$ terms, while the right side has $\dfrac{n+1}{2}$ terms. Therefore,

$\left(\dfrac{n-1}{4}(a+d+a+(n-2)d)\right) - \left(\dfrac{n+1}{4}(a+a+(n-1)d)\right) = 501\\ \left(\dfrac{(n-1)(2a-d+nd)}{4}\right) - \left(\dfrac{(n+1)(2a+nd-d)}{4}\right) = 501\\ \dfrac{(2a+nd-d)(n-1-(n+1))}{4} = 501\\ \dfrac{-2(2a+nd-d)}{4} = 501\\ \dfrac{-(2a+nd-d)}{2} = 501\\ \dfrac{-(a+a+(n-1)d)}{2} = 501\\ \dfrac{-(T_1+T_n)}{2} = 501$

Now, we know that $T_1$ and $T_n$ are positive integers. This means that the LHS of the equation will be negative, which makes this equation impossible.

Case : If $n$ is even

We get

$\left(T_2 + T_4 + T_6 + T_8 + \ldots + T_n\right) - \left(T_1 + T_3 + T_5 + T_7 + \ldots + T_{n-1}\right) = 501\\ \left((a+d) + (a+3d) + \ldots + (a+(n-1)d)\right) - \left(a+a+d+ \ldots + (a+(n-2)d)\right) = 501$

Both sides of the equation have $\dfrac{n}{2}$ terms. Therefore,

$\left(\dfrac{n}{4}(a+d+a+(n-1)d)\right) - \left(\dfrac{n}{4}(a+a+(n-2)d)\right) = 501\\ \dfrac{n}{4}\left((2a+nd) - (2a+nd-2d)\right) = 501\\ \dfrac{2nd}{4} = 501\\ nd=1002$

Both $n$ and $d$ are positive, therefore this equation is possible.

This means that $n$ is even. Which leaves us with two possibilities:

$n \equiv 0 \pmod{4}$ or $n \equiv 2 \pmod{4}$

Now, we know that the terms of the progression alternate between odd and even integers. This implies that $d$ is an odd integer.

From that conclusion, we can say that $1002 \equiv p \pmod{4}$ since whether $1002$ is divisible by $4$ or not depends entirely on $n$

It is found that $1002 \equiv 2 \pmod{4}$

Therefore, $p=\boxed{2}$

• Lets say first term is 'K' which is an odd number.

• Let the difference is 'D'.

• Here series will go like this K, K + D, K + 2D, K + 3D.. and so on. This implies Tn = K + (n-1)D.

• Now T(even) - T(odd) = T2 - T1 + T4 - T3... = K + D - K + K + 3D - K + 2D... = n * D / 2 (n=total terms).

• Now 1st case: 501 = 3 * 167 . Here if n > D then n / 2 = 167, So n = 334. Now (334 mod 4) = 2

• 2nd case: 501 = 1 * 501. Here if n > D then n / 2 = 501, So n = 1002.Now (1002 mod 4) = 2.

• 3rd case: 501 = 3 * 167. Here if n < D then n / 2 = 3, So n = 6.Now (6 mod 4) = 2.

• 4th case : 501 = 1 * 501. Here if n < D then n / 2 = 1, So n = 2. Now (2 mod 4) = 2.

• Here regardless of cases we can prove. If n is even and no factors of 501 is even, the lets say n= 2 * (2 * M - 1) = 4M - 2 which in any case will give remainder of -2, and so 2.