Let us take $a_1 = a$. We have
$$a^2 = a + kd, \quad (a + pd)^2 = a + ld, \quad (a + qd)^2 = a + md.$$
Thus we have
$$a + ld = (a + pd)^2 = a^2 + 2pad + p^2d^2 = a + kd + 2pad + p^2d^2.$$
Since we have non-constant AP, we see that $d \neq 0$. Hence we obtain $2pa + p^2d = l - k$. Similarly, we get $2qa + q^2d = m - k$. Observe that $p^2q - pq^2 \neq 0$. Otherwise $p = q$ and $\text{gcd}(p, q) = p > 1$ which is a contradiction to the given hypothesis that $\text{gcd}(p, q) = 1$. Hence we can solve the two equations for $a, d$:
$$a = \frac{p^2(m-k) - q^2(l-k)}{2(p^2q - pq^2)}, \quad d = \frac{q(l-k) - p(m-k)}{p^2q - pq^2}.$$
It follows that $a, d$ are rational numbers. We also have
$$p^2a^2 = p^2a + kp^2d.$$
But $p^2d = l - k - 2pa$. Thus we get
$$p^2a^2 = p^2a + k(l - k - 2pa) = (p - 2k)pa + k(l - k).$$
This shows that $pa$ satisfies the equation
$$x^2 - (p - 2k)x - k(l - k) = 0.$$
Since $a$ is rational, we see that $pa$ is rational. Write $pa = w/z$, where $w$ is an integer and $z$ is a natural number such that $\text{gcd}(w, z) = 1$. Substituting in the equation, we obtain
$$w^2 - (p - 2k)wz - k(l - k)z^2 = 0.$$
This shows $z$ divides $w$. Since $\text{gcd}(w, z) = 1$, it follows that $z = 1$ and $pa = w$ an integer. (In fact any rational solution of a monic polynomial with integer coefficients is necessarily an integer.) Similarly, we can prove that $qa$ is an integer. Since $\text{gcd}(p, q) = 1$, there are integers $u$ and $v$ such that $pu + qv = 1$. Therefore $a = (pa)u + (qa)v$. It follows that $a$ is an integer. But $p^2d = l - k - 2pa$. Hence $p^2d$ is an integer. Similarly, $q^2d$ is also an integer. Since $\text{gcd}(p^2, q^2) = 1$, it follows that $d$ is an integer. Combining these two, we see that all the terms of the AP are integers.

Alternatively, we can prove that $a$ and $d$ are integers in another way. We have seen that $a$ and $d$ are rationals; and we have three relations:
$$a^2 = a + kd, \quad p^2d + 2pa = n_1, \quad q^2d + 2qa = n_2,$$
where $n_1 = l - k$ and $n_2 = m - k$. Let $a = u/v$ and $d = x/y$ where $u, x$ are integers and $v, y$ are natural numbers, and $\text{gcd}(u, v) = 1$, $\text{gcd}(x, y) = 1$. Putting this in these relations, we obtain
$$u^2 y = uvy + kxv^2, \qquad (1)$$
$$2puy + p^2vx = vyn_1, \qquad (2)$$
$$2quy + q^2vx = vyn_2. \qquad (3)$$
Now (1) shows that $v|u^2y$. Since $\text{gcd}(u, v) = 1$, it follows that $v|y$. Similarly (2) shows that $y|p^2vx$. Using $\text{gcd}(y, x) = 1$, we get that $y|p^2v$. Similarly, (3) shows that $y|q^2v$. Therefore $y$ divides $\text{gcd}(p^2v, q^2v) = v$. The two results $v|y$ and $y|v$ imply $v = y$, since both $v, y$ are positive. Substitute this in (1) to get
$$u^2 = uv + kxv.$$
This shows that $v|u^2$. Since $\text{gcd}(u, v) = 1$, it follows that $v = 1$. This gives $v = y = 1$. Finally $a = u$ and $d = x$ which are integers.