For the following two (02) items: Let (6 + 10 + 14 + up to m terms)...

The sum of the first $m$ terms of the arithmetic progression $6, 10, 14, \ldots$ is given by:

$S_m = \dfrac{m}{2}[2(6) + (m - 1)4] = \dfrac{m}{2}[12 + 4m - 4] = 2m^2 + 4m$

The sum of the first $n$ terms of the arithmetic progression $1, 3, 5, 7, \ldots$ is given by:

$S_n = \dfrac{n}{2}[2(1) + (n - 1)2] = n^2$

Equating the two sums, we get:

$2m^2 + 4m = n^2$

$2(m^2 + 2m + 1) - 2 = n^2$

$2(m + 1)^2 - n^2 = 2$

Let $u = m + 1$. The equation becomes $2u^2 - n^2 = 2$. This implies that $n$ must be an even number. Let $n = 2v$. Substituting this into the equation yields:

$2u^2 - 4v^2 = 2$

$u^2 - 2v^2 = 1$

This is a standard Pell's equation. The fundamental solution is $u_1 = 3$ and $v_1 = 2$.

For $u_1 = 3$ and $v_1 = 2$, we have $m + 1 = 3 \Rightarrow m = 2$ and $n = 2(2) = 4$. Both $m < 25$ and $n < 25$ are satisfied.

The next solution is generated by $(u_1 + v_1\sqrt{2})^2 = (3 + 2\sqrt{2})^2 = 17 + 12\sqrt{2}$.

Thus, $u_2 = 17$ and $v_2 = 12$, which gives $m + 1 = 17 \Rightarrow m = 16$ and $n = 2(12) = 24$. Both $m < 25$ and $n < 25$ are satisfied.

The third solution is generated by $(3 + 2\sqrt{2})^3 = 99 + 70\sqrt{2}$.

Thus, $u_3 = 99$ and $v_3 = 70$, which gives $m + 1 = 99 \Rightarrow m = 98$ and $n = 2(70) = 140$. This does not satisfy the given conditions $m < 25$ and $n < 25$.

Therefore, the only possible values for $m$ are $2$ and $16$. There are exactly two possible values.

Answer: Two