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

The first series is an arithmetic progression with first term $a = 6$ and common difference $d = 4$. The sum up to $m$ terms is given by:

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

$S_m = \dfrac{m}{2}[12 + 4m - 4] = \dfrac{m}{2}[4m + 8] = 2m(m + 2)$

The second series is an arithmetic progression with first term $a = 1$ and common difference $d = 2$ (which is the sum of the first $n$ odd natural numbers). The sum up to $n$ terms is given by:

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

Equating the two sums, we get:

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