If the sums of n terms of two AP.'s are in the ratio (3n + 2) : (2n + 3),

Let the first terms of the two A.P.'s be a and a'; and their common difference be d and d'.

Now,

$\frac{S_{n}}{S_{n}{ }^{\prime}}=\frac{(3 n+2)}{(2 n+3)}$

$\Rightarrow \frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}\left[2 a^{\prime}+(n-1) d^{\prime}\right]}=\frac{(3 n+2)}{(2 n+3)}$

$\Rightarrow \frac{[2 a+(n-1) d]}{\left[2 a^{\prime}+(n-1) d^{\prime}\right]}=\frac{(3 n+2)}{(2 n+3)}$

Let $n=23$

$\Rightarrow \frac{2 a+(23-1) d}{2 a^{\prime}+(23-1) d^{\prime}}=\frac{3 \times 23+2}{2 \times 23+3}$

$\Rightarrow \frac{2 a+22 d}{2 a^{\prime}+22 d^{\prime}}=\frac{69+2}{46+3}$

$\Rightarrow \frac{2(a+11 d)}{2\left(a^{\prime}+11 d^{\prime}\right)}=\frac{71}{49}$

$\therefore \frac{a_{12}}{a_{12}}=\frac{71}{49}$

So, the ratio of their 12th terms is 71 : 49.