If the sum of first n terms of two A.P.’s are in the ratio 3n + 8 : 7n + 15, then the ratio of 12th term is

$$\begin{gathered} \frac{{\displaystyle \frac{\mathrm{n}}{2}}\left(2{\mathrm{a}}_1+(\mathrm{n}-1){\mathrm{d}}_1\right)}{{\displaystyle \frac{\mathrm{n}}{2}}\left(2{\mathrm{a}}_2+(\mathrm{n}-1){\mathrm{d}}_2\right)}=\frac{3\mathrm{n}+8}{7\mathrm{n}+15} \\ \frac{{\mathrm{a}}_1+{\displaystyle \frac{\mathrm{n}-1}{2}}{\mathrm{d}}_1}{{\mathrm{a}}_2+{\displaystyle \frac{\mathrm{n}-1}{2}}{\mathrm{d}}_2}=\frac{3\mathrm{n}+8}{7\mathrm{n}+15} \\ \Rightarrow \frac{\mathrm{n}-1}{2}=11 \\ \mathrm{n}=23 \\ \frac{{\mathrm{a}}_1+11{\mathrm{d}}_1}{{\mathrm{a}}_2+11{\mathrm{d}}_2}=\frac{69+8}{161+15}=\frac{77}{176}=\frac{7}{16} \end{gathered}$$