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$\\Solution:\\\\ \text{Let the first terms } a_1,\ \ a_2 \ \ \text{ and the common differences } d_1,\ d_2 \ \text{ of first and second APs respectively} \\\\ Sum\ of\ n\ the\ terms\ in\ 1^{st}\ AP=\frac{n}{2}(2a_1+(n-1)d_1)\\\\\\ Sum\ of\ n\ the\ terms\ in\ 2^{nd}\ AP=\frac{n}{2}(2a_2+(n-1)d_2)\\\\\\ \text{Given: Ratio of the sums of n terms of two AP}=\frac{3n+4}{5n +6}\\\\\\ \Rightarrow \frac{\frac{n}{2}(2a_1+(n-1)d_1)}{\frac{n}{2}(2a_2+(n-1)d_2)}=\frac{3n+4}{5n +6}\\\\\\ \Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{3n+4}{5n +6}\\\\\\ \Rightarrow \frac{a_1+(\frac{n-1}{2})d_1}{a_2+(\frac{n-1}{2})d_2}=\frac{3n+4}{5n +6}\ \ \ \ \rightarrow (1)\\\\\\ Let,\ \frac{n-1}{2}=m-1\\\\ \Rightarrow n-1=2(m-1)\\\\ \Rightarrow n=2m-2+1\\\\ \Rightarrow n=2m-1\\\\ Substitute\ above\ values\ in\ equation\ (1)\\\\ \Rightarrow \frac{a_1+(m-1)d_1}{a_2+(m-1)d_2}=\frac{3(2m-1)+4}{5(2m-1)+6}$

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$\begin{array}{l} S_{n}=\sum\left(2 n^{2}+3 n\right) \\ \Rightarrow S_{n}=2 \sum n^{2}+3 \sum n \\ \Rightarrow S_{n}=2\left(\frac{n(n+1)(2 n+1)}{6}\right)+3\left(\frac{n(n+1)}{2}\right) \\ \Rightarrow S_{n}=[n(n+1)] \times\left[\frac{2 n+1}{3}+\frac{3}{2}\right] \\ \Rightarrow S_{n}=[n(n+1)] \times\left[\frac{4 n+2+9}{6}\right] \\ \Rightarrow S_{n}=[n(n+1)] \times\left[\frac{4 n+11}{6}\right] \\ \Rightarrow S_{n}=\frac{n(n+1)(4 n+11)}{6} \end{array}$

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