Solution: \n \nTo solve the problem step by step, we first determine the $n^{\text{th}}$ term of the given arithmetic progression (A.P.) and then find the sum of the first $n$ terms of the new A.P. formed with the same first term and double the common difference.\n \n\subsection*{Step 1: Identify the given sum of the first $n$ terms}\n \nThe sum of the first $n$ terms of the A.P. is given by\n\n\[\n\nS_n = 2n + 3n^2\n\n\]\n \n\subsection*{Step 2: Find the $n^{\text{th}}$ term of the A.P.}\n \nThe $n^{\text{th}}$ term is obtained using the formula:\n\n\[\n\nT_n = S_n - S_{n-1}\n\n\]\n \nFirst, find $S_{n-1}$:\n\n\[\n\nS_{n-1} = 2(n-1) + 3(n-1)^2\n\n\]\n \nExpanding:\n\n\[\n\nS_{n-1} = 2n - 2 + 3(n^2 - 2n + 1)\n\n\]\n\n\[\n\nS_{n-1} = 2n - 2 + 3n^2 - 6n + 3\n\n\]\n\n\[\n\nS_{n-1} = 3n^2 - 4n + 1\n\n\]\n \nNow compute $T_n$:\n\n\[\n\nT_n = (2n + 3n^2) - (3n^2 - 4n + 1)\n\n\]\n\n\[\n\nT_n = 2n + 3n^2 - 3n^2 + 4n - 1\n\n\]\n\n\[\n\nT_n = 6n - 1\n\n\]\n \n\subsection*{Step 3: Identify the first term and common difference}\n \nThe standard form of the $n^{\text{th}}$ term of an A.P. is:\n\n\[\n\nT_n = a + (n - 1)d\n\n\]\n \nGiven:\n\n\[\n\nT_n = 6n - 1\n\n\]\n \nFor $n = 1$:\n\n\[\n\na = 6(1) - 1 = 5\n\n\]\n \nHence, the common difference is:\n\n\[\n\nd = 6\n\n\]\n \n\subsection*{Step 4: Form the new A.P.}\n \nThe new A.P. has:\n\n\[\n\na = 5, \quad d' = 2d = 2 \times 6 = 12\n\n\]\n \n\subsection*{Step 5: Find the sum of the first $n$ terms of the new A.P.}\n \nThe sum of the first $n$ terms is:\n\n\[\n\nS_n' = \frac{n}{2} \left[ 2a + (n - 1)d' \right]\n\n\]\n \nSubstituting the values:\n\n\[\n\nS_n' = \frac{n}{2} \left[ 2(5) + (n - 1)(12) \right]\n\n\]\n\n\[\n\nS_n' = \frac{n}{2} (10 + 12n - 12)\n\n\]\n\n\[\n\nS_n' = \frac{n}{2} (12n - 2)\n\n\]\n\n\[\n\nS_n' = n(6n - 1)\n\n\]\n \n\subsection*{Final Result}\n \n\[\n\n\boxed{S_n' = 6n^2 - n}\n\n\]\n\n