1. **Stating the problem:** We have two sequences: - A geometric progression (GP) with first term $a_1=3$ and common ratio $r=3$. - An arithmetic progression (AP) with first term $A_1=3$ and common difference $d=3$. A new sequence is formed by adding the corresponding terms of the GP and AP. We need to find: (a) The first four terms of the new sequence. (b) The $n^{th}$ term of the new sequence. (c) The sum of the first 10 terms of the new sequence. 2. **Formulas and rules:** - The $n^{th}$ term of a GP is given by: $$g_n = a_1 r^{n-1}$$ - The $n^{th}$ term of an AP is given by: $$a_n = A_1 + (n-1)d$$ - The new sequence's $n^{th}$ term is the sum of the corresponding terms: $$t_n = g_n + a_n$$ - The sum of the first $n$ terms of an AP is: $$S_n = \frac{n}{2} [2A_1 + (n-1)d]$$ - The sum of the first $n$ terms of a GP is: $$S_n = a_1 \frac{r^n - 1}{r - 1}$$ 3. **Calculations:** (a) First four terms of the new sequence: - For GP: $$g_1 = 3 \times 3^{0} = 3$$ $$g_2 = 3 \times 3^{1} = 9$$ $$g_3 = 3 \times 3^{2} = 27$$ $$g_4 = 3 \times 3^{3} = 81$$ - For AP: $$a_1 = 3$$ $$a_2 = 3 + (2-1) \times 3 = 6$$ $$a_3 = 3 + (3-1) \times 3 = 9$$ $$a_4 = 3 + (4-1) \times 3 = 12$$ - New sequence terms: $$t_1 = 3 + 3 = 6$$ $$t_2 = 9 + 6 = 15$$ $$t_3 = 27 + 9 = 36$$ $$t_4 = 81 + 12 = 93$$ (b) $n^{th}$ term of the new sequence: $$t_n = g_n + a_n = 3 \times 3^{n-1} + [3 + (n-1) \times 3] = 3 \times 3^{n-1} + 3n$$ (c) Sum of the first 10 terms of the new sequence: - Sum of first 10 terms of GP: $$S_{10}^{GP} = 3 \times \frac{3^{10} - 1}{3 - 1} = 3 \times \frac{3^{10} - 1}{2}$$ - Sum of first 10 terms of AP: $$S_{10}^{AP} = \frac{10}{2} [2 \times 3 + (10 - 1) \times 3] = 5 [6 + 27] = 5 \times 33 = 165$$ - Sum of first 10 terms of new sequence: $$S_{10} = S_{10}^{GP} + S_{10}^{AP} = 3 \times \frac{3^{10} - 1}{2} + 165$$ **Final answers:** - (a) First four terms: $6, 15, 36, 93$ - (b) $n^{th}$ term: $$t_n = 3 \times 3^{n-1} + 3n$$ - (c) Sum of first 10 terms: $$S_{10} = 3 \times \frac{3^{10} - 1}{2} + 165$$