16. In an arithmetic sequence the 17 th term is 9 times the first term and t 9 th term is 6 less than 3 times the 3 rd term. Find the first 3 terms the sequence.

Let's solve the problem step by step. **Given:** - Arithmetic sequence with first term \( a \) and common difference \( d \). - The 17th term is 9 times the first term. - The 9th term is 6 less than 3 times the 3rd term. **Step 1: Express the terms in terms of \( a \) and \( d \).** - The \( n \)-th term of an arithmetic sequence is given by: \[ t_n = a + (n-1)d \] - The 17th term: \[ t_{17} = a + 16d \] According to the problem: \[ a + 16d = 9a \implies 16d = 8a \implies 2d = a \quad \text{(Equation 1)} \] - The 9th term: \[ t_9 = a + 8d \] - The 3rd term: \[ t_3 = a + 2d \] According to the problem: \[ t_9 = 3t_3 - 6 \implies a + 8d = 3(a + 2d) - 6 \] Simplifying: \[ a + 8d = 3a + 6d - 6 \implies 2d = 2a - 6 \quad \text{(Equation 2)} \] **Step 2: Substitute \( a \) from Equation 1 into Equation 2.** \[ 2d = 2(2d) - 6 \implies 2d = 4d - 6 \implies -2d = -6 \implies d = 3 \] \[ a = 2d = 2 \times 3 = 6 \] **Step 3: Find the first three terms.** \[ t_1 = a = 6 \] \[ t_2 = a + d = 6 + 3 = 9 \] \[ t_3 = a + 2d = 6 + 2 \times 3 = 12 \] **Answer:** The first three terms of the sequence are 6, 9, and 12.