9. \( 3 x+1 ; 2 x ; 3 x-7 \) are the first three terms of an arithmetic sequence. Calculate the value of 3 10. The first and second terms of an arithmetic sequence are 10 and 6 respectively. a) Calculate the \( 11^{\text {th }} \) term of the sequence. b) The sum of the first n terms of this sequence is -560 . Calculate \( n \). 11. Determine \( 3+6+12+24+\ldots \) to 10 terms.

Let's address each of your questions step by step. --- ### **Problem 9** **Given:** The first three terms of an arithmetic sequence are: - \( T_1 = 3x + 1 \) - \( T_2 = 2x \) - \( T_3 = 3x - 7 \) **To Find:** The value of \( x \). **Solution:** In an **arithmetic sequence**, the difference between consecutive terms is constant. Therefore: \[ T_2 - T_1 = T_3 - T_2 \] Plugging in the given terms: \[ 2x - (3x + 1) = (3x - 7) - 2x \] Simplify both sides: \[ 2x - 3x - 1 = 3x - 2x - 7 \] \[ - x - 1 = x - 7 \] Now, solve for \( x \): \[ - x - 1 = x - 7 \\ -1 + 7 = x + x \\ 6 = 2x \\ x = 3 \] **Answer:** \( x = 3 \) --- ### **Problem 10** **Given:** - First term (\( a_1 \)) = 10 - Second term (\( a_2 \)) = 6 **To Find:** a) The 11th term of the sequence. b) The number of terms (\( n \)) when the sum of the first \( n \) terms (\( S_n \)) is -560. --- #### **Part a: Calculate the 11th Term** **Solution:** First, find the common difference (\( d \)) of the arithmetic sequence: \[ d = a_2 - a_1 = 6 - 10 = -4 \] The \( n^{th} \) term of an arithmetic sequence is given by: \[ a_n = a_1 + (n - 1) \times d \] For the 11th term (\( a_{11} \)): \[ a_{11} = 10 + (11 - 1) \times (-4) \\ a_{11} = 10 + 10 \times (-4) \\ a_{11} = 10 - 40 \\ a_{11} = -30 \] **Answer:** The 11th term is **-30**. --- #### **Part b: Calculate \( n \) When \( S_n = -560 \)** **Solution:** The sum of the first \( n \) terms of an arithmetic sequence is: \[ S_n = \frac{n}{2} \times [2a_1 + (n - 1) \times d] \] Plugging in the known values: \[ -560 = \frac{n}{2} \times [2 \times 10 + (n - 1) \times (-4)] \] Simplify inside the brackets: \[ -560 = \frac{n}{2} \times [20 - 4(n - 1)] \\ -560 = \frac{n}{2} \times [20 - 4n + 4] \\ -560 = \frac{n}{2} \times (24 - 4n) \\ \] Multiply both sides by 2 to eliminate the fraction: \[ -1120 = n \times (24 - 4n) \\ -1120 = 24n - 4n^2 \\ \] Rearrange the equation: \[ 4n^2 - 24n - 1120 = 0 \] Divide the entire equation by 4 to simplify: \[ n^2 - 6n - 280 = 0 \] Solve the quadratic equation using the quadratic formula: \[ n = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times (-280)}}{2 \times 1} \\ n = \frac{6 \pm \sqrt{36 + 1120}}{2} \\ n = \frac{6 \pm \sqrt{1156}}{2} \\ n = \frac{6 \pm 34}{2} \] This gives two possible solutions: 1. \( n = \frac{6 + 34}{2} = \frac{40}{2} = 20 \) 2. \( n = \frac{6 - 34}{2} = \frac{-28}{2} = -14 \) (Discarded since \( n \) must be positive) **Answer:** The number of terms \( n \) is **20**. --- ### **Problem 11** **Given:** The series to determine is: \[ 3 + 6 + 12 + 24 + \ldots \quad \text{(10 terms)} \] **To Find:** The sum of the first 10 terms. **Solution:** This is a **geometric series** where each term is multiplied by a common ratio (\( r \)). Identify the first term (\( a \)) and the common ratio (\( r \)): \[ a = 3 \\ r = \frac{6}{3} = 2 \] The formula for the sum of the first \( n \) terms of a geometric series is: \[ S_n = a \times \frac{r^n - 1}{r - 1} \] Plugging in the values: \[ S_{10} = 3 \times \frac{2^{10} - 1}{2 - 1} \\ S_{10} = 3 \times (1024 - 1) \\ S_{10} = 3 \times 1023 \\ S_{10} = 3069 \] **Answer:** The sum of the first 10 terms is **3069**. ---