Example 2 The common difference of an arithmetic sequence is 10 and the 5 th term is 49 . Find the general formula for the sequence. Example 3 In an arithmetic sequence, the 2 nd term is 1 and the 7 th term is 16 . Find the 1st three terms of the sequence. 2) In an arithmetic sequence, \( T_{3}=-2 \) and \( T_{8}=23 \). Determine the first term and the constant difference. 3) Find the number of terms in the arithmetic sequence \[ -5 ;-11 ;-17 ; \ldots ;-491 \] 4) The first three terms of an arithmetic sequence are, \( x-8 ; x ; 2 x-5 \). Determine a) The value of \( x \). b) The general term. c) The value of the \( 115^{\text {th }} \) term Example 4 Consider the following arithmetic sequence: \[ 2 k-1 ; k+7 ; 5 k-5 ; \ldots \] (a) Find the value of \( k \). (b) Find the 1st three terms of the sequence. (c) Find \( T_{n} \).

Let's solve each example step by step. ### Example 2 **Problem:** The common difference of an arithmetic sequence is 10 and the 5th term is 49. Find the general formula for the sequence. **Known Conditions:** - Common difference \( d = 10 \) - 5th term \( T_5 = 49 \) **Step 1:** The general formula for the \( n \)-th term of an arithmetic sequence is given by: \[ T_n = a + (n-1)d \] where \( a \) is the first term. **Step 2:** For the 5th term: \[ T_5 = a + 4d \] Substituting the known values: \[ 49 = a + 4(10) \] \[ 49 = a + 40 \] \[ a = 49 - 40 = 9 \] **Step 3:** Now we can write the general formula: \[ T_n = 9 + (n-1) \cdot 10 \] \[ T_n = 9 + 10n - 10 \] \[ T_n = 10n - 1 \] ### Example 3 **Problem 1:** In an arithmetic sequence, the 2nd term is 1 and the 7th term is 16. Find the first three terms of the sequence. **Known Conditions:** - \( T_2 = 1 \) - \( T_7 = 16 \) **Step 1:** Using the general formula: \[ T_n = a + (n-1)d \] For the 2nd term: \[ T_2 = a + d = 1 \quad (1) \] For the 7th term: \[ T_7 = a + 6d = 16 \quad (2) \] **Step 2:** Subtract equation (1) from equation (2): \[ (a + 6d) - (a + d) = 16 - 1 \] \[ 5d = 15 \implies d = 3 \] **Step 3:** Substitute \( d \) back into equation (1): \[ a + 3 = 1 \implies a = -2 \] **Step 4:** The first three terms are: \[ T_1 = a = -2, \quad T_2 = 1, \quad T_3 = a + 2d = -2 + 6 = 4 \] Thus, the first three terms are \( -2, 1, 4 \). **Problem 2:** In an arithmetic sequence, \( T_3 = -2 \) and \( T_8 = 23 \). Determine the first term and the constant difference. **Known Conditions:** - \( T_3 = a + 2d = -2 \quad (1) \) - \( T_8 = a + 7d = 23 \quad (2) \) **Step 1:** Subtract equation (1) from equation (2): \[ (a + 7d) - (a + 2d) = 23 + 2 \] \[ 5d = 25 \implies d = 5 \] **Step 2:** Substitute \( d \) back into equation (1): \[ a + 2(5) = -2 \implies a + 10 = -2 \implies a = -12 \] ### Example 4 **Problem:** Consider the following arithmetic sequence: \( 2k-1, k+7, 5k-5 \). **Step 1:** Set the common differences equal: \[ (k + 7) - (2k - 1) = (5k - 5) - (k + 7) \] Simplifying both sides: \[ k + 7 - 2k + 1 = 5k - 5 - k - 7 \] \[ - k + 8 = 4k - 12 \] \[ 8 + 12 = 4k + k \implies 20 = 5k \implies k = 4 \] **Step 2:** Find the first three terms: \[ T_1 = 2(4) - 1 = 8 - 1 = 7 \] \[ T_2 = 4 + 7 = 11 \] \[ T_3 = 5(4) - 5 = 20 - 5 = 15 \] Thus, the first three terms are \( 7, 11, 15 \). **Step 3:** The general term \( T_n \): \[ T_n = a + (n-1)d \] Where \( a = 7 \) and \( d = 4 \): \[ T_n = 7 + (n-1) \cdot 4 \] \[ T_n = 7 + 4n - 4 = 4n + 3 \] ### Summary of Results - Example 2: \( T_n = 10n - 1 \) - Example 3 (1): First three terms are \( -2, 1, 4 \) - Example 3 (2): First term \( a = -12 \), common difference \( d = 5 \) - Example 4: \( k = 4 \), first three terms \( 7, 11, 15 \), general term \( T_n = 4n + 3 \)