-The second difference is 10 - The first two terms are equa \( T_{1}=T_{2} \) \( T_{1}+T_{2}+T_{3}=28 \)

Let's extract the known conditions from the problem: 1. The second difference of a sequence is 10. 2. The first two terms are equal: \( T_1 = T_2 \). 3. The sum of the first three terms is 28: \( T_1 + T_2 + T_3 = 28 \). ### Step 1: Define the terms Since \( T_1 = T_2 \), we can denote both terms as \( T_1 = T_2 = a \). ### Step 2: Express the third term From the sum of the first three terms, we have: \[ T_1 + T_2 + T_3 = a + a + T_3 = 2a + T_3 = 28 \] This simplifies to: \[ T_3 = 28 - 2a \] ### Step 3: Understand the second difference The second difference being 10 implies that the sequence is quadratic. The second difference of a quadratic sequence is constant. Let’s denote the first term as \( T_1 = a \), the second term as \( T_2 = a \), and the third term as \( T_3 = 28 - 2a \). The first differences can be expressed as: - \( D_1 = T_2 - T_1 = a - a = 0 \) - \( D_2 = T_3 - T_2 = (28 - 2a) - a = 28 - 3a \) The second difference is: \[ \Delta D = D_2 - D_1 = (28 - 3a) - 0 = 28 - 3a \] Setting this equal to the known second difference of 10: \[ 28 - 3a = 10 \] ### Step 4: Solve for \( a \) Now, we can solve for \( a \): \[ 28 - 3a = 10 \\ -3a = 10 - 28 \\ -3a = -18 \\ a = 6 \] ### Step 5: Find \( T_3 \) Now that we have \( a \), we can find \( T_3 \): \[ T_3 = 28 - 2a = 28 - 2(6) = 28 - 12 = 16 \] ### Final Terms Thus, the terms are: - \( T_1 = 6 \) - \( T_2 = 6 \) - \( T_3 = 16 \) ### Conclusion The first three terms of the sequence are \( T_1 = 6 \), \( T_2 = 6 \), and \( T_3 = 16 \).