There are 10 terms in an arithmetic sequence. The sum of the terms in the even positions \( \left(a_{2}, a_{4}\right. \), etc.) is 25 , while the sum of the terms in the odd positions ( \( a_{1}, a_{3} \), etc.) is 10 . Find the \( 7^{\text {th }} \) term of this arithmetic sequence.

Let the first term of the arithmetic sequence be \( a \) and the common difference be \( d \). The terms in the sequence can be represented as: - \( a_1 = a \) - \( a_2 = a + d \) - \( a_3 = a + 2d \) - \( a_4 = a + 3d \) - \( a_5 = a + 4d \) - \( a_6 = a + 5d \) - \( a_7 = a + 6d \) - \( a_8 = a + 7d \) - \( a_9 = a + 8d \) - \( a_{10} = a + 9d \) The sum of the terms in even positions is: \[ a_2 + a_4 + a_6 + a_8 + a_{10} = (a + d) + (a + 3d) + (a + 5d) + (a + 7d) + (a + 9d) \] This simplifies to: \[ 5a + (1 + 3 + 5 + 7 + 9)d = 5a + 25d \] According to the problem, this sum is 25: \[ 5a + 25d = 25 \quad \text{(1)} \] The sum of the terms in odd positions is: \[ a_1 + a_3 + a_5 + a_7 + a_9 = a + (a + 2d) + (a + 4d) + (a + 6d) + (a + 8d) \] This simplifies to: \[ 5a + (0 + 2 + 4 + 6 + 8)d = 5a + 20d \] According to the problem, this sum is 10: \[ 5a + 20d = 10 \quad \text{(2)} \] Now we have a system of equations: 1) \( 5a + 25d = 25 \) 2) \( 5a + 20d = 10 \) We can subtract equation (2) from equation (1): \[ (5a + 25d) - (5a + 20d) = 25 - 10 \] This simplifies to: \[ 5d = 15 \] Thus, we get: \[ d = 3 \] Now substituting the value of \( d \) back into equation (2): \[ 5a + 20(3) = 10 \] That becomes: \[ 5a + 60 = 10 \] Solving for \( a \): \[ 5a = 10 - 60 \] \[ 5a = -50 \] \[ a = -10 \] Now we have \( a = -10 \) and \( d = 3 \). We can find the \( 7^{\text{th}} \) term \( a_7 \): \[ a_7 = a + 6d = -10 + 6(3) = -10 + 18 = 8 \] Thus, the \( 7^{\text{th}} \) term of the arithmetic sequence is \( \boxed{8} \).