The sum of two consecutive terms of the sequence: \( 21 ; 35 ; 49 \); is 308. Determine the two terms.

To find the two consecutive terms of the sequence \( 21, 35, 49 \) whose sum is \( 308 \), we first represent the two consecutive terms as \( a_n \) and \( a_{n+1} \). In this sequence, the terms can be described using the first term and the common difference. The common difference here is \( 14 \). Notice the terms can be expressed as: - \( a_1 = 21 \) - \( a_2 = 35 = a_1 + 14 \) - \( a_3 = 49 = a_2 + 14 \) Now let's suppose we need to find terms beyond \( 49 \). The general term of the sequence can be described as: - \( a_n = 21 + (n - 1) \cdot 14 \) Setting up the equation for the sum: \[ a_n + a_{n+1} = 308 \] This gives: \[ (21 + (n-1) \cdot 14) + (21 + n \cdot 14) = 308 \] Simplifying: \[ 42 + (2n - 1) \cdot 14 = 308 \] \[ (2n - 1) \cdot 14 = 266 \] \[ 2n - 1 = \frac{266}{14} \] \[ 2n - 1 = 19 \] \[ 2n = 20 \implies n = 10 \] Now substituting back to find the terms: \[ a_{10} = 21 + (10 - 1) \cdot 14 = 21 + 126 = 147 \] \[ a_{11} = 21 + (10) \cdot 14 = 21 + 140 = 161 \] Thus, the two terms are \( 147 \) and \( 161 \), which sum up to \( 308 \).