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📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 $a + (a + 2d) = 14$ → $2a + 2d = 14$ → $a + d = 7$

📌 $a + d = 5$

📌 But $a + d = 7$ from (1) and $a + d = 5$ from (2) is a contradiction. Recheck: The second term is $a + d = 5$, and first + third is $a + (a + 2d) = 2a + 2d = 14$ → $a + d = 7$.

📌 But $a + d = 5$ and $a + d = 7$ is impossible. Therefore, no such sequence exists unless the problem is misinterpreted.

📌 Wait: The second term is $a + d = 5$, and $a + (a + 2d) = 14$ → $2a + 2d = 14$ → $a + d = 7$.

📌 Contradiction: $a + d = 5$ and $a + d = 7$ cannot both hold. Thus, no solution.

📌 But perhaps the problem meant the first term is $a$, third term is $a + 2d$, and their sum is 14, and second term is $a + d = 5$. Then:

📌 From $a + d = 5$, $a = 5 - d$. Substitute into $a + (a + 2d) = 14$:

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