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Home » We know $ an 60^\circ = \sqrt3 $, so: - Silent Sales Machine

We know $ an 60^\circ = \sqrt3 $, so: - Silent Sales Machine

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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Solution: The terms are $a$, $a + d$, $a + 2d$, $a + 3d$, $a + 4d$.
$a + (a + 2d) = 14$ → $2a + 2d = 14$ → $a + d = 7$
$a + d = 5$
But $a + d = 7$ from (1) and $a + d = 5$ from (2) is a contradiction....
But $a + d = 5$ and $a + d = 7$ is impossible. Therefore, no such sequ...
Wait: The second term is $a + d = 5$, and $a + (a + 2d) = 14$ → $2a +...
Contradiction: $a + d = 5$ and $a + d = 7$ cannot both hold. Thus, no...
But perhaps the problem meant the first term is $a$, third term is $a...
From $a + d = 5$, $a = 5 - d$. Substitute into $a + (a + 2d) = 14$:
$(5 - d) + (5 - d + 2d) = 14$ → $5 - d + 5 + d = 10 = 14$ → $10 = 14$,...

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