Three numbers a, b, c, non-zero, form an arithmetic progression. Increasing a by 1 or increasing c by 2 results in a geometric progression. Then b equals:
Explanation:
a, b, c form an AP. ∴ 2b = a + c Increasing a by 1 or c by 2 results in a GP
∴ b2 = (a + 1)c …(1) and b2 = a(c + 2) …(2)
∴ (a + 1)c = a(c + 2) ∴ ac + c = ac + 2a ∴ c = 2a
Now, 2b = a + c ∴ 2b = a + 2a ∴ b = 3a/2
Putting this in (1), we get (9a2)/4 = (a + 1)2a ∴ 9a/4 = 2a + 2 ∴ 9a = 8a + 8 ∴ a = 8
∴ b = 3a/2 = (3×8)/2 = 12
Hence, option (c).
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