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1.	If log (a + c), log (c – a) and log (a – 2b + c) are in A.P, then(a) a, b, c are in A.P. (b) a2, b2, c2 are in A.P. (c) a, b, c are in G.P. (d) a, b, c are in H.P.
Answer» (d) a, b, c are in H.P. log (a + c), log (c – a) and log (a – 2b + c) are in A.P. ⇒ 2 log (c – a) = log (a + c) + log (a – 2b + c) ⇒ log (c – a)² = log [(a + c) (a – 2b + c)] ⇒ c² + a² – 2ac = a² + ca – 2ba – 2bc + ac + c² ⇒ 2ab + 2bc = 4ac ⇒ b (a + c) = 2ac ⇒ b = \(\frac{2ac}{a+c}\) ⇒ a, b, c are in H.P.

If log (a + c), log (c – a) and log (a – 2b + c) are in A.P, then(a) a, b, c are in A.P. (b) a2, b2, c2 are in A.P. (c) a, b, c are in G.P. (d) a, b, c are in H.P.

Answer»

(d) a, b, c are in H.P.

log (a + c), log (c – a) and log (a – 2b + c) are in A.P.

⇒ 2 log (c – a) = log (a + c) + log (a – 2b + c)

⇒ log (c – a)² = log [(a + c) (a – 2b + c)]

⇒ c² + a² – 2ac = a² + ca – 2ba – 2bc + ac + c²

⇒ 2ab + 2bc = 4ac ⇒ b (a + c) = 2ac

⇒ b = \(\frac{2ac}{a+c}\) ⇒ a, b, c are in H.P.

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