Question

log₃ 2, log₆ 2, log₁₂ 2 are in

(a) A.P. 

(b) G.P. 

(c) H.P. 

(d) None of these

Answer 1

(c) H.P.

Since log₂ 3 + log₂ 12 = log₂ (3 x 12) = log₂ 36 = log₂ 6² 

= 2 log₂ 6, therefore, 

log₂ 3, log₂ 6 and log₂ 12 in A.P

⇒ \(\frac{1}{log_2\,3}\), \(\frac{1}{log_2\,6}\), \(\frac{1}{log_2\,12}\) and in H.P.

⇒ log₃ 2, log₆ 2, log₁₂ 2 and in H.P \(\bigg[\)Using log_a x = \(\frac{1}{log_x\,a}\bigg]\)