(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 loga x = \(\frac{1}{log_x\,a}\bigg]\)