If a, b and c are increasing terms of a G.P., with r as the common ratio, find…

Concept: For a Geometric Progression (GP), if the middle term is b, the two neighbouring terms can always be written as b/r and b·r for some common ratio r. This makes the product of three consecutive GP terms always equal to the cube of the middle term — a × b × c = b3 — no matter what r is.

Also, a sum of logarithms such as (log a + log b + log c) collapses to log(abc) by the product rule, and dividing by log 6 converts the condition into a base-6 logarithm equation, log6(abc) = 6.

Applying this to the question:

1. Rewrite the given condition: (log a + log b + log c)/log 6 = log6(abc) = 6, so abc = 66.

2. Since a, b, c are consecutive GP terms with common ratio r, write them as x/r, x, xr; their product is x3.

3. So x3 = 66. Since 66 = (62)3, x = 62 = 36. Hence b = 36, a = 36/r, c = 36r.

4. So (c − b) = 36r − 36 = 36(r − 1).

5. Because a, b, c are positive (needed for the logs to exist) and strictly increasing, r must be greater than 1 — so (c − b) = 36(r − 1) is always positive.

6. Important caveat: the question places NO restriction on r other than r > 1 ("r can be any real number"), so r can be taken arbitrarily close to 1, driving (c − b) arbitrarily close to 0. Strictly speaking, (c − b) therefore has no attained minimum over the reals — its infimum is 0, never reached. As an MCQ, the question intends the minimum among the four values actually offered as choices, i.e. which of 36, 24, 18, 12 is the smallest value that (c − b) can equal for some valid r > 1.

Cross-check: At r = 4/3: a = 36 × 3/4 = 27, b = 36, c = 36 × 4/3 = 48 — the terms 27, 36, 48 form an increasing GP, and log6(27 × 36 × 48) = log6(66) = 6, confirming the given condition; (c − b) = 48 − 36 = 12. Each of the other three offered values (36, 24, 18) is also attainable at r = 2, 5/3, 3/2 respectively, but 12 is the smallest of the four.

Result: So, among the four choices given, the minimum value of (c − b) is 12 (keyed answer). Note this is the smallest of the offered values, not a rigorously attained global minimum, since (c − b) can in principle be made smaller still for r closer to 1.