If a, b and c are forming increasing terms of G.P., r is the common ratio then find the minimum value of (c-b), given that (log a+log b+log c)/log 6 = 6.Note that r can be any real no.	Choose the most appropriate option:
A) 23
B) 16
C) 17
D) 12
E) 62
Explanation:   a,b,c are in G.P. so let the first term of G.P. = arar, and common ratio = r.   Therefore, a = arar, b = aa, c = arar   Given, loga+logb+logclog6=6log⁡a+log⁡b+log⁡clog⁡6=6   ⇒logabclog6=6⇒log⁡abclog⁡6=6   ⇒log6abc=6⇒abc=66⇒log6abc=6⇒abc=66   put the value of a,b,c in gp format   ⇒ar×a×ar=66⇒ar×a×ar=66   ⇒a3=66⇒a=36⇒a3=66⇒a=36   Now a = 36r36r, b = 36, c = 36r.   We have to find the minimum value of c - b = 36r - 36.   r can be any number. So for r &lt; 0, we get c - b negative.   When r = 1, c - b = 0   But none of the options are not representing it.   From the given options, r = 4/3, then c = 48. So option d satisfies this.   Answer: 12
<result>D</result>
If a, b and c are forming increasing terms of G.P., r is the common ratio then find the minimum value of (c-b), given that (log a+log b+log c)/log 6 = 6.Note that r can be any real no.