If logea,logeb,logec are in an A.P. and logea− loge2b,loge2b−loge3c,loge3c−logea are also in an A.P, then a:b:c is equal to

The problem is asking for the ratio of a:b:c given that the logarithms of a, b, and c are in an arithmetic progression (A.P.) and that the differences of the logarithms of a, b, and c with respect to the logarithms of 2b, 3c, and a respectively are also in an A.P.

Step 1: Given that logea,logeb,logec are in an A.P., we can write this as: logeb = (logea + logec)/2

Step 2: We can simplify this using the properties of logarithms to get: b = sqrt(ac)

Step 3: Similarly, given that logea− loge2b,loge2b−loge3c,loge3c−logea are also in an A.P, we can write this as: loge2b - logea = loge3c - loge2b

Step 4: Simplifying this using the properties of logarithms gives us: 2b/a = 3c/2b

Step 5: Substituting b = sqrt(ac) from step 2 into the equation from step 4 gives us: 2sqrt(ac)/a = 3c/2sqrt(ac)

Step 6: Solving this equation for c gives us: c = 2sqrt(a)

Step 7: Substituting c = 2sqrt(a) into the equation b = sqrt(ac) from step 2 gives us: b = sqrt(a*2sqrt(a)) = sqrt(2a^2) = sqrt(2)*a

Step 8: Therefore, the ratio a:b:c is 1:sqrt(2):2.