If  log a, log b, and log c are in A.P and also loga−log2b,log2b−log3c,log3c−loga  are  in A.P .,then:

log  a,log b,log c   are in A.P ⇒2log b=loga+logc⇒logb2=log(ac) ⇒b2=ac⇒a,b,c  are   in   G.P. loga−log2b,log2b−log3c,log3c−loga    are   in    A.P ⇒2(log2b−log3c)=(loga−log2b)+(log3c−loga) ⇒3log2b=3log3c⇒2b=3c Now,b2=ac⇒b2=a.2b3⇒b=2a3,c=4a9 i.e.  a= a. b=2a3,c=4a9⇒a:b:c=1:23:49=9:6:4 since, sum of any two is greater than the 3rd,a,b,c, form a triangle.