So to solve this problem we suppose that the 3 consecutive nos are n, n+1, n+2. These numbers are a,ar^p and ar^q respectively (i.e. we have assumed that the 3 consecutive numbers can form terms of G.P.

 

So now we see that (n+1)/n = r^p and (n+2)/n = r^q .

 

                       ((n+1)/n)^q = ((n+2)/n)^p = r^pq

 

                       (n+1)^q = (n+2)^p. n^(q-p)                 ... (1)

 

 Now we claim that no two consecutive integers can divide each other except that one is 1 or -1.

 

Let us see why?

The two numbers be n and n+1.

So if n divdes n+1 then n+1/n is integer i.e 1/n is integer which is possible only if n is 1 or -1.

 

n cannot be -1 because then n+1 will become 0 and zero cannot be term of a GP.

 

Also if n = 1 then eq 1 becomes

 

                  2^q   = 3^p which is impossible because 2 and 3 are primes and they have no factor in common. So how can two numbers which have no common factors be equal.

 

Again returning to eq 1. Now we have this fact that n+1 cannot divide n+2 as well as n.

 

So now the LHS of eq1 when divided by n+1 will give an integer whereas the RHS will not and thus eq 1 cannot hold.

 

This clearly means that three consecutive numbers cannot form 3 terms of G.P.

 

I hope I have cleared your doubt ........ even if doubt persist write back.