The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a GP. Find the numbers.

Let a - d, a, a + d be three numbers in A.P. Since their sum is 21
∴    a - d + a + a + d = 21
                        3a = 21     a = 7
∴   Numbers are 7 - d, 7, 7 + d
It is given that if the second number is reduced by 1 and third is increased by 1 then resulting three numbers form G.P.
∴  7 - d, 7 - 1, 7 + d + 1 are in G.P. or 7 - d, 6, 8 + d are in G.P.

                        (8 + d) (7 - d) = 36
     56 - 8d + 7d - d² = 36               

                 d = 
When a = 7 and d = 4, numbers are 7 - 4,  7 + 4 or 3, 7, 11
when a = 7 and d = -5, numbers are 7 - (-5), 7, 7 - 5  or 12, 7, 2