The sum of three numbers which are consecutive terms 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 G.P. Find the numbers.
Let the first term of an A.P. is a and its common difference be d.
Now, according to the questiong :
a,a+d−1 and a+2d+1 are in G.P.
Now, putting a = 2, 12 in equation (i), we get d = 5, -5, respectively.
Thus, for a = 2 and d = 5, the numbers are 2, 7 and 12.
And, for a = 12 and d = -5, the numbers are 12, 7 and 2.