The sum of three numbers which are consecutive terms of an A.P. is 21.If the second number is reduced by 1 while third number is increased by 1,three consecutive terms of a G.P. result.Find numbers.

let three no. AP are 

 a-d , a , a+d

then their sum = a-d +a+ a+d = 21

 3a =21

 a=7

 now acc to ques . no. becomes

 a-d , a-1 , a+d+1

i.e 7-d, 6 , 8 +d are in gp

therefore 6/ 7-d = 8+d/6

  36= (7-d)(8+d)

  36= 56 + 7d -8d-d²  

d² +d -20=0 --->  d² + 5d - 4d- 20=0  --->

d(d+5 ) -4(d+5)=0

d=4,-5

hence no. are 3,7,11 & 12,7,2