Question

The sum of three numbers in G.P is 56 . if we subtract 1,7,21 from these numbers in this order we obtain an A.P .the numbers are

Answer 1

let the middle number is x
common ratio is r
then numbers in GP will be x/r, x, xr

we have x/r + x + xr = 56  ................................(1)

now as per ques x/r-1, x-7, xr-21 are in AP

=> (x-7) - (x/r - 1) = (xr-21) - (x-7)

=>      -xr + 2x - x/r = -8  ..............................(2)

from (1) and (2) , we get 
      
                 x = 16

substituting in (1), 

16/r + 16 + 16r = 56 
=> 1/r +1+r = 56/16
 => (1+r^2)/r = 40/16 = 5/2
=> 2r^2 - 5r + 2 = 0 

solving we get

r = 1/2 and 2

for x = 16 & r = 2 
we get the numbers as 4, 8, 16 

for x = 16 & r = 1/2

we get the required number 32, 16, 8 which add up to 56.

hence the answer is 32, 16, 8