Let three numbers in GP are a , ar , ar².
Given, a + ar + ar² = 56
a(1 + r + r²) = 56 --------------------(1)
A/C to question,
(a - 1) , (ar - 7), ( ar² - 21) are in AP
so, common difference is always constant.
e.g (ar² - 21) - (ar - 7) = (ar - 7) - (a - 1)
ar² - ar - 21 + 7 = ar - a - 7 + 1
ar² - 2ar + a - 8 = 0
a( r² - 2r + 1) = 8 --------------------(2)
dividing equation (1) by equation (2) , we get
a( 1 + r + r²)/a(r² - 2r + 1) = 56/8
(1 + r + r²)/(1 - 2r + r²) = 7
1 + r + r² = 7 - 14r + 7r²
6r² - 15r + 6 = 0
2r² - 5r + 2 = 0
2r² - 4r - r + 2 = 0
2r( r - 2) -(r - 2) = 0
(2r - 1)(r - 2) = 0
r = 2, 1/2
if r = 2 then, from equation (1)
a + 2a + 4a = 56
7a = 56
a = 8
Then, numbers are a = 8
ar = 8 × 2 = 16
ar² = 8 × 2² = 32
hence, 8 , 16 , 32
similarly if r = 1/2 then from equation (1)
a + a/2 + a/4 = 56
7a/4 = 56
a = 32
then, numbers are 32 , 32 × 1/2 , 32 × 1/2²
32 , 16 , 8