The sum of three numbers m GP is 56. If we subtract 1.7,21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Answer

N/a

Explanation

Let 3 numbers in G.P. are `a, ar, ar^(2)`.
Given that, `a + ar + ar^(2) = 56`
`rArr` `a(1 + r + r^(2)) = 56" "` …(1)
and `a-1, ar - 7, ar^(2) - 21` are in A.P.
`rArr` `2(ar - 7) = (a -1) + (ar^(2) - 21)`
`rArr` `a(1 + r^(2) - 2r) = 8`
`rArr (56)/(1 + r + r^(2)) (1 + r^(2) - 2r) = 8` [from equation (1)]
`rArr` `7(1 + r^(2) - 2r) = 1 + r + r^(2)`
`rArr` `6r^(2) - 15r + 6 = 0`
`rArr` `2r^(2) - 5r + 2 = 0 " " rArr " " (r - 2)(2r - 1) = 0`
`rArr` `r = 2 or r = (1)/(2)`
If r = 2, then `a = (56)/(1 + 2 + 2^(2)) = 8`
and numbers = 8, 16, 32.
If `r = (1)/(2)`, then `a = (56)/(1 + (1)/(2) + (1)/(2^(2))) = 31`
and numbers = 32, 16, 8