the sum of three numbers in GP is 56. if we subtract 1, 7, 21 from these numbers in that order, we obtain an AP. find the numbers.
Answers
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