The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Explanation

Let the three numbers be a , ar, `ar^(2)`
Given , ` a + ar + ar^(2) = 21 ` or ` a (1 + r + r^(2)) = 21]` …(i)
and ` a^(2) + a^(2) r^(2) + a^(2) r^(4) = 189 ` or ` a^(2) (1 + r^(2) + r^(4) ) = 189` …(ii)
Squaring (i) and dividing (ii) by it , we get
`rArr (a^(2)(1 + r^(2) + r^(4)))/(a^(2)(1 + r+ r^(2))^(2)) = (189)/(441) `" "`rArr (1 + 2r^(2) + r^(4) - r^(2))/(( 1 + r + r^(2))^(2)) = (3)/(2)`
`rArr ((1 + r^(2)) - r^(2))/(1 + r + r^(2))^(2) = (3)/(7) ` " "`rArr (1+r^(2) -r)/(1 + r +r^(2)) = (3)/(7)`
`rArr 7 (1 + r^(2) - r ) = 3 (1 + r + r^(2)) `" " `rArr 4r^(2) - 10r + 4 = 0 `
`rArr 2r^(2) - 5r + 2 = 0 ` " "`rArr 2r^(2) - 4r - r + 2 = 0`
`rArr 2r (r - 2) - 1 (r - 2) = 0 ` " "`rArr (2r -1) (r-2) = 0 `
`therefore r = (1)/(2) , 2 `
Putting ` r = (1)/(2) ` in (i) , we get a `(21)/(1+(1)/(2) + (1)/(4) )=(21xx4)/(4 + 2 + 1) = 12 `
` therefore ar = 6 , ar^(2) = 3 `
Putting ` r = 2 ` in (i) , we get a ` (12)/(1 + 2 + 4) = 3 `
` therefore ar = 6, ar^(2) = 12 `
Hence , the three numbers are 3, 6 and 12 .