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The sum of the squares of three distinct...

The sum of the squares of three distinct real numbers, which are in GP, is `S^(2)`. If their sum is `a S`, then show that
`a^(2) in ((1)/(3), 1) uu (1,3)`

Text Solution

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Let three numbers in GP be `a, ar, ar^(2)`
`:. a^(2) + a^(2) r^(2) + a^(2) r^(4) = S^(2)`...(i)
and `a + ar + ar^(2) = a S`....(ii)
On dividing Eq. (i) by Eq. (ii) after squaring it, we get
`(a^(2) (1 + r^(2) + r^(4)))/(a^(2)(1 + r + r^(2))^(2)) = (S^(2))/(a^(2)S^(2))`
`rArr ((1 + r^(2))^(2) - r^(2))/((1 + r + r^(2))^(2)) = (1)/(a^(2))`
`rArr ((1 + r^(2) -r))/((1 + r^(2) + r)) = (1)/(a^(2))`
`rArr a^(2) = (r + (1)/(r) + 1)/(r + (1)/(r) - 1)`
Put `r + (1)/(r) = y`
`:. (y + 1)/(y -1) = a^(2)`
`rArr y + 1 = a^(2) y -a^(2)`
`rArr y = (a^(2) +1)/(a^(2) -1) " " [ :' |y| = |r + (1)/(r) gt 2]`
`rArr |a^(2) +1| gt |a^(2) -1|`
`rArr (a^(2) + 1)^(2) - {2(a^(2) -1)}^(2) gt0`
`rArr {(a^(2) + 1) -2 (a^(2) -1)} {(a^(2) +1) + 2(a^(2) -1)} gt 0`
`rArr(-1^(2) + 3) (3a^(2) -1) gt 0`
`:. (1)/(3) lt a^(2) lt 3`
`:. a^(2) in ((1)/(3), 1)uu(1,3) " " [ :' a^(2) != 1]`
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