If three unequel numbers are in HP and their squares are in AP, show that they are in the ratio `1+sqrt(3):-2:1-sqrt(3) " or " 1-sqrt(3):-2:1+sqrt(3)`.

Explanation

Let a,b,c (unequal number) are in HP.
`:. b =(2ac)/(a+c)`
`implies (b)/(2)=(ac)/(a+c)=lambda " " [" say "]`
`implies b =2lambda" and " ac=lambda (a+c)" " "……..(i)"`
Now, `a^(2)b^(2)c^(2)` are in AP
So, `b^(2)=(a^(2)c^(2))/(2) implies 2b^(2)=a^(2)+c^(2)`
`implies 2(2lambda)^(2)=(a+c)^(2)-2ac`
`implies =(a+c)^(2)-2lambda(a+c)-8lambda^(2)=0`
`implies =(a+c-4lambda)(a+c+2lambda)=0`
`implies a+c=4lambda" or " a+c=-2lambda`
Case I If `a+c=4lambda`
`:. ac=4lambda^(2)" " [" fromEq.(i) "]`
`implies (a-c)^(2)=(a+c)^(2)-4ac`
`implies (a-c)^(2)=16lambda^(2)-16lambda^(2)`
`implies (a-c)^(2)=0 implies a=c`
Let given that a,b,c are distinct, so `a+c=4lambda` is not valid.
Case II If `a+c=-2lambda`
`implies ac=-2lambda^(2)" " [" fromEq. (i) "]`
`:.(a-c)^(2)=(a+c)^(2)-4ac`
`implies (a-c)^(2)=4lambda^(2)+8lambda^(2) implies (a-c)=pm2sqrt(3lambda)" " ".......(ii)"`
If `a-c=2sqrt(3lambda)" " ".......(iii)'`
then `a+c=2lambda`
From Eqs. (ii) and (iii), we get
`a=(sqrt(3-1))lambda " and "c=-(1+sqrt(3))lambda)`
`:.a:b:c=(sqrt(3-1))lambda:2lambda:-(sqrt(3)+1)lambda`
`a:b:c=(sqrt(3-1)):2:-(sqrt(3)+1)`
`impliesa:b:c=(1-sqrt(3)):-2:(sqrt(3)+1)`
If `a-c=-2sqrt(3lambda)" " "........(iv)"`
then `a+c=-2lambda " " ".......(v)"`
From Eqs. (iv) and (v), we get
`a=-(sqrt(3)+1)lambda" and " c=(sqrt(3)-1)lambda`
`:.a:b:c=-(sqrt(3)+1)lambda:2lambda:(sqrt(3)-1)lambda`
`impliesa:b:c=(1+sqrt(3)):-2:(1-sqrt(3))`.