If `a , b , c ,` are in `A P ,a^2,b^2,c^2` are in HP, then prove that either `a=b=c` or `a , b ,-c/2` from a GP (2003, 4M)

Explanation

Given, `a,b,c` are in AP.
`therefore " " b=(a+c)/(2) " " "……(i)`
and `a^(2),b^(2),c^(2)` in HP.
`therefore " " b^(2)=(2a^(2)c^(2))/(a^(2)+c^(2)) " " "………(ii)`
From Eq. (ii) `b^(2){(a+c)^(2)-2ac}=2a^(2)c^(2)`
` implies b^(2){(2b)^(2)-2ac}=2a^(2)c^(2) " " [" from Eq. (i)"]`
`implies 2b^(4)-acb^(2)-a^(2)c^(2)=0`
`implies (2b^(2)+ac)(b^(2)-ac)=0`
`implies 2b^(2)+ac =0` or `b^(2)-ac=0`
If `2b^(2)+ac=0`, than `b^(2)=-(1)/(2)ac` or `-(a)/(2),b,c` are in GP
and if `b^(2)-ac =0 implies a,b,c` are in GP.
But given, `a,b,c` are in AP.
Which is possible only when `a=b=c`.