As a,b and c are in GP, we have b^2=ac.
Further as a,b,c are distinct numbers, a!=b!=c!=a
If a+x, b+x and c+x are in HP, then
1/(a+x),1/(b+x) and 1/(c+x) are in AP and therefore
1/(c+x)-1/(b+x)=1/(b+x)-1/(a+x)
i.e. (b+x-c-x)/((c+x)(b+x))=(a+x-b-x)/((b+x)(a+x))
i.e. (b-c)/(bc+bx+cx+x^2)=(a-b)/(ba+bx+ax+x^2)
or (b-c)(ba+bx+ax+x^2)=(a-b)(bc+bx+cx+x^2)
or b^2a+b^2x+abx+bx^2-abc-bcx-acx-cx^2=abc+abx+acx+ax^2-b^2c-b^2x-bcx-bx^2
or b^2a+2b^2x+2bx^2-2abc-2acx-cx^2-ax^2+b^2c=0
or x^2(2b-c-a)+x(2b^2-2ac)+b^2a-2abc+b^2c=0
As b^2=ac, this becomes x^2(2b-c-a)+b^2(a-2b+c)=0,
or x^2(2b-c-a)-b^2(2b-c-a)=0,
rArr (2b-c-a)(x^2-b^2)=0.
rArr {(b-c)+(b-a)}}(x-b)(x+b)=0.
Since, anebnec, (b-c)ne0,(b-a)ne0, :.(b-c)+(c-a)ne0.
Also, x+bne0.
Clearly, x=b.