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.#
