As a,ba,b and cc are in GP, we have b^2=acb2=ac.
Further as a,b,ca,b,c are distinct numbers, a!=b!=c!=aa≠b≠c≠a
If a+xa+x, b+xb+x and c+xc+x are in HP, then
1/(a+x),1/(b+x)1a+x,1b+x and 1/(c+x)1c+x are in AP and therefore
1/(c+x)-1/(b+x)=1/(b+x)-1/(a+x)1c+x−1b+x=1b+x−1a+x
i.e. (b+x-c-x)/((c+x)(b+x))=(a+x-b-x)/((b+x)(a+x))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)b−cbc+bx+cx+x2=a−bba+bx+ax+x2
or (b-c)(ba+bx+ax+x^2)=(a-b)(bc+bx+cx+x^2)(b−c)(ba+bx+ax+x2)=(a−b)(bc+bx+cx+x2)
or b^2a+b^2x+abx+bx^2-abc-bcx-acx-cx^2=abc+abx+acx+ax^2-b^2c-b^2x-bcx-bx^2b2a+b2x+abx+bx2−abc−bcx−acx−cx2=abc+abx+acx+ax2−b2c−b2x−bcx−bx2
or b^2a+2b^2x+2bx^2-2abc-2acx-cx^2-ax^2+b^2c=0b2a+2b2x+2bx2−2abc−2acx−cx2−ax2+b2c=0
or x^2(2b-c-a)+x(2b^2-2ac)+b^2a-2abc+b^2c=0x2(2b−c−a)+x(2b2−2ac)+b2a−2abc+b2c=0
As b^2=acb2=ac, this becomes x^2(2b-c-a)+b^2(a-2b+c)=0,x2(2b−c−a)+b2(a−2b+c)=0,
or x^2(2b-c-a)-b^2(2b-c-a)=0,x2(2b−c−a)−b2(2b−c−a)=0,
rArr (2b-c-a)(x^2-b^2)=0.⇒(2b−c−a)(x2−b2)=0.
rArr {(b-c)+(b-a)}}(x-b)(x+b)=0.⇒{(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.