May I please get the full solution of question no. 8 please! It's urgent!

hi paridhi, lemme first point a slight mistake in the question. it should be “a. b, z are in HP” and not “1/a, 1/b, 1/z in HP” because that would imply a,b,z being in HP hence x= z resulting in the question making no sense whatsoever.

now, a,b, x are in AP

so, 2b=a+x

for GP

b^2=ay or 4ay= (a+x)^2=a^2+x^2+2ax

and for HP

2az=b(a+z) or 4az=(a+x)(a+z) or a^2+a(x – 3z)+xz=0

subtracting and rearranging, we get a=x(z – x)/(x+3z – 4y)

now, substituting this value of a in (a+x)^2=4ay, we get

4x(z – y)^2=y(z – x)(x+3z – 4y)

open all the brackets, and adding like terms, we get

yx^2+4zy^2+4xz^2=3z^2y+6xyz

or y(x^2+z^2 – 2zx)= 4z(yz+xy – xz – y^2)

or y(x – z)^2=4z(y – z)(x – y)

hence proved :)