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Grade: 11


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

2 years ago

Answers : (1)

Aditya Gupta
2069 Points
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
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 :)
2 years ago
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