Question

if a, b, c are in GP show that 1/a^2 , 1/b^2, 1/c^2 are in GP​

Answer 1

Step-by-step explanation:

b^2=ac

therefore,

LHS

=(1/b^2)^2

=1/b^4

while,

RHS

=1/((a^2)(c^2))

=1/((b^2)^2)

=1/b^4

LHS=RHS

HENCE PROVED

this is the most easiest way that can be used in boards

Answer 2

Answer:

we know that,

b^2 = ac

squaring on both sides we get

(b^2)^2 = (ac)^2

1/(ac) ^2 = 1/(b^2) ^2

(1/a ^2).(1/c ^2) = (1/b^2) ^2

assume that

1/a^2 = x, 1/b^2 = y and 1/c^2 = z

then,

x z = y^2

Hence we prove that

1/a^2 , 1/b^2, 1/c^2 are in GP

Hope that it is helpful to you

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