Question

if a, b, c are in gp then prove that:1/(a²-b²) +1/(b²) =1/(b²-c²) ​

Answer 1

Answer:

Step-by-step explanation:

Let the nos.be a, ar and ar²

Therefore, LHS=1/(a²-b²) + 1/b²=1/(a²-a²r²) + 1/a²r²

= 1/a²r²(1-r²)

RHS=1/(b²-c²)=1/(a²r²-a²r⁴)=1/a²r²(1-r²)

Therefore, LHS=RHS

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Answer 2

Answer:

Given a, b, c are in GP

let 'r' be common ratio of this sequence

So b/a = c/b = r

Therefore we can write that

b = ar and c=br=ar²

L.H.S.=1/(a²-b²) + 1/b²

=1/(a²-a²r²) + 1/a²r²

=1/a²r²(1-r²)

=1/(a²r²-a²r4)

=1/((ar)²-(ar²)²)

=1/(b²-c²) = R.H.S.