Question
If $(b-c)^{2},(c-a)^{2},(a-b)^{2}$ are in A.P. then prove that $\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}$ are also in A.P.
Step 1
P. This implies that the differences of consecutive terms are equal. Therefore, we have: \[(c-a)^{2} - (b-c)^{2} = (a-b)^{2} - (c-a)^{2}\] Show more…
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