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If (b-c)^2,(c-a)^2,(a-b)^2 are in A.P., ...

If `(b-c)^2,(c-a)^2,(a-b)^2` are in A.P., then prove that `1/(b-c),1/(c-a),1/(a-b)` are also in A.P.

Text Solution

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`1/(c-a)-1/(b-c)=1/(a-b)-1/(c-a)`
`((a+b-2ac))/(b-c)=((c+b-2a))/(a-b)`
or (a-b)(a+b-2c)=(b-c)(b+c-2a) (1)
Above is true by given condition as shown below by (2). We are given
`(c-a)^(2)-(b-c)^(2)=(a-b)^(2)-(c-a)^(2)`
or (c-a+b-c)(c-a-b+c)
=(a-b+c-a)(a-b-c+a)
or (b-a)(2c-a-b)=(c-b)(2a-b-c)
or (b-c)(b+c-2a)=(a-b)(a+b-2c)
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