If $$1b$$−c,$$1c$$−a,$$1a$$−bbe consecutive terms of an AP, then $$(b$$−$$c)2$$,$$(c$$−$$a)2$$,$$(a$$−$$b)2$$ will be in

Question

If $$1b$$−c,$$1c$$−a,$$1a$$−bbe consecutive terms of an AP, then $$(b$$−$$c)2$$,$$(c$$−$$a)2$$,$$(a$$−$$b)2$$ will be in

Infinity Learn · Accepted Answer

Now,  we assume (b$$−$$c)2$$,$$(c$$−$$a)2$$,$$(a$$−$$b)2$$ are in AP, then we $$have(c$$−$$a)2$$−$$(b$$−$$c)2=(a$$−$$b)2$$−$$(c$$−$$a)2$$⇒ $$(b$$−$$a)(2c$$−a−$$b)=(c$$−$$b)(2a$$−b−$$c)    ......(i)Also$$,if  $$1b$$−c,$$1c$$−a,$$1a$$−b are in,AP, $$then1c$$−a−$$1b$$−$$c=1a$$−b−$$1c$$−$$ab+a$$−$$2c(c$$−$$a)(b$$−$$c)=c+b$$−$$2a(a$$−$$b)(c$$−$$a)⇒ (a$$−$$b)(b+a$$−$$2c)=(b$$−$$c)(c+b$$−$$2a)⇒ (b$$−$$a)(2c$$−a−$$b)=(c$$−$$b)(2a$$−b−$$c)which$$ is equal to Eq.$$(i), so our hypothesis is true.

If $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ be consecutive terms of an AP, then $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ will be in

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If 1b−c,1c−a,1a−bbe consecutive terms of an AP, then (b−c)2,(c−a)2,(a−b)2 will be in

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If $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ be consecutive terms of an AP, then $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ will be in