If a, b, c are in A.P.,then frac{(a - c)^2}{( | vedclass

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(D) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $b = \frac{a + c}{2}$.Substituting this into the expression $\frac{(a - c)^2

Vedclass · Accepted Answer

$4$

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(D) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $b = \frac{a + c}{2}$.Substituting this into the expression $\frac{(a - c)^2}{(b^2 - ac)}$:$\frac{(a - c)^2}{(\frac{a + c}{2})^2 - ac} = \frac{(a - c)^2}{\frac{a^2 + c^2 + 2ac}{4} - ac}$$= \frac{(a - c)^2}{\frac{a^2 + c^2 + 2ac - 4ac}{4}} = \frac{4(a - c)^2}{a^2 + c^2 - 2ac}$$= \frac{4(a - c)^2}{(a - c)^2} = 4$.Alternatively,let $a = 1, b = 2, c = 3$. These are in $A.P.$The expression becomes $\frac{(1 - 3)^2}{(2^2 - 1 	imes 3)} = \frac{(-2)^2}{4 - 3} = \frac{4}{1} = 4$.

If $a, b, c$ are in $A.P.$,then $\frac{(a - c)^2}{(b^2 - ac)} = $

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