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

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Question

(d) If $a,\;b,\;c$ are in $A.P.$ $ \Rightarrow $$2b = a + c$

So, $\frac{{{{(a - c)}^2}}}{{({b^2} - ac)}} = \frac{{{{(a - c)}^2}}}{{\left\{ {{{\left

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(d) If $a,\;b,\;c$ are in $A.P.$ $ \Rightarrow $$2b = a + c$

So, $\frac{{{{(a - c)}^2}}}{{({b^2} - ac)}} = \frac{{{{(a - c)}^2}}}{{\left\{ {{{\left( {\frac{{a + c}}{2}} \right)}^2} - ac} \right\}}}$

$ = \frac{{{{(a - c)}^2}4}}{{[{a^2} + {c^2} + 2ac - 4ac]}} = \frac{{4{{(a - c)}^2}}}{{{{(a - c)}^2}}} = 4$.

Trick : Put $a = 1,\;b = 2,\;c = 3$,

then the required value is $\frac{4}{1} = 4$.

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

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