MCQ
If $a,\;b,\;c$ are in $A.P.$, then $\frac{{{{(a - c)}^2}}}{{({b^2} - ac)}} = $
- A$1$
- B$2$
- C$3$
- ✓$4$
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 $\frac{4}{1} = 4$.
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$(A)$ $a+b=4$
$(B)$ $a-b=2$
$(C)$ The length of the diagonal $P R$ of the parallelogram $P Q R S$ is $4$
$(D)$ $\overrightarrow{ w }$ is an angle bisector of the vectors $\overrightarrow{ PQ }$ and $\overrightarrow{ PS }$