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If $a, b, c$ are in A.P., then $\frac{(a-c)^2}{\left(b^2-a c\right)}=$
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Verified Answer
The correct answer is:
$4$
If $a, b, c$ are in A.P. $\Rightarrow 2 b=a+c$
$\frac{(a-c)^2}{\left(b^2-a c\right)}$ $=\frac{(a-c)^2}{\left\{\left(\frac{a+c}{2}\right)^2-a c\right\}}$
so, $=\frac{(a-c)^2 4}{\left[a^2+c^2+2 a c-4 a c\right]}$ $=\frac{4(a-c)^2}{(a-c)^2}=4$
Trick : Put $a=1, b=2, c=3$, then the required value is $\frac{4}{1}=4$
$\frac{(a-c)^2}{\left(b^2-a c\right)}$ $=\frac{(a-c)^2}{\left\{\left(\frac{a+c}{2}\right)^2-a c\right\}}$
so, $=\frac{(a-c)^2 4}{\left[a^2+c^2+2 a c-4 a c\right]}$ $=\frac{4(a-c)^2}{(a-c)^2}=4$
Trick : Put $a=1, b=2, c=3$, then the required value is $\frac{4}{1}=4$
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