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Let $f(x)=a x^{2}+b x+c$ such that $f(1)=f(-1)$ and $a, b, c$ are in Arithmetic Progression.
What is the value of b?
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What is the value of b?
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The correct answer is:
0
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c} \\ & \therefore \mathrm{f}(1)=\mathrm{a}+\mathrm{b}+\mathrm{c} \\ & \text { and } \mathrm{f}(-1)=\mathrm{a}-\mathrm{b}+\mathrm{c} \\ & \therefore \mathrm{f}(1)=\mathrm{f}(-1) \\ & \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{a}-\mathrm{b}+\mathrm{c} \Rightarrow \mathrm{b}=0 \end{aligned}$
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