# If $a-b=b-c$ the value of $a^2-2b^2+c^2$

If $a-b=b-c$ .How to find the value of $a^2-2b^2+c^2$

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Note that $a-b=b-c$ is equivalent to $c=2b-a$.

Substitute for $c$ in $a^2-2b^2+c^2$. We get $a^2-2b^2+(2b-a)^2$.

Expand the square. We get $a^2-2b^2+(4b^2-4ab+a^2)$.

This simplifies to $2a^2-4ab+2b^2$, which simplifies to $2(a-b)^2$.

Further simplification is not possible, since we have only one fact (equation), so cannot expect to eliminate more than one variable.

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Thanks ,but if i was given choices$(120,68,112,45,128)$. How is this going to help which one is correct ? – user61454 Feb 8 '13 at 6:30
If these are the only choices, then the answer is $128$, since it is the only choice that is twice a perfect square. Of course, all the others are valid choices, since $120=2(\sqrt{60})^2$. But if $a$ and $b$ are integers, the only possible answer is $128$. You did not give the full problem in your post. This slowed down the process o getting an answer. – André Nicolas Feb 8 '13 at 6:37
Please note I left out the word "twice" in my first writing the comment above, and only caught it on edit. – André Nicolas Feb 8 '13 at 6:39
Thanks "André Nicolas"! now i understand how to eliminate the other answers .I was supposed to put the the full problem in the post . – user61454 Feb 8 '13 at 6:49

\begin{align} a-b&=b-c\\ a+c&=2b\\ (a+c)^2&=(2b)^2\\ \end{align}

\begin{align}a^2+c^2+2ac&=4b^2\\a^2+c^2-2b^2&=2b^2-2ac\\\end{align}

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$a-b=b-c$ means $b=(a+c)/2$ therefore $a^2-2b^2+c^2=a^2+c^2-2(\frac{a+c}{2})^2=\frac{2a^2+2c^2-(a+c)^2}{2}=\frac{a^2+c^2-2ac}{2}=\frac{(a-c)^2}{2}$

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$a-b=b-c\implies a,b,c$ are in A.P.

$a^2-2b^2+c^2=(a^2-b^2)-(b^2-c^2)=(a-b)(a+b)-(b-c)(c+b)=(a-b)(a+b)-(a-b)(c+b)=(a-b)(a+b-c-b)=(a-b)(a-c)$

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