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for any sequence of real number $A= \{ a_1, a_2, a_3,\dots\}$ a sequence $\Delta A$ is defined as $\Delta A=\{a_2 - a_1, a_3 - a_2,\dots\}$ suppose that all terms of sequence $\Delta(\Delta A)$ are $1$ and that $a_{19}=a_{92}=0$ then find the digit at unit place of $a_3$ .
I got the following equations but can't understand how to proceed further.
$$a_{21} - 2 a_{20}= 1$$ $$a_{17} - 2 a_{18}= 1$$ $$a_{18} + a_{20}= 1$$

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    $\begingroup$ please read this page immediately. $\endgroup$ – MAN-MADE Aug 3 '17 at 2:50
  • $\begingroup$ This might work: let $a_1$ be $x$, let $a_2$ be $y$, then find a formula for $a_3$ in terms of $x$ and $y$, then a formula for $a_4$, then $a_5$, keep going until you see a pattern, prove that the pattern you have found really does persist throughout the sequence. Then you'll be able to use the information about $a_{19}$ and $a_{92}$ to find $x$ and $y$ and $a_3$. $\endgroup$ – Gerry Myerson Aug 3 '17 at 3:48
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Using a little theory, if the second difference is constant, the sequence is quadratic. If the constant is 1, the coefficient of the $n^2$-term is $1/2$. If $a_{19}=a_{92}=0$, then the quadratic has the factors $n-19$ and $n-92$. Now just put it all together.

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