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For any sequence of real numbers $A=\{a_1,a_2,a_3\dots\}$, define $ΔA$ to be the sequence $\{a_2-a_1,a_3-a_2,a_4-a_3,\dots\}$ whose $n$th term is $a_{n+1}-a_n$. Suppose that all of the terms of the sequence $Δ(ΔA)$ are 1, and that $a_{19}=a_{94}=0$. Find $a_1$.

I'm seeing that the first series must be $\{1,2,4,7,\dots\}$ which is strictly increasing, but the series has two zeros.

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  • $\begingroup$ Tried anything? $\endgroup$ – Parcly Taxel Sep 14 '16 at 12:38
  • $\begingroup$ yeh, used some nth term tried to manipulate it didnt get far, then worked that the first series must be like this; 1, 2,4,7..., but if it is like that i dont understand how a19 and a94 can be zero, as the first series must be increasing? $\endgroup$ – grigori Sep 14 '16 at 12:42
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We have $$\Delta(\Delta A)=\{1,1,1,\dots\}$$ $$\Delta A=\{b,b+1,b+2,\dots\}$$ $$A=\{c,c+b,c+2b+1,c+3b+3,\dots\}$$ For a fixed $b$ and $c$ we have $$a_n=c+(n-1)b+\frac{(n-1)(n-2)}2$$ The condition that $a_{19}=a_{94}=0$ translates to $$a_{19}=c+18b+153=0$$ $$a_{94}=c+93b+4278=0$$ Subtract the first equation from the second: $$75b+4125=0;\ b=-55$$ $$c=-153-18b=-153-18(-55)=837$$ Hence $a_1=c=837$.

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  • $\begingroup$ how did you work out that nth term is (n-1)(n-2)/2?, i know its easy to work out from looking at the sequence 1,3,6,10... but is there a more systematic approach? $\endgroup$ – grigori Sep 15 '16 at 10:55
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    $\begingroup$ @grigori In $\Delta A$ the constant part of the expression (after the $b$) goes 0, 1, 2, 3 and so on. Hence the constant part in $A$ must go 0, 0+0, 0+0+1, 0+1+2 and so on, and the constant in $a_n$ is the sum of all natural numbers from 1 to $n-2$. A formal derivation of the triangular numbers formula can be found on Wikipedia. $\endgroup$ – Parcly Taxel Sep 15 '16 at 12:09

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