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In the sequence {$a_n$}, $$a_n=\frac{a_1+a_2+...+a_{n-1}}{n-1} \, , \quad n \ge 3$$ If $a_1+a_2 \ne 0$ and the sum of the first $N$ terms is $12(a_1+a_2)$, find $N$.

Kind of lost on where to start with this one. My initial thought was, $$\sum^{N}_{n=3}\frac{a_1+a_2+...+a_{n-1}}{n-1}=12(a_1+a_2)$$ $$\sum^{N}_{n=3}\frac{a_1+a_2+...+a_{n-1}}{n-1}=\frac{1}{2}(a_1+a_2)+\frac{1}{3}(a_1+a_2+a_3)...$$but someting seems wrong here, or I do not know where to go from here.

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    Hint: Look at writing the first few terms in terms of $a_1+a_2$ and see if you can find a pattern. For instance, clearly $a_3=\frac{a_1+a_2}{2}$; now $a_4=\frac13(a_1+a_2+a_3) =$...? – 2017-02-17
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    I see that in the first few terms but I dont see how to associate it with $12(a_1+a_2)$ – 2017-02-17

2 Answers 2

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\begin{align*} S_2 &= a_1+a_2 \\ S_n &= a_1+a_2+\ldots+a_n \\ a_{n+1} &= \frac{a_1+\ldots+a_{n}}{n} \\ &= \frac{S_n}{n} \\ S_{n+1} &= S_{n}+\frac{S_n}{n} \\ &= \frac{n+1}{n} S_{n} \\ &= \frac{n+1}{n} \cdot \frac{n}{n-1} \cdot \ldots \cdot \frac{4}{3} \cdot \frac{3}{2} S_{2} \\ &= \frac{n+1}{2} S_2 \\ S_{N} &= \frac{N}{2} S_2 \\ &= \frac{N(a_1+a_2)}{2} \\ \end{align*}

Also $\forall n\ge 3$, $$a_{n}=\frac{a_1+a_2}{2}$$

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    N is supposed to have an actual integral value that I am supposed to be able to find from the information I have been given. – 2017-02-17
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    Solve $\dfrac{N}{2}=12$ – 2017-02-17
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    Oh! I see! Thank you so much! – 2017-02-17
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    elegant approach ! – 2017-02-17
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    Wouldn't $S_3=a_1+a_2$? and $S_n=a_1+a_2+...+a_{n-1}$? or am I mistaken? – 2017-02-17
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    $$S_3=a_1+a_2+a_3=a_1+a_2+\frac{a_1+a_2}{2}=\frac{3}{2}(a_1+a_2)$$ – 2017-02-17
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$a_1$
$a_2$
$a_3=\dfrac{a_1+a_2}2$
$a_4=\dfrac {a_1+a_2+(a_1+a_2)}2$
$a_4=\dfrac{a_1+a_2}2$

Similarly, $a_3=a_4=a_5=a_6=\ldots$

Sum up to $n$ numbers is $a_1+a_2+a_3+a_4+\cdots+a_n$

$s=a_1+a_2+(n-2)\cdot a_3$

$s=\dfrac n2(a_1+a_2)$

So $n=12\cdot2=24$

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    Please use [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) in your answers. – 2017-02-17