# Find Nth element where difference between elements is in AP [closed]

Let I have a sequence as follows,

$$a_1,a_2,a_3,a_4,a_5...$$

where $(a_2-a_1),(a_3-a_2),(a_4-a_3),...$ are in arithmetic progression.

How can I find the Nth element ($a_N$) of the series $a_1,a_2,a_3,a_4,...,a_N,..$

## closed as off-topic by Tom-Tom, Shailesh, N. F. Taussig, Leucippus, Hans EnglerFeb 8 '16 at 4:07

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• It's a quadratic. Let $$a_n = xn^2 + yn + z$$ and choose any three elements of the sequence, solve the system of linear equations. – Zubin Mukerjee Feb 7 '16 at 21:17

Hint:

Set $\forall n\in\mathbb N,\,u_n=a_{n+1}-a_n$. $(u_n)_{n\in\mathbb N}$ is an AP so you know the general formula for $\sum_{k=1}^{N-1}u_k$. But this sum is a telescoping one so $\sum_{k=1}^{N-1}u_k=\sum_{k=1}^{N-1}(a_{k+1}-a_k)=a_N-a_1$.

Because the common differences in the sequence are equal, the relationship between $n$ and $a_{n}$ must be quadratic. For a quadratic, $3$ terms are enough to find the exact relation. We let $a_{1} = x + y + z,$ $a_{2} = 4x + 2y + z,$ and $a_{3} = 9x + 3y + z.$

We see that $3x + y = a_{2} - a_{1}$ and $5x + y = a_{3} - a_{2}.$ This leaves $2x = a_{1} + a_{3},$ so $x = \frac{1}{2}\left(a_{1} + a_{3}\right).$

Knowing $x$ and using the first of the equations found previously, we have $$\frac{3}{2}\left(a_{1} + a_{3}\right) + y = a_{2} - a_{1}$$ $$y = a_{2} - \frac{5}{2}a_{1} - \frac{3}{2}a_{3}.$$

Finally, plugging in $x$ and $y$ into the first of the main equations yields $$a_{1} = \frac{1}{2}\left(a_{1} + a_{3}\right) + a_{2} - \frac{5}{2}a_{1} - \frac{3}{2}a_{3} + z$$ $$z = -a_{1} - a_{2} + a_{3}.$$

The relation, using our findings, is $$\boxed{a_{n} = \frac{1}{2}\left(a_{1} + a_{3}\right)n^{2} + \left(a_{2} - \frac{5}{2}a_{1} - \frac{3}{2}a_{3}\right)n - a_{1} - a_{2} + a_{3}}.$$