Arithmetic Progression with a Sum Property Question I had a question in a exam, and I had no idea how to do it.
The question is:
Let $(a_k)_{k\geq1}$ a sequence of not null real numbers such that:
$$\displaystyle \sum_{k=1}^{n-1}\frac{1}{a_k\cdot a_{k+1}}=\frac{n-1}{a_1\cdot a_n} \, \forall \,n\geq 2$$
Show that this sequence is an arithmetic progression.

I know that to show that this is an arithmetic progression, I have to show that $a_n-a_{n-1}$ is constant. But I don't know how to do it with the sum hypothesis.
Appreciate your help.

Let $r = a_2 - a_1$. To prove that $(a_n)$ is arithmetic is equivalent to prove that:
$$\forall n \geq 1, a_n = (n - 1) r + a_1$$
Let's proceed by strong induction.

*

* It's clear, it's true for $n = 1$.

* Assume it's true for $k \in \{1, \ldots, n - 1\}$ then :
$$\begin{array}{lcl}
\displaystyle \dfrac{n - 1}{a_1 a_n} & = & \displaystyle \sum_{k = 1}^{n - 1} \dfrac{1}{a_k a_{k + 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \sum_{k = 1}^{n - 2} \dfrac{1}{a_k a_{k + 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \sum_{k = 1}^{n - 2} \dfrac{a_{k + 1} - a_k}{a_k a_{k + 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \sum_{k = 1}^{n - 2} \left(\dfrac{1}{a_k} - \dfrac{1}{a_{k + 1}}\right) \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \left(\dfrac{1}{a_1} - \dfrac{1}{a_{n - 1}}\right) \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \dfrac{a_{n - 1} - a_1}{a_1 a_{n - 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \dfrac{(n - 2) r + a_1 - a_1}{a_1 a_{n - 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{n - 2}{a_1 a_{n - 1}}
\end{array}$$
We deduce that :
$$\begin{array}{lcl}
\dfrac{n - 2}{a_1 a_{n - 1}} & = & \dfrac{1}{a_n} \left(\dfrac{n - 1}{a_1} - \dfrac{1}{a_{n - 1}}\right) \\[3mm]
& = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) a_{n - 1} - a_1}{a_1 a_{n - 1}}\right) \\[3mm]
& = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) ((n - 2) r + a_1) - a_1}{a_1 a_{n - 1}}\right) \\[3mm]
& = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) (n - 2) r + (n - 2) a_1}{a_1 a_{n - 1}}\right) \\[3mm]
& = & \dfrac{n - 2}{a_n} \left(\dfrac{(n - 1) r + a_1}{a_1 a_{n - 1}}\right) \\[3mm]
\end{array}$$ 
then :
$$\dfrac{(n - 1) r + a_1}{a_n} = 1$$
so $a_n = (n - 1) r + a_1$. 


By induction : 
$$\forall n \geq 1,a_n = (n - 1) r + a_1$$ 
so $(a_n)$ is arithmetic.
