Writo's Arithmetic Progression

$\dfrac{1}{d}\left( \dfrac{a_{2} - a_{1}}{a_{2}a_{1}} + \dfrac{a_{3} - a_{2}}{a_{3}a_{2}} + \cdots + \dfrac{a_{4001} - a_{4000}}{a_{4001}a_{4000}}\right) = 10$

$= \dfrac{1}{d}\left( \dfrac{1}{a_{1}} - \dfrac{1}{a_{2}} + \dfrac{1}{a_{2}} - \dfrac{1}{a_{3}} + \cdots + \dfrac{1}{a_{4000}} - \dfrac{1}{a_{4001}}\right) = 10$

$= \dfrac{1}{d}\left( \dfrac{1}{a_{1}} - \dfrac{1}{a_{4001}}\right) = 10$

$= \dfrac{1}{d}\left( \dfrac{a_{4001} - a_{1}}{a_{1}a_{4001}}\right) = 10$

$= \dfrac{4000}{a_{1}a_{4001}} = 10$ ( $as~a_{4001} = a_{1} + 4000d$ )

$a_{1}a_{4001} = 400$

$a_{2} + a_{4000} = 50$

$\implies (a_{1} + d) + (a_{1} + 3999d) = 50$

$\implies a_{1} + a_{4001} = 50$

$(a_{1} - a_{4001})^2 = (a_{1} + a_{4001})^2 - 4a_{1}a_{4001}$

$|a_{1} - a_{4001}| = 30$

The answer is (+30) or (-30)

Every term $\frac{1}{a_{n}a_{n + 1}}$ can be written as $\frac{1}{r} * (\frac{1}{a_{n}} - \frac{1}{a_{n+1}})$ , where $r$ is the common difference of the progression. Thus, we can rewrite the sum as:

$\frac{1}{r}*(\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + \ldots + \frac{1}{a_{4000}} - \frac{1}{a_{4001}}) = 10$

It follows that: $\frac{1}{a_{1}} - \frac{1}{a_{4001}} = 10r$

$\frac{4000r}{a_{1}*a_{4001}} = 10r$

$a_{1}*(a_{1} + 4000r) = 400$ (Equation $I$ )

Now, from the condition $a_{2} + a_{4000} = 50$ , we derive the following:

$2a_{1} + 4000r = 50$

$4000r = 50 - 2a_{1}$ (Equation $II$ )

Plugging the result of $II$ back in $I$ , we get:

$a_{1}*(50 - a_{1}) = 400$

$a_{1}^{2} - 50a_{1} + 400 = 0$

Solving for $a_{1}$ yields: $a_{1} = 40$ or $a_{1} = 10$

If $a_{1} = 40$ , then $4000r = -30$ and so $a_{4001} = 10$

If $a_{1} = 10$ , then $4000r = 30$ and so $a_{4001} = 40$

In either case, $|a_{1} - a_{4001}| = 30$ .

Let $N=4000$ as a shorthand. We have an arithmetic progression $a_k=a_1+(k-1)d$ and need to calculate $|x|:=|a_1-a_{N+1}|=N|d|$ . The first equation yields a telescope sum:

$\begin{aligned} 10&=\sum_{k=1}^{N}\frac{1}{a_ka_{k+1}}=\sum_{k=1}^{N}\frac{1}{(a_1+(k-1)d)(a_1+kd)}\underset{\text{part. frac.}}{=}\sum_{k=1}^{N}\frac{\frac{1}{d}}{a_1+(k-1)d}-\frac{\frac{1}{d}}{a_1+kd}\\ &=\frac{1}{d}\sum_{k=1}^N\frac{1}{a_k}-\frac{1}{d}\sum_{k=2}^{N+1}\frac{1}{a_k}=\frac{1}{da_1}-\frac{1}{da_{N+1}}=\frac{1}{da_1}-\frac{1}{a_1+Nd}=\frac{N}{a_1(a_1+Nd)}\\ \Rightarrow\quad\frac{N}{10}&=a_1(a_1+Nd) \end{aligned}$

The second equation yields

$\begin{aligned} 50&=a_2+a_{N}=2a_1+Nd\quad\Rightarrow\quad a_1=25-\frac{Nd}{2} \end{aligned}$

Inserting the second in the first equation and using $|x|=N|d|$ , we get

$\begin{aligned} \frac{N}{10}&=\left(25-\frac{Nd}{2}\right)\left(25+\frac{Nd}{2}\right)\underset{\text{3. Binomi}}{=}625-\frac{x^2}{4}&\Rightarrow&&|x|=\sqrt{2500-\frac{2N}{5}}\underset{N=4000}{=}\fbox{30} \end{aligned}$