Mantel attack on handsome boys: Sega continues

Actually, this problem is flawed. If $M = N$ , it can get infinitely many answers for $MN$ th term. The problem should have the condition that $M \neq N$ .

Let $x_{i} = a + d\times i$ for any $a,d,i$ ( $i$ is positive number.)

We want to find $x_{MN} = a + d\times MN$ $(*)$

Setting $i = M$ and $i = N$ we get

$\displaystyle \frac{1}{N} = a + d\times M$ (1)

$\displaystyle \frac{1}{M} = a + d\times N$ (2)

Subtract both equations we get $\displaystyle \frac{1}{N} - \frac{1}{M} = d(M-N)$

$\displaystyle (M-N)(d - \frac{1}{MN}) = 0$

We have $M = N$ or $\displaystyle d = \frac{1}{MN}$ .

Case 1: $M = N$ (it's flawed here)

$\displaystyle \frac{1}{N} = a + d\times N$

$1 = a + d\times MN$

Case 2: $\displaystyle d = \frac{1}{MN}$

Substitute into (1) or (2) we get $a = 0$ .

Which leads to $\displaystyle x_{MN} = 0 + \frac{MN}{MN} = \boxed{1}$ from $(*)$