Reciprocals of each other.

Let the first term and the common difference of the AP be $a$ and $d$ respectively. Then we have:

$\begin{cases} a_{11} = \dfrac{1}{13} \\ a_{13} = \dfrac{1}{11} = a_{11} + 2 d \quad \Rightarrow d = \dfrac {\frac{1}{11} - \frac{1}{13}}{2} = \dfrac{1} {143} \end{cases}$

Now $a_{11} = a + 10d = a + \dfrac{10}{143} = \dfrac{1}{13} = \dfrac{11}{143} \quad \Rightarrow a = \dfrac{1}{143}$

$S_{mn} = S_{143} = \dfrac {mn}{2} \left(2a + (mn-1)d \right) = \dfrac {143}{2} \left(\dfrac{2}{143} + \dfrac{142}{143} \right) = \boxed{72}$

We have $a+(m-1)d=\frac { 1 }{ n } ,\quad a+(n-1)d=\frac { 1 }{ m }$

Subtracting 2nd eqn from 1st we get

$\frac { 1 }{ n } -\frac { 1 }{ m } =(m-1)d-(n-1)d=(m-n)d=\frac { m-n }{ nm }$

$\Rightarrow \frac { 1 }{ mn } =d$

Therefore we have

$a+(m-1)\frac { 1 }{ mn } =\frac { 1 }{ n }$

$\Rightarrow a=\frac { 1 }{ mn }$

Now we have

${ S }_{ mn }=\frac { mn }{ 2 } \left( 2a+\left( mn-1 \right) d \right)$

${ S }_{ mn }=\frac { mn }{ 2 } \left( \frac { 2 }{ mn } +\left( mn-1 \right) \frac { 1 }{ mn } \quad \right)$

${ S }_{ mn }=1+\frac { mn-1 }{ mn } \times \frac { mn }{ 2 } =\frac { 1 }{ 2 } \left( mn+1 \right)$

Now evaluating $\frac { 1 }{ 2 } \left( mn+1 \right)$ ) at $m=13,n=11$ we get answer as $72$ .

We have $a+(m-1)d=\frac { 1 }{ n } ,\quad a+(n-1)d=\frac { 1 }{ m }$

Subtracting 2nd eqn from 1st we get

$\frac { 1 }{ n } -\frac { 1 }{ m } =(m-1)d-(n-1)d=(m-n)d=\frac { m-n }{ nm }$

$\Rightarrow \frac { 1 }{ mn } =d$

Therefore we have

$a+(m-1)\frac { 1 }{ mn } =\frac { 1 }{ n }$

$\Rightarrow a=\frac { 1 }{ mn }$

Now we have

${ S }_{ mn }=\frac { mn }{ 2 } \left( 2a+\left( mn-1 \right) d \right)$

${ S }_{ mn }=\frac { mn }{ 2 } \left( \frac { 2 }{ mn } +\left( mn-1 \right) \frac { 1 }{ mn } \quad \right)$

${ S }_{ mn }=1+\frac { mn-1 }{ mn } \times \frac { mn }{ 2 } =\frac { 1 }{ 2 } \left( mn+1 \right)$

Now evaluating $\frac { 1 }{ 2 } \left( mn+1 \right)$ ) at $m=13,n=11$ we get answer as $72$ .

$a_{13}=\frac 1{11}= \frac {13}{143}\\\\ a_{11}=\frac 1{13}=\frac {11}{143}\\\\ \Rightarrow a_n=\frac n{143}\\\\ \Rightarrow S_n=\frac 1{143} \cdot \frac {n(n+1)}2\\\\ S_{143}=\frac 1{143}\cdot \frac {143(144)}2=72.$

Notice that $143 = 11\cdot13$ , so

$a_{11} = \frac{1}{13} = \frac{11}{143}$ and $a_{13} = \frac{1}{11} = \frac{13}{143}$ .

From this it is easy to see that $d = \frac{1}{143}$ and $a_{143} = \frac{143}{143} = 1$

$S_{143} = \frac{143(\frac{1}{143} + 1)}{2} = \frac{144}{2} = 72$

We create a second arithmetic progression, $\left\{b_n\right\}$ , such that for all $i,\,b_i=143 a_i$ . Then $b_{11}=11 \text{ and } b_{13} = 13$ . So clearly $\left\{b_n\right\} = \mathbb{N}$ .

Let $B_k$ be the sum of the first $k$ terms of $\left\{b_n\right\}$ . Then $B_{143} = 1 + 2 + 3 + \dots + 143 = \dfrac{(143)(144)}{2}$ .

But clearly, for all $k,\,B_k = 143 S_k$ . Thus $S_k = \dfrac{144}{2} = \boxed{72}$ .