The general term for an Arithmetic Series is given by $\rm t_{n} = a + (n - 1)d$, where the symbols have their usual meanings.

The sum of the first n terms of an Arithmetic Series is given by $\rm S_{n} = \frac{n}{2} \cdot (2a + (n- 1) d)$.

Given,

$\rm t_{7} = 70$

$\rm or, a + (7 - 1) d = 70$

$\rm or, a + 6d = 70$

To find: $\rm S_{11} = ?$

$\rm or, S_{13} = \frac{13}{2} \cdot \left( 2 \cdot a + (13 - 1) d \right )$

$\rm = \frac{13}{2} \left ( 2a + 12 d \right )$

$\rm = \frac{13}{2} \left ( 2 (a + 6d ) \right )$

$\rm = \frac{13}{2} \cdot 2 \cdot (a + 6d)$

$\rm = 13 \cdot (a + 6d)$

From above, we have $\rm a + 6d = 70$. On substituting the value of this expression, we get,

$\rm = 13 \cdot 70$

$\rm \therefore S_{13} = 910$

Hence, the required sum of the first 13 terms of the given Arithmetic Series is 910.