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,

$\rm t_{6} = 64$

$\rm or, a + (6 - 1) d = 64$

$\rm or, a + 5d = 64$

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

$\rm or, S_{11} = \frac{11}{2} \cdot \left( 2 \cdot a + (11 - 1) d \right )$

$\rm = \frac{11}{2} \left ( 2a + 10 d \right )$

$\rm = \frac{11}{2} \left ( 2 (a + 5d ) \right )$

$\rm = \frac{11}{2} \cdot 2 \cdot (a + 5d)$

From above, we have $\rm a + 5d = 64$. On substituting the value of this expression, we get,

$\rm = 11 \cdot 64$

$\rm \therefore S_{11} = 704$

Hence, the required sum of the first 11 terms of the given Arithmetic Series is 704.