Find ${\mathrm{a}}_1$ and ${\mathrm{S}}_{\mathrm{n}}$ for a geometric sequence with the values given below.
$$r=-\frac{1}{2},a_n=\frac{11}{32},n=6$$
$a_1=-11$ (Simplify your answer.)
${\mathrm{S}}_{\mathrm{n}}=◻$ (Simplify your answer.)

Step 1 ：Given that the common ratio $r=-\frac{1}{2}$, the nth term $a_n=\frac{11}{32}$, and $n=6$.

Step 2 ：We can find the first term $a_1$ of the geometric sequence using the formula $a_1=\frac{a_n}{r^{n-1}}$.

Step 3 ：Substituting the given values into the formula, we get $a_1=\frac{11/32}{(-1/2{)}^{6-1}}$.

Step 4 ：Solving the above equation, we find that $a_1=-11$.

Step 5 ：We can find the sum $S_n$ of the first n terms of the geometric sequence using the formula $S_n=\frac{a_1(1-r^n)}{1-r}$.

Step 6 ：Substituting the given values into the formula, we get $S_n=\frac{-11(1-(-1/2{)}^6)}{1-(-1/2)}$.

Step 7 ：Solving the above equation, we find that $S_n=-\frac{231}{32}$.

Step 8 ：So, the first term of the geometric sequence is $\boxed{{\displaystyle -11}}$ and the sum of the first 6 terms is $\boxed{{\displaystyle -\frac{231}{32}}}$.