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.) $\mathrm{S}_{\mathrm{n}}=\square$ (Simplify your answer.)
Solution
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 \(\boxed{-11}\) and the sum of the first 6 terms is \(\boxed{-\frac{231}{32}}\).
