In general, a geometric sequence can be expressed as shown below
[tex]\begin{gathered} a_n=a_1(r)^{n-1} \\ a_n\rightarrow\text{ n-th term} \\ r\rightarrow\text{ common ratio} \end{gathered}[/tex]
Thus, in our case, given that r=-2 and the 7th term is equal to 11,
[tex]\begin{gathered} a_7=11 \\ \Rightarrow11=a_1(-2)^{7-1} \\ \Rightarrow11=a_1(-2)^6 \\ \Rightarrow a_1=\frac{11}{(-2)^6}=\frac{11}{64} \\ \Rightarrow a_1=\frac{11}{64} \end{gathered}[/tex]
Then,
[tex]\Rightarrow a_n=\frac{11}{64}(-2)^{n-1}[/tex]
Set n=10,
[tex]\begin{gathered} \Rightarrow a_{10}=\frac{11}{64}(-2)^{10-1}=\frac{11}{64}(-2)^9=\frac{11}{64}(-512)=-88 \\ \Rightarrow a_{10}=-88 \end{gathered}[/tex]
Hence, the 10th term of the sequence is equal to -88.
