Since $a_i=2x_i-x_{i-1}+m$, this means $2^{i-1}a_i=2^ix_i-2^{i-1}x_{i-1}+2^{i-1}m$, so adding for $i=1$ to $i=n$ gives $a_1+2a_2+4a_3+\cdots+2^{n-1}a_n=(2^n-1)x_n+(2^n-1)m$, so you can solve for $x_n$.
Since $a_i=2x_i-x_{i-1}+m$, this means $2^{i-1}a_i=2^ix_i-2^{i-1}x_{i-1}+2^{i-1}m$, so adding for $i=1$ to $i=n$ gives $a_1+2a_2+4a_3+\cdots+2^{n-1}a_n=(2^n-1)x_n+(2^n-1)m$, so you can solve for $x_n$.