## Intermediate Algebra for College Students (7th Edition)

$1727$
Here, $a_1=5, d=7$ Sum of $22$nd term of an arithmetic sequence can be calculated as: $S_n=\dfrac{n}{2}[2a_1+(n-1)d]$ Now, $S_{22}=\dfrac{22}{2}[2(5)+(22-1)7]=11[10+(21)7]=1727$