Chapter 11 - Further Topics in Algebra - 11.2 Arithmetic Sequences and Series - 11.2 Exercises - Page 1024: 27

$a_n=\frac{9}{2}n-39$ $a_8=-3$

$a_{10}=6, a_{11}=10.5$ Then $d=a_{11}-a_{10}=10.5-6=4.5$ Determine $a_1$: $a_n=a_1+(n-1)d$ $a_{10}=a_1+(10-1)d$ $6=a_1+9*4.5$ $6=a_1+40.5$ $a_1=-34.5$ $a_n=a_1+(n-1)d$ $a_n=-34.5+(n-1)*4.5$ $a_n=-34.5+4.5n-4.5$ $a_n=4.5n-39$ $\boxed{a_n=\frac{9}{2}n-39}$ $a_8=\frac{9}{2}*8-39$ $a_8=36-39$ $\boxed{a_8=-3}$