Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Mixed Review - Page 825: 51

$a_n=17.5-1.5n$

Here, we have $a_n= a_1+d(n-1)$ for the Arithmetic series. The common difference between the successive terms is $d=-1.5$ and $a_9=4$ Here, we have $a_9= a_1+(9-1)d$ or, $a_9=a_1-8d \implies a_1=4-8(-1.5)=16$ Now, $a_n=16+(-1.5) \times (n-1)=16-1.5n+1.5$ Hence, $a_n=17.5-1.5n$