Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Mixed Review - Page 825: 49

$a_n=-7n+41=41-7n$

Here, we have $a_n= a_1+d(n-1)$ for the Arithmetic series. The common difference between the successive terms is $d=6$ Here, we have $a_7= a_1-7d$ or, $-8=a_1-7d \implies a_1=34$ Now, $a_n=34+(6) \times (n-1)=198-18n+18$ Hence, $a_n=-7n+41=41-7n$