## College Algebra 7th Edition

The fifth term is: $a_5=1150$
RECALL: The $n^{th}$ $a_n$ of an arithmetic sequence is given by the formula: $a_n = a + d(n-1)$ where $a$ = first term $d$ = common difference The 100th term is $-750$ and the common difference is $-20$. Thus, $a_{100} = -750$ and the common difference is $d=-20$. Substitute these values to the formula above to obtain: $a_n = a + d(n-1) \\a_{100} = a + d(100-1) \\-750 = a + (-20)(99) \\-750 = a + (-1980) \\-750=a - 1980 \\-750+1980=a \\1230=a$ This means that the $n^{th}$ term $a_n$ of the sequence is given by the formula: $a_n=1230+(-20)(n-1)$ To solve for the fifth term, substitute $5$ to $n$ to obtain: $a_5 =1230 + (-20)(5-1) \\a_5=1230+(-20)(4) \\a_5=1230+(-80) \\a_5=1150$