## College Algebra (10th Edition)

$a_1=-3$; common difference $= 2$; recursive formula: $a_n=2+a_{n-1};$ formula for the $n^{th}$ term: $a_n=-3+2(n-1)$
RECALL: (1) To find the next term of an arithmetic sequence, the common difference is added to the current term. This means that to find the 4th term if the 2nd term is known, the common difference must be added twice to the 2nd term. (2) The nth term of an arithmetic is given by the formula: $a_n = a_1 + d(n-1)$ where $a_1$ = first term and $d$ = common difference (3) The recursive formula for an arithmetic sequence is: $a_n = d+ a{n-1}$ $\bf{\text{Solve for$d$}}:$ The given arithmetic sequence has: $a_4=3$ and $a_{20} = 35$ Note that from $a_4$, the common difference must be added 16 times to get the value of $a_{20}$ Thus, $a_{20} = a_4 + 16(d)$ Substitute the values of $a_4$ and $a_{20}$ to obtain: $a_{20}=a_4+16d \\35=3+16d \\35-3 = 16d \\32=16d \\\frac{32}{16} = d \\2=d$ $\bf{\text{Solve for the first term:}}$ Note that from $a_1$, the common difference must be added 3 times to find the value of $a_4$. Thus, $a_4=a_1 + 3d$ Substitute the values of $d$ and $a_4$ to obtain: $a_4=a_1+3d \\3=a_1+3(2) \\3=a_1+6 \\3-6=a_1 \\-3=a_1$ $\bf{\text{Write the formulas:}}$ The given sequence has $a_1=-3$ and $d=2$. Therefore: recursive formula: $a_n=2+a_{n-1}; a_1=-3$ formula for the $n^{th}$ term: $a_n=-3+2(n-1)$