Find the 98th term of the arithmetic sequence 14,10,6,ldots Answer Attemptiout of square

To find the 98th term of the arithmetic sequence, we first need to determine the common difference between consecutive terms.<br /><br />Given the sequence: $14, 10, 6, \ldots$<br /><br />The common difference, $d$, can be found by subtracting any term from the previous term:<br />$d = 10 - 14 = -4$<br /><br />Now, we can use the formula for the $n$th term of an arithmetic sequence:<br />$a_n = a_1 + (n-1)d$<br /><br />Where:<br />- $a_n$ is the $n$th term<br />- $a_1$ is the first term<br />- $n$ is the term number<br />- $d$ is the common difference<br /><br />Substituting the values into the formula:<br />$a_{98} = 14 + (98-1)(-4)$<br />$a_{98} = 14 + 97(-4)$<br />$a_{98} = 14 - 388$<br />$a_{98} = -374$<br /><br />Therefore, the 98th term of the arithmetic sequence is $\boxed{-374}$.