Arithmetic Sequence II

Problem 221: Find the $12^{th}$ term in this sequence: $8, 14, 20, 26, 32, \dots, :$

Solution: The $n^{th}$ term of an arithmetic sequence is given by

(1) $\begin{equation*} a_n = a_0 + d(n-1)\end{equation*}$

where $a_0$ is the first term in the sequence and $d$ is the common difference. To find the common difference, we just take the difference between the terms.

(2) $\begin{align*} 14 - 8 & =6 \\20 - 14 & = 6 \\ 26 - 20 & = 6 \\ 32-26 & = 6 \end{align*}$

This means that $d = 6$ . We are looking for $a_{12}$ . That is,

(3) $\begin{align*}a_n & = a_0 + d(n-1) \\ a_{12} & = 8 + 6(12-1) \\ & = 8 + 6(11) \\ & = 8 + 66 \\ & = 74. \end{align*}$

Thus $a_{12}= 74$ .

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