Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The first four terms are,${{a}_{1}}=1,{{a}_{2}}=11,{{a}_{3}}=21\text{ and }{{a}_{4}}=31$ and the $8\text{th}$ term is ${{a}_{8}}=71$, and the $12\text{th}$ term is ${{a}_{12}}=111$.
${{a}_{n}}=10n-9$ …… (1) To find the value of the first term, ${{a}_{1}}$ put $n=1$ in equation (1) \begin{align} & {{a}_{n}}=10n-9 \\ & {{a}_{1}}=10\cdot 1-9 \\ & =10-9 \\ & =1 \end{align} For the second term, ${{a}_{2}}$ put $n=2$ in equation (1) \begin{align} & {{a}_{n}}=10n-9 \\ & {{a}_{2}}=10\cdot 2-9 \\ & =20-9 \\ & =11 \end{align} For the third term, ${{a}_{3}}$ put $n=3$ in equation (1) \begin{align} & {{a}_{n}}=10n-9 \\ & {{a}_{3}}=10\cdot 3-9 \\ & =30-9 \\ & =21 \end{align} For the fourth term, ${{a}_{4}}$ put $n=4$ in equation (1) \begin{align} & {{a}_{n}}=10n-9 \\ & {{a}_{4}}=10\times 4-9 \\ & =40-9 \\ & =31 \end{align} For the $8\text{th}$ term, ${{a}_{8}}$ put $n=8$ in equation (1) \begin{align} & {{a}_{n}}=10n-9 \\ & {{a}_{8}}=10\times 8-9 \\ & =80-9 \\ & =71 \end{align} For the $\text{12th}$ term, ${{a}_{12}}$ put $n=12$ in equation (1) \begin{align} & {{a}_{n}}=10n-9 \\ & {{a}_{12}}=10\times 12-9 \\ & =120-9 \\ & =111 \end{align}