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![\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=1100\\ d=35\\ a_n=3725 \end{cases} \\\\\\ 3725=1100+(n-1)35\implies 3725=1100+35n-35 \\\\\\ 3725=1065+35n\implies 2660=35n\implies \cfrac{2660}{35}=n\implies 76=n](/tpl/images/0071/9833/a4c9b.png)
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How many terms are in the sequence: 1100, 1135, ?...
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