Question

- In the arithmatic sequence 3,5,7,9, 11, ..., how many terms appear after the term 315 but before the term 639?

Answer 1

We have the following:

`a=3,5,7,9,11\ldots`

Which means that it increases by 2 for each term that advances. Therefore we must subtract 639 and 315, we know the number of whole numbers that exist between the two terms and then we divide by 2, since this is how it increases

`\begin{gathered} d=(639-315)/(2) \\ d=(324)/(2) \\ d=162 \end{gathered}`

that is to say that among those numbers there are a total of 162 terms