please explain how to do so i can do others

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find the nth term of each arithmetic sequence described

a1=6, d=3, n=12

Aug 30th, 2015

Thank you for the opportunity to help you with your question!

Each term in an arithmetic sequence a[n] (which looks like a[1], a[2], a[3, ...) 
differs by a fixed amount d (in this case, d = 6). 

The problem states that a[1] = 6, so a[2] = a[1] + 3 = 9. Then a[3] = a[2] + 3 = 12. What is the pattern? THIS IS THE KEY.

a[1] = 6, a[2] = a[1]+3, 

a[2] = a[1]+3, and a[3] = a[2]+3.

Now a[3] = a[2] + 3 = (a[1] + 3) + 3, so a[3] = a[1] + 2 * 3.

Likewise, a[4] = a[3] + 3, so from the previous line, a[4] = (a[1] + 2 * 3) + 3 = a[1] + 3 * 3. 

In general, then, a[k] = a[1] + (k-1) * 3, so (plugging in n=12)

a[12] = a[1] + (12-1) * 3 = 6 + 11 * 3 = 6 + 33 = 39.  

You might also look at https://www.mathsisfun.com/algebra/sequences-sums-arithmetic.html Please let me know if you need any clarification. I'm always happy to answer your questions.
Aug 30th, 2015

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