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Mathematics 7 Online
OpenStudy (anonymous):

Geometric Sequence help anyone? List the next four terms of a geometric sequence with t1 = 4 and tn = -3tn-1. A. 12, 36, 108, 324 B. 12, 36, 48, 60 C. -12, 36, -108, 324 D. -12, 36, -48, 60 *If you'd rather walk me through the steps versus giving me the answer I'd appreciate that also!

jimthompson5910 (jim_thompson5910):

"tn = -3tn-1" is really written \[\Large t_{n} = -3t_{n-1}\] I'm assuming

OpenStudy (anonymous):

I'm still confused on how I go about solving this problem

jimthompson5910 (jim_thompson5910):

If that assumption is correct, it means that the nth term \(\Large t_{n}\) is found by multiplying the previous term \(\Large t_{n-1}\) by -3

jimthompson5910 (jim_thompson5910):

t1 is the first term t1 = 4 so the first term is 4

jimthompson5910 (jim_thompson5910):

t2 is the second term you find t2 by multiplying the first term (t1) by -3 because of that rule that defines the nth term

jimthompson5910 (jim_thompson5910):

so what is t2?

jimthompson5910 (jim_thompson5910):

In other words, \[\Large t_{2} = -3*t_{1}\] what is \(\Large t_{2}\) ?

OpenStudy (anonymous):

what is the value of t alone again? 4?

jimthompson5910 (jim_thompson5910):

t has no value but when we say t1, t2, t3, etc etc...we're talking about the first term, second term, third term, etc etc...

jimthompson5910 (jim_thompson5910):

initially, we are given \[\Large t_{1} = 4\] this is the first term

OpenStudy (anonymous):

How do I find the second term? I'm sorry, it's been a long day my brain is working slow

jimthompson5910 (jim_thompson5910):

you multiply -3 by the previous term to get the next term

jimthompson5910 (jim_thompson5910):

that's exactly what this rule means \[\Large t_{n} = -3t_{n-1}\]

jimthompson5910 (jim_thompson5910):

\(\Large t_{n}\) is the nth term \(\Large t_{n-1}\) is the term just before the nth term (the previous term of the nth term)

OpenStudy (anonymous):

Oh okay, that makes more sense!

OpenStudy (anonymous):

I think I got it from here, thank you so much for your help!

jimthompson5910 (jim_thompson5910):

you're welcome

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