anonymous
  • anonymous
The sequence 1,4,7,10....34 has 12 terms. Evaluated the related series
Algebra
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Hi bro I heard u need help on Damon algebra
anonymous
  • anonymous
First, you want to examine a pattern. I will use \(t_n\) to denote the "\(n\)"th term (Such that \(t_1=1, t_2=4, t_3=7...\) One may notice that the difference between each term is the same. That is, to say,: \[t_4-t_3=t_3-t_2=t_2-t_1...\] So we notice that the common difference, \(d\) is \(3\) This means that the sequence is a arithmetic series. The formula for a arithmetic sequence is: \(t_n=a_0+(n-1)d\), where \(a_0\) is the starting term and \(d\) is the common difference. So since with this sequence, the common difference is \(3\), and the starting value is \(1\) So the formula is: \[t_n=1+(n-1)(3)=1+3n-3=3n-2\] Therefore the twelfth term is: \[t_12=3(12)-2=36-2=34\] In general. for the formula for the sum of a sequence from the first term to the nth term is: \[S_n=n\left(\frac{t_1+t_n}{2}\right)\] So, the first term is 1, the twelfth term is 34 and \(n\) is 12. So the sum is: \[S_{12}=12\left(\frac{1+34}{2}\right)=12\left(\frac{35}{2}\right)=12(17.5)=210\]
anonymous
  • anonymous
Another way is that a sum formula from \(t_1\) to \(t_n\) using the formula we know with our a and d values is: \[S_n=\frac{n[2a+(n-1)d]}{2}\] \[S_{12}=\frac{12[2+(11)(3)]}{2}=\frac{12[2+33]}{2}=\frac{12[35]}{2}=210\] Eitherway, we reach the same conclusion

Looking for something else?

Not the answer you are looking for? Search for more explanations.