Question

The sequence 1,4,7,10....34 has 12 terms. Evaluated the related series

Answer 1

Hi bro I heard u need help on Damon algebra

Answer 2

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:
\[S_{12}=12\left(\frac{1+34}{2}\right)=12\left(\frac{35}{2}\right)=12(17.5)=210\]

Answer 3

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