The series is given to be:
[tex]2+8+14+20+26+...[/tex]The nth term is given to be 1704.
The formula to calculate the nth term is given to be:
[tex]a_n=a_1+(n-1)d[/tex]where a₁ is the first term and d is the common difference.
The sum of the sequence can be calculated using the formula:
[tex]S_n=\frac{n}{2}[2a+(n-1)d][/tex]From the series, we have:
[tex]\begin{gathered} a_1=2 \\ d=8-2=6 \end{gathered}[/tex]Therefore, the sum of the nth term can be calculated as shown below:
[tex]\begin{gathered} 1704=\frac{n}{2}[2\cdot2+(n-1)6] \\ \mathrm{Multiply\:both\:sides\:by\:}2 \\ 1704\cdot \:2=\frac{n}{2}\left[2\cdot \:2+\left(n-1\right)\cdot \:6\right]\cdot \:2 \\ 3408=n\left(4+6\left(n-1\right)\right) \\ Expanding\text{ }parentheses \\ 3408=6n^2-2n \\ \mathrm{Subtract\:}3408\mathrm{\:from\:both\:sides} \\ 6n^2-2n-3408=3408-3408 \\ 6n^2-2n-3408=0 \end{gathered}[/tex]Using the quadratic formula to solve, we have:
[tex]\begin{gathered} n_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:6\left(-3408\right)}}{2\cdot \:6} \\ Solving,\text{ }we\text{ }have \\ n=24,\:n=-\frac{71}{3} \end{gathered}[/tex]Since the number cannot be negative or a decimal/fraction, the term number will be 24.
