Question

Find the term number (n) from the arithmetic series, such that 2+8+14+20+26…. Sn=1704

Answer 1

The series is given to be:

`2+8+14+20+26+...`

The nth term is given to be 1704.

The formula to calculate the nth term is given to be:

`a_n=a_1+(n-1)d`

where a₁ is the first term and d is the common difference.

The sum of the sequence can be calculated using the formula:

`S_n=(n)/(2)[2a+(n-1)d]`

From the series, we have:

`\begin{gathered} a_1=2 \\ d=8-2=6 \end{gathered}`

Therefore, the sum of the nth term can be calculated as shown below:

`\begin{gathered} 1704=(n)/(2)[2\cdot2+(n-1)6] \\ \mathrm{Multiply\:both\:sides\:by\:}2 \\ 1704\cdot \:2=(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}`

Using the quadratic formula to solve, we have:

`\begin{gathered} n_(1,\:2)=(-\left(-2\right)\pm √(\left(-2\right)^2-4\cdot \:6\left(-3408\right)))/(2\cdot \:6) \\ Solving,\text{ }we\text{ }have \\ n=24,\:n=-(71)/(3) \end{gathered}`

Since the number cannot be negative or a decimal/fraction, the term number will be 24.