Question

How do you find the sum of the arithmetic sequence $2+5+8+...+56$?

Answer 1

Determine the sum of the arithmetic sequence:

Given: $2+5+8+...+56$

The sum of $n$ numbers in an arithmetic sequence is:

$S_n=\frac{\mathrm{n}}{2}(2\mathrm{a}+(\mathrm{n}-1\left)\mathrm{d}\right)$

The value of the first term is $\mathrm{a}=2$.

Now we will find the common difference, $\mathrm{d}$.

$\mathrm{d}=5-2=8-5=3$

Also, the last term is $56$

Now we have to find the number of terms by using the ${\mathrm{n}}^{\mathrm{th}}$ term formula.

$$\begin{gathered} \mathrm{a}+(\mathrm{n}-1)\mathrm{d}={\mathrm{a}}_{\mathrm{n}} \\ \Rightarrow 2+(\mathrm{n}-1)\cdot 3=56 \\ \Rightarrow (\mathrm{n}-1)\cdot 3=54 \\ \Rightarrow \mathrm{n}=\frac{54}{3}+1=19 \end{gathered}$$

Therefore,

${\mathrm{S}}_{19}=\frac{19}{2}(2\times 2+(19-1)\times 3)$

$\Rightarrow {\mathrm{S}}_{19}=\frac{19}{2}\times \left(58\right)$

$=551$.

Hence, the arithmetic sequence is $551$.