# What is the sum of the arithmetic sequence 4, 11, 18 …, if there are 26 terms?

What is the sum of the arithmetic sequence 4, 11, 18 …, if there are 26 terms?
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DoryErrofbi
This sequence can be written as:
${a}_{n+1}={a}_{n}+7$
$\forall n\in \left[0,26\right]$ (for all n from 0 to 26, inclusive)
with ${a}_{0}=4$.
You can also write this as:
$\sum _{n=1}^{26}7n-3$
...which is easier to solve, even by hand.
$4+11+18+...$
$=\left(7\left(1\right)-3\right)+\left(7\left(2\right)-3\right)+\left(7\left(3\right)-3\right)+...$
$=7\left(1\right)+7\left(2\right)+7\left(3\right)+...-3-3-3-...$
$=7\left(1+2+3+...+26\right)-3\left(26\right)$
$=7\left(\frac{26\cdot 27}{2}\right)-78$
$=7\left(13\cdot 27\right)-78$
$=7\left(3\cdot 27+10\cdot 27\right)-78$
$=7\left(351\right)-78$
$=2457-78$
$=2379$
So, in general:
$\sum _{n=1}^{N}an±b=\frac{a}{2}\cdot n\left(n+1\right)±bn$
$\frac{7}{2}\cdot \left(26\cdot 27\right)-3\cdot 26=2379$