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

Jun 11, 2015

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:

${\sum}_{n = 1}^{26} 7 n - 3$

...which is easier to solve, even by hand.

$4 + 11 + 18 + \ldots$

$= \left(7 \left(1\right) - 3\right) + \left(7 \left(2\right) - 3\right) + \left(7 \left(3\right) - 3\right) + \ldots$

$= 7 \left(1\right) + 7 \left(2\right) + 7 \left(3\right) + \ldots - 3 - 3 - 3 - \ldots$

$= 7 \left(1 + 2 + 3 + \ldots + 26\right) - 3 \left(26\right)$

$= 7 \left(\frac{26 \cdot 27}{2}\right) - 78$

$= 7 \left(13 \cdot 27\right) - 78$

$= 7 \left(3 \cdot 27 + 10 \cdot 27\right) - 78$

$= 7 \left(351\right) - 78$

$= 2457 - 78$

$= 2379$

So, in general:

${\sum}_{n = 1}^{N} a n \pm b = \frac{a}{2} \cdot n \left(n + 1\right) \pm b n$

$\frac{7}{2} \cdot \left(26 \cdot 27\right) - 3 \cdot 26 = 2379$