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

Sariah Mcguire 2022-10-25 Answered
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:
$\sum _{n=1}^{N}an±b=\frac{a}{2}\cdot n\left(n+1\right)±bn$
$\frac{7}{2}\cdot \left(26\cdot 27\right)-3\cdot 26=2379$