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

Answered 2022-10-26
Author has **12** answers

This sequence can be written as:

${a}_{n+1}={a}_{n}+7$

$\forall n\in [0,26]$ (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.

$4+11+18+...$

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

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

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

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

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

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

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

$=2457-78$

$=2379$

So, in general:

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

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

${a}_{n+1}={a}_{n}+7$

$\forall n\in [0,26]$ (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.

$4+11+18+...$

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

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

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

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

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

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

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

$=2457-78$

$=2379$

So, in general:

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

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

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