# Question 7f344

Sep 11, 2016

${a}_{1} = 5 , {a}_{2} = 9 , {a}_{3} = 13 , {a}_{4} = 17$

$\delta = 4$

#### Explanation:

The first $n$ arithmetic series sum is given by

${\sum}_{k = 1}^{n} {a}_{k} = n {a}_{0} + \left({\sum}_{k = 1}^{n} k\right) \delta = n {a}_{0} + \frac{n \left(n + 1\right)}{2} \delta$

where ${a}_{0}$ is the base element and $\delta$ is the common difference: ${a}_{k} = {a}_{0} + k \delta$

So we have

$n \left(2 n + 3\right) = n {a}_{0} + \frac{n \left(n + 1\right)}{2} \delta$

This equality must be true for all $n$ then grouping the coefficients of the powers of $n$ we have the conditions

{ (2 a_0 + delta -6 = 0), (delta-4=0):}#

Solving for ${a}_{0} , \delta$ we obtain

${a}_{0} = 1 , \delta = 4$

and

${a}_{1} = 5 , {a}_{2} = 9 , {a}_{3} = 13 , {a}_{4} = 17$