Question #7f344

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Question #7f344

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Answer 1 · 2016-09-11T19:07:28

$a_{1} = 5, a_{2} = 9, a_{3} = 13, a_{4} = 17$ $a_1 = 5, a_2 = 9, a_3 = 13, a_4 = 17$

$δ = 4$ $delta = 4$

Explanation:

The first $n$ $n$ arithmetic series sum is given by

$n \sum k = 1 a_{k} = n a_{0} + (n \sum k = 1 k) δ = n a_{0} + \frac{n (n + 1)}{2} δ$ $sum_(k=1)^n a_k = n a_0+(sum_(k=1)^n k) delta = na_0+(n(n+1))/2delta$

where $a_{0}$ $a_0$ is the base element and $δ$ $delta$ is the common difference: $a_{k} = a_{0} + k δ$ $a_k = a_0 + k delta$

So we have

$n (2 n + 3) = n a_{0} + \frac{n (n + 1)}{2} δ$ $n(2n+3)=na_0+(n(n+1))/2delta$

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

$a_1 = 5, a_2 = 9, a_3 = 13, a_4 = 17$

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Question #7f344

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