Question

Question #7f344

Answer 1

`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+(sum_(k=1)^n k) delta = na_0+(n(n+1))/2delta`

where `a_0` is the base element and `delta` is the common difference: `a_k = a_0 + k delta`

So we have

`n(2n+3)=na_0+(n(n+1))/2delta`

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`