Question #7f344

1 Answer
Sep 11, 2016

Answer:

#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#