Answer
First three terms: $4,2,0$
last term: $-74$
Sum of the an arithmetic series: $-1400$
Work Step by Step
First three terms can be found as:$a_1=-2+6=4;a_2=-4+6=2;a_3=-6+6=0$
Last term can be found as: $a_{40}=-2(40)+6=-74$
Sum of the an arithmetic series can be calculated by using formula $S_n=\dfrac{n}{2}(a_1+a_n)$
Here $n=40$
Thus,
$S_{40}=\dfrac{40}{2}(4-74)=-1400$
our results are:
First three terms: $4,2,0$
last term: $-74$
Sum of the an arithmetic series: $-1400$
