Answer
$a_{1}=(-1)^{1+1}(3(1)-5)=(-1)^2(-2)=-2$
$a_{2}=(-1)^{2+1}(3(2)-5)=(-1)^3(1)=-1$
$a_{3}=(-1)^{3+1}(3(3)-5)=(-1)^4(4)=4$
$a_{4}=(-1)^{4+1}(3(4)-5)=(-1)^5(7)=-7$
$a_{10}=(-1)^{10+1}(3(10)-5)=(-1)^{11}(25)=-25$
$a_{15}=(-1)^{15+1}(3(15)-5)=(-1)^{16}(40)=40$
Work Step by Step
If we want to find a term, we have to substitute $n$ by its index:
$a_{1}=(-1)^{1+1}(3(1)-5)=(-1)^2(-2)=-2$
$a_{2}=(-1)^{2+1}(3(2)-5)=(-1)^3(1)=-1$
$a_{3}=(-1)^{3+1}(3(3)-5)=(-1)^4(4)=4$
$a_{4}=(-1)^{4+1}(3(4)-5)=(-1)^5(7)=-7$
$a_{10}=(-1)^{10+1}(3(10)-5)=(-1)^{11}(25)=-25$
$a_{15}=(-1)^{15+1}(3(15)-5)=(-1)^{16}(40)=40$
