Let [tex]a_k[/tex] denote the kth term of the sequence. Then
[tex]a_k=a_1+d(k-1)[/tex]
where d is the common difference between consecutive terms in the sequence and a₁ is the first term.
The sum of the first n terms is
[tex]S_n=\displaystyle\sum_{k=1}^na_k=a_1+a_2+\cdots+a_{n-1}+a_n[/tex]
From the formula for [tex]a_k[/tex], we get
[tex]S_n=\displaystyle\sum_{k=1}^n(a_1+d(k-1))=a_1\sum_{k=1}^n1+d\sum_{k=1}^n(k-1)[/tex]
[tex]S_n=\displaystyle na_1+d\sum_{k=0}^{n-1}k[/tex]
[tex]S_n=na_1+\dfrac{d(n-1)n}2[/tex]
[tex]S_n=\dfrac n2(2a_1-d+dn)[/tex]
So we have [tex]d=-5[/tex], and [tex]2a_1-d=16[/tex] so that [tex]a_1=\frac{11}2[/tex].
Then the nth term in the sequence is
[tex]a_n=\dfrac{11}2-5(n-1)=\boxed{\dfrac{21-10n}2}[/tex]
