Question

What is the nth term?

Answer 1

Let $a_k$ denote the kth term of the sequence. Then

$a_k=a_1+d(k-1)$

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

$S_n=\displaystyle\sum_{k=1}^na_k=a_1+a_2+\cdots+a_{n-1}+a_n$

From the formula for $a_k$, we get

$S_n=\displaystyle\sum_{k=1}^n(a_1+d(k-1))=a_1\sum_{k=1}^n1+d\sum_{k=1}^n(k-1)$

$S_n=\displaystyle na_1+d\sum_{k=0}^{n-1}k$

$S_n=na_1+\dfrac{d(n-1)n}2$

$S_n=\dfrac n2(2a_1-d+dn)$

So we have $d=-5$, and $2a_1-d=16$ so that $a_1=\frac{11}2$.

Then the nth term in the sequence is

$a_n=\dfrac{11}2-5(n-1)=\boxed{\dfrac{21-10n}2}$