Question

Find an for the arithmetic series with s16 = -288 And a1 =-21

Answer 1

If $d$ is the common difference between terms in the sequence $\{a_n\}$, then

$a_1=-21$
$a_2=a_1+d=-21+d$
$a_3=a_2+d=-21+2d$
...
$a_n=a_{n-1}+d=\cdots=-21+(n-1)d$

You're told that $S_{16}=-288$ (the sum of the first 16 terms in the sequence, presumably). Well, we know that

$S_{16}=\displaystyle\sum_{n=1}^{16}a_n=\sum_{n=1}^{16}(-21+(n-1)d)$
$S_{16}=\displaystyle(-21-d)\sum_{n=1}^{16}1+d\sum_{n=1}^{16}n$

Recall that

$\displaystyle\sum_{n=1}^kn=\frac{k(k+1)}2$

so that we have

$-288=16(-21-d)+\dfrac{16(16+1)}2d\implies d=\dfrac25$

So we get

$\begin{cases}a_1=-21\\\\a_n=-21+\dfrac25(n-1)\end{cases}$