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If [tex]d[/tex] is the common difference between terms in the sequence [tex]\{a_n\}[/tex], then

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

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

[tex]S_{16}=\displaystyle\sum_{n=1}^{16}a_n=\sum_{n=1}^{16}(-21+(n-1)d)[/tex]
[tex]S_{16}=\displaystyle(-21-d)\sum_{n=1}^{16}1+d\sum_{n=1}^{16}n[/tex]

Recall that

[tex]\displaystyle\sum_{n=1}^kn=\frac{k(k+1)}2[/tex]

so that we have

[tex]-288=16(-21-d)+\dfrac{16(16+1)}2d\implies d=\dfrac25[/tex]

So we get

[tex]\begin{cases}a_1=-21\\\\a_n=-21+\dfrac25(n-1)\end{cases}[/tex]

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