Answer:
[tex]a_{89}=1128[/tex]
Step-by-step explanation:
The given arithmetic progression is -16, -3, 10,...
We observe that the first term of this progression is [tex]a_1=-16[/tex].
The constant difference is
[tex]d=-3--16\\d=-3+16=13[/tex]
The 89th term of an arithmetic sequence is given by:
[tex]a_{89}=a_1+88d[/tex]
We substitute the first term and the common difference to obtain:
[tex]a_{89}=-16+88(13)[/tex]
We simplify to get:
[tex]a_{89}=-16+1144[/tex]
This simplifies to:
[tex]a_{89}=1128[/tex]
