SL 1.2—Sequences and Sigma Notation

We are given $u_1=-8$, $d=5$, and we need $u_{15}$, so $n=15$.

The general term formula is $u_n=u_1+(n-1)d$. Substituting: $u_{15}=-8+(15-1)(5)$

$u_{15}=-8+14\times 5=-8+70=62$

The 15th term is $62$.

We need the 15th term of an arithmetic sequence with $u_1=-8$ and $d=5$.

The option $67$ would require $-8+75=67$, meaning $75=15\times 5$. This uses $n$ steps instead of $(n-1)$, which is the classic off-by-one error. Eliminated.

The option $57$ would give $-8+65=57$, meaning $65=13\times 5$, corresponding to only 13 steps — incorrect. Eliminated.

The option $72$ would require $-8+80=72$, i.e., $80=16\times 5$ — too many steps. Eliminated.

Only $62$ remains: $-8+(14)(5)=-8+70=62$. This is correct.

Using $u_n=u_1+(n-1)d$: $u_4=u_1+3d=13(1)$ $u_9=u_1+8d=33(2)$

$(u_1+8d)-(u_1+3d)=33-13$ $5d=20$

$d=\frac{20}{5}=4$

The common difference is $d=4$.

Between the 4th and 9th terms, there are $9-4=5$ steps. The total increase is $33-13=20$.

If $d=5$, five steps gives $5\times 5=25\ne 20$. Eliminated.

If $d=3$, five steps gives $5\times 3=15\ne 20$. Eliminated.

If $d=6$, five steps gives $5\times 6=30\ne 20$. Eliminated.

Testing $d=4$: $5\times 4=20$ ✓. The common difference is $d=4$.