## Precalculus: Mathematics for Calculus, 7th Edition

$a_{n}$ = $\sqrt 3$ + $\sqrt 3$(n-1) = formula for the nth term 10th term: 10$\sqrt 3$
Use the formula $a_{n}$ = a + d(n-1) and plug in the given values to find a formula for the nth term. For the 10th term, plug in 10 for n and solve: $a_{10}$ = $\sqrt 3$ + $\sqrt 3$(10-1) = $\sqrt 3$ + $\sqrt 3$(9) = $\sqrt 3$ + 9$\sqrt 3$ = 10$\sqrt 3$