Simplify a continued fraction Continued fractions are expressions that describe fractions iteratively. They can be represented graphically:
$$
a_0 +\cfrac 1 
 {a_1 + \cfrac 1
  {a_2 + \cfrac 1
   {\ddots + \cfrac 1
    {a_n}
   }
  }
 }
$$
Or they can be represented as a list of values: \$[a_0 ; a_1, a_2, \dots, a_n]\$
The challenge:
take a base number: \$a_0\$ and a list of denominator values: \$[a_1, a_2, \dots, a_n]\$ and simplify the continued fraction to a simplified rational fraction: return or print numerator and denominator separately.
Examples:

\$\sqrt 19\$: [4;2,1,3,1,2]: 170/39
\$e\$: [1;0,1,1,2,1,1]: 19/7
\$\pi\$: [3;7,15,1,292,1]: 104348/33215
\$ϕ\$: [1;1,1,1,1,1]: 13/8

Example implementation: (python)
def foo(base, sequence):
    numerator = 1
    denominator = sequence[-1]
    for d in sequence[-2::-1]:
        temp = denominator
        denominator = d * denominator + numerator
        numerator = temp
    return numerator + base * denominator, denominator
Winning: shortest code in bytes: --no builtins that do the entire problem allowed--
J, 8 5 bytes
Same as this, but uses a build-in for rationals.
Argument is {a0,a1,a2,a3,...} as a list of J extended precision rational numbers. Result is the fraction as a J extended precision rational number.
(+%)/

(+%) the plus-the-reciprocal-of
/ reduction over
Try it online!
-3 thanks to miles.