Here is the question,
If the tenth number in a Fibonacci-type sequence of increasing positive integers is 301, what is the fourth number?
How do I go about solving this without overly complex math?
EDIT | Fibonacci-type: the 2 previous numbers sum to the next in the sequence. EX: 4,5,9
Since you know the sequence satisfies the same recurrence as Fibonacci, its solution $g_n$ can be written as $$ g_n = g_1 F_n + (g_2-g_1) F_{n-1} = g_1 F_{n-2} + g_2 F_{n-1},\quad n=1,2,\dots, 10 $$ where $F_n$ is the usual Fibonacci. So $$ 301 = 55 g_1 + 34 (g_2 -g_1) = 21 g_1 + 34 g_2 $$ and you know $g_1,g_2$ are positive integers. So the only solutions are $(g_1,g_2)=(3,7)$, giving the fourth term $g_4=3F_2+7F_3=17$.