Given that $:\begin{array}{}\;f_{n+1}(x)&=f_n(x)+1\;\text{if n is a multiple of 3}\\ &=f_n(x)-1\;\text{otherwise.}\end{array}$
and $n = 0,f_1(1)=0$
Now $n = 1,f_{2}(1) = f_{1}(1) -1 = 0 -1 = -1$
$n = 2,f_{3}(1) = f_{2}(1) -1 = -1-1 = -2$
$n = 3,f_{4}(1) = f_{3}(1) + 1 = -2 + 1 = -1$
$n = 4,f_{5}(1) = f_{4}(1) -1 = -1-1 = -2$
$n = 5,f_{6}(1) = f_{5}(1) -1 = -2-1 = -3$
$n = 6,f_{7}(1) = f_{6}(1) +1 = -3+1 = -2$
$n = 7,f_{8}(1) = f_{7}(1) -1 = -2-1 = -3$
$n = 8,f_{9}(1) = f_{8}(1) -1 = -3-1 = -4$
$n = 9,f_{10}(1) = f_{9}(1) +1 = -4+1 = -3$
Now, we got the pattern , every multiple of $’3’$, the number decreases by $’-1’.$
- $n=3,f_{4}(1)\implies -1$
- $n=6,f_{7}(1)\implies -2$
- $n=9,f_{10}(1)\implies -3$
- $n=12,f_{13}(1)\implies -4$
- $n=15,f_{16}(1)\implies -5$
- $n=18,f_{19}(1)\implies -6$
- $n=21,f_{22}(1)\implies -7$
- $n=24,f_{25}(1)\implies -8$
- $n=27,f_{28}(1)\implies -9$
- $n=30,f_{31}(1)\implies -10$
- $n=33,f_{34}(1)\implies -11$
- $n=36,f_{37}(1)\implies -12$
- $n=39,f_{40}(1)\implies -13$
- $n=42,f_{43}(1)\implies -14$
- $n=45,f_{46}(1)\implies -15$
- $n = 48,f_{49}(1)\implies -16$
- $n = 49,f_{50}(1) = f_{49}(1)-1 = -16-1 = -17.$
$$\text{(OR)}$$
$3\implies -1\Leftrightarrow 48 \implies -16$
Now, $n = 49,f_{50}(1) = f_{49}(1)-1 = -16-1 = -17.$
So, the correct answer is $(C).$